Algebra II & Trigonometry

Why linear equations matter

Almost every quantitative question you will ever ask - how long a trip takes, how much a plan costs, when two savings accounts hold equal amounts - eventually turns into an equation. The simplest and most common kind is the linear equation, in which the unknown appears only to the first power, never squared, cubed, or hidden inside a root. A linear equation in one variable can always be rearranged into the form ax + b = c, where a, b, and c are numbers and a is not zero. This lesson builds the complete toolkit for solving linear equations, then extends it to inequalities and to expressions wrapped in absolute value bars. These skills are the foundation for every later chapter, so it is worth mastering them until they feel automatic.

The balancing principle

Think of an equation as a balance scale. The equals sign is the pivot, and the two sides are pans that currently hold the same weight. As long as you do exactly the same thing to both pans, the scale stays balanced. If you add 5 to the left, you must add 5 to the right. If you divide the left by 3, you must divide the right by 3. This is the single idea behind every step of solving: use inverse operations to peel away everything attached to the variable until the variable stands alone.

Inverse operations come in pairs that undo each other. Addition and subtraction are inverses; multiplication and division are inverses. To move a term to the other side, apply its inverse. Because +7 is undone by -7, and multiplication by 4 is undone by division by 4, you always have a legal move available. The art is choosing the order that keeps the arithmetic clean, and the standard order is to undo addition and subtraction first, then undo multiplication and division.

A basic solve, narrated

Solve 5x - 8 = 27. The variable x is multiplied by 5, and then 8 is subtracted. To isolate x, undo those operations in reverse order. First undo the subtraction of 8 by adding 8 to both sides: 5x - 8 + 8 = 27 + 8, which simplifies to 5x = 35. Now undo the multiplication by 5 by dividing both sides by 5: 5x/5 = 35/5, giving x = 7. Always confirm by substitution: 5(7) - 8 = 35 - 8 = 27, which matches the right side, so x = 7 is correct.

Worked example 1: a multi-step equation with variables on both sides

Solve 3(x - 2) + 4 = 2x + 9. When the variable appears on both sides and there are parentheses, the reliable plan has four stages: distribute, combine like terms on each side, collect all variable terms on one side, then isolate.

1. Distribute the 3 across (x - 2): 3x - 6 + 4 = 2x + 9.
2. Combine like terms on the left, since -6 + 4 = -2: 3x - 2 = 2x + 9.
3. Subtract 2x from both sides to gather the variable on the left: 3x - 2x - 2 = 9, so x - 2 = 9.
4. Add 2 to both sides: x = 11.

Check: the left side is 3(11 - 2) + 4 = 3(9) + 4 = 27 + 4 = 31; the right side is 2(11) + 9 = 22 + 9 = 31. Both equal 31, so x = 11 is confirmed.

Clearing fractions with the least common denominator

Fractions make arithmetic error-prone, so a powerful first move is to eliminate them entirely. Multiply every single term on both sides by the least common denominator (LCD), the smallest number that all the denominators divide into evenly. This scales every fraction up to a whole number in one clean stroke.

Worked example 2: an equation full of fractions

Solve x/2 + 1/3 = x/6 + 2. The denominators are 2, 3, and 6, and the LCD is 6. Multiply each term by 6, being careful to include the constant term 2:

1. 6 · (x/2) = 3x, 6 · (1/3) = 2, 6 · (x/6) = x, and 6 · 2 = 12.
2. The equation becomes 3x + 2 = x + 12.
3. Subtract x from both sides: 2x + 2 = 12.
4. Subtract 2: 2x = 10, then divide by 2: x = 5.

Check: the left side is 5/2 + 1/3. Writing both over 6, that is 15/6 + 2/6 = 17/6. The right side is 5/6 + 2 = 5/6 + 12/6 = 17/6. They match, so x = 5.

Special cases: no solution and all real numbers

Not every linear equation has exactly one answer. Occasionally the variable cancels out completely. If what remains is a false statement, the equation has no solution. For instance, solving 2x + 3 = 2x - 5 by subtracting 2x from both sides leaves 3 = -5, which is never true, so no value of x works. If instead the variable cancels and what remains is a true statement, then every real number is a solution; the equation is called an identity. Solving 2(x + 1) = 2x + 2 gives 2x + 2 = 2x + 2, then 2 = 2, which is always true, so the solution set is all real numbers. Recognizing these outcomes prevents the frustration of hunting for a single answer that does not exist.

Linear inequalities

An inequality compares two expressions using one of the symbols < (less than), > (greater than), ≤ (less than or equal to), or ≥ (greater than or equal to). Instead of a single value, the solution is usually a whole range of values. You solve an inequality with the same inverse-operation steps as an equation, with one crucial extra rule.

The sign-flip rule: when you multiply or divide both sides of an inequality by a negative number, you must reverse the direction of the inequality symbol. The reason is easy to see with numbers: 3 < 5 is true, but if you multiply both sides by -1 without flipping you would write -3 < -5, which is false. The correct statement is -3 > -5. Multiplying by a negative reflects the number line, reversing the order of everything, so the symbol must reverse too.

Worked example 3: an inequality that requires a flip

Solve -2x + 3 > 11. Subtract 3 from both sides: -2x > 8. Now divide both sides by -2, and because -2 is negative, reverse the symbol: x < -4. The solution is every number less than -4. In interval notation that is written as the open interval from negative infinity up to -4.

Compound inequalities

Sometimes a variable is trapped between two bounds, as in -1 ≤ 2x + 3 < 7. This is really two inequalities at once, and you solve it by performing the same operation on all three parts. Subtract 3 from each part: -4 ≤ 2x < 4. Divide each part by 2: -2 ≤ x < 2. The solution is every number from -2 (included) up to 2 (not included).

Absolute value: distance from zero

The absolute value of a number is its distance from zero on the number line, and distance is never negative. We write it with vertical bars, so |5| = 5 and |-5| = 5. Because two different numbers can sit the same distance from zero, absolute value equations frequently have two answers.

Absolute value equations

The equation |x| = 5 asks which numbers are exactly 5 units from zero, and both x = 5 and x = -5 qualify. In general, an equation of the form |expression| = k (with k ≥ 0) splits into two separate equations: expression = k or expression = -k. If k is negative, as in |x| = -3, there is no solution, because a distance can never be negative.

Worked example 4: an absolute value equation

Solve |2x - 1| = 7. Split into two cases. Case one: 2x - 1 = 7 gives 2x = 8, so x = 4. Case two: 2x - 1 = -7 gives 2x = -6, so x = -3. The solutions are 4 and -3. Checking case one: |2(4) - 1| = |7| = 7. Checking case two: |2(-3) - 1| = |-7| = 7. Both work.

Absolute value inequalities

Absolute value inequalities come in two flavors, and it helps to translate them into plain English about distance.

• A less than inequality like |x| < 3 means "within 3 units of zero." That is an and statement squeezing x between two bounds: -3 < x < 3. Think of it as a single connected interval around zero.
• A greater than inequality like |x| > 3 means "more than 3 units from zero." That is an or statement pointing outward in two directions: x < -3 or x > 3. The solution is two separate rays.

A memory aid: less-than gives you a bounded region (the "and" case, sometimes remembered as "less thand"), while greater-than gives you two outward pieces (the "or" case, "greator").

Worked example 5: an absolute value inequality

Solve |3x + 2| ≤ 8. Because it is a less-than-or-equal statement, write the compound inequality -8 ≤ 3x + 2 ≤ 8. Subtract 2 from all three parts: -10 ≤ 3x ≤ 6. Divide all three parts by 3: -10/3 ≤ x ≤ 2. The solution is every number from -10/3 to 2, inclusive.

Real-world application: budgeting and tolerances

These ideas appear constantly outside the classroom. Suppose a phone plan charges a 20 dollar base fee plus 0.10 dollars per gigabyte of data, and you want to keep your bill at or below 35 dollars. Letting g be gigabytes, the inequality is 20 + 0.10g ≤ 35. Subtract 20: 0.10g ≤ 15. Divide by 0.10: g ≤ 150. You can use up to 150 gigabytes and stay within budget. Absolute value inequalities model manufacturing tolerances: if a machined part must be 50 millimeters "give or take" 0.2 millimeters, the acceptable lengths L satisfy |L - 50| ≤ 0.2, which unpacks to 49.8 ≤ L ≤ 50.2.

Common mistakes

• Forgetting to flip the inequality sign when dividing or multiplying by a negative number. This single slip reverses the entire answer.
• Distributing to only part of a group. In 3(x - 2) you must multiply the 3 by both the x and the -2, giving 3x - 6, not 3x - 2.
• Multiplying only some terms by the LCD. When clearing fractions, every term must be multiplied, including whole-number terms with no visible denominator.
• Dropping a case in an absolute value problem. Both the positive and negative versions must be solved; skipping one loses half the solution.
• Treating a "no solution" as if it must have an answer. If the variable cancels and the leftover statement is false, stop and report no solution rather than inventing one.
• Mixing up the and/or logic for absolute value inequalities. Less-than makes a bounded "and" interval; greater-than makes an outward "or" pair.

Recap

To solve a linear equation, use inverse operations to isolate the variable, keeping both sides balanced; distribute and combine like terms first, and clear fractions with the LCD when they appear. Watch for the special outcomes of no solution (a false leftover statement) and all real numbers (a true leftover statement). Inequalities follow the same steps but require flipping the symbol whenever you multiply or divide by a negative. Absolute value measures distance from zero: an equation |expr| = k splits into two cases, a less-than inequality becomes a bounded "and" statement, and a greater-than inequality becomes an outward "or" statement. Always finish by checking your answer in the original equation.

Sources

• OpenStax, Algebra and Trigonometry 2e, Chapter 2 (Equations and Inequalities), sections on linear equations and absolute value. Available free at openstax.org.
• OpenStax, Intermediate Algebra 2e, Chapter 2 (Solving Linear Equations and Inequalities). Available free at openstax.org.
• Khan Academy, "Linear equations and inequalities" and "Absolute value equations" units, khanacademy.org.
• Paul's Online Math Notes, Lamar University, "Algebra - Solving Equations and Inequalities," tutorial.math.lamar.edu.

The idea of a function

The word function is one of the most important in all of mathematics, and the concept is surprisingly simple once you see it clearly. A function is a rule that takes an input and produces exactly one output. That "exactly one" is the whole point. A vending machine that returns your chosen snack every time is like a function; a broken machine that sometimes gives one snack and sometimes gives two for the same button is not. This lesson defines functions precisely, introduces the notation used to work with them, and shows how to find the set of legal inputs (the domain) and the resulting outputs (the range).

Relations versus functions

A relation is any set of ordered pairs (x, y), with no restriction at all. A function is a special kind of relation in which every input (every x-value) is paired with exactly one output (one y-value). The key test is: could a single input ever produce two different outputs? If yes, it is not a function.

Consider the relation {(1, 2), (2, 4), (3, 6)}. Each input appears once, and each is tied to a single output, so this is a function. Now consider {(1, 2), (1, 5), (3, 6)}. The input 1 is paired with both 2 and 5, so a single input has two outputs, and this relation is not a function. Notice that repeated outputs are perfectly fine: {(1, 4), (2, 4), (3, 4)} is a function, because although the output 4 repeats, no input is ever reused with a different output.

The function machine analogy

A helpful mental picture is a machine. You drop a number in the input slot, gears turn according to the rule, and a single number drops out the output slot. Because a well-built machine behaves the same way every time, the same input always yields the same output. This is why a function cannot send one input to two different outputs: the machine would have to do two contradictory things at once. Keeping this picture in mind clarifies almost every question about functions.

The vertical line test

When a relation is drawn as a graph, there is a quick visual check called the vertical line test: a graph represents a function if and only if no vertical line crosses it more than once. The reasoning is direct. A vertical line collects all points sharing one x-value. If such a line hits the graph twice, that x-value has two y-values, which violates the definition. A parabola opening up or down passes the test, so it is a function. A full circle fails, because a vertical line through its interior meets it at both a top and a bottom point, meaning one x maps to two y-values.

Function notation

Instead of always writing y, we give functions names and use function notation. We write f(x), read aloud as "f of x," to mean the output of the function named f when the input is x. This is a critical point of confusion for beginners: f(x) does not mean f multiplied by x. The parentheses here signal "the function f evaluated at x," not multiplication.

The notation is powerful because it names both the rule and the specific input at once. If f(x) = 2x + 1, then f(3) asks for the output when the input is 3: f(3) = 2(3) + 1 = 7. Likewise f(0) = 2(0) + 1 = 1 and f(a) = 2a + 1. You can even substitute an entire expression: f(x + 1) = 2(x + 1) + 1 = 2x + 3. Different letters name different functions, so g(x) and h(x) are simply other machines with their own rules.

Worked example 1: evaluate a function at a negative input

Given f(x) = x² - 4x + 5, find f(-2). The safest habit is to substitute the input inside parentheses everywhere x appears, which prevents sign errors:

1. f(-2) = (-2)² - 4(-2) + 5.
2. Evaluate each piece: (-2)² = 4, and -4(-2) = +8.
3. = 4 + 8 + 5.
4. = 17.

The most common error here is writing -2² and getting -4. Squaring a negative gives a positive, which is exactly why the parentheses matter.

Worked example 2: evaluate at an algebraic input

Given g(x) = 3x - 7, find g(a + 2). Substitute the whole expression a + 2 for x: g(a + 2) = 3(a + 2) - 7 = 3a + 6 - 7 = 3a - 1. Being comfortable substituting expressions, not just numbers, is essential for later topics like composition and calculus.

Domain and range

The domain of a function is the set of all allowed input values; the range is the set of all output values the function can actually produce. In this course, two situations restrict a domain, and you should scan every function for them:

• Division by zero is forbidden. Any input that makes a denominator zero must be excluded.
• The square root of a negative number is not real. Any input that makes the quantity under an even root negative must be excluded (until complex numbers are introduced later).

If a function has neither a variable in a denominator nor a variable under an even root - for example any polynomial like f(x) = 3x - 7 or f(x) = x² + 1 - then its domain is all real numbers.

Worked example 3: a domain restricted by division

Find the domain of g(x) = 1/(x - 3). Set the denominator not equal to zero: x - 3 ≠ 0, which means x ≠ 3. The domain is all real numbers except 3. In words, you may input any number except 3, because dividing by zero is undefined.

Worked example 4: a domain restricted by a square root

Find the domain of h(x) = √(x - 2). The quantity under the square root, called the radicand, must be greater than or equal to zero: x - 2 ≥ 0, so x ≥ 2. The domain is all real numbers greater than or equal to 2. The range here is all values greater than or equal to 0, since a square root never returns a negative number.

Worked example 5: a domain with both restrictions

Find the domain of k(x) = √(x - 1)/(x - 4). Two conditions must hold at once. First, the radicand must be nonnegative: x - 1 ≥ 0, so x ≥ 1. Second, the denominator must be nonzero: x - 4 ≠ 0, so x ≠ 4. Combining, the domain is all x ≥ 1 except x = 4.

Reading domain and range from a graph

Graphically, the domain is the set of x-values the graph covers, read by sweeping your eyes left to right, and the range is the set of y-values, read by sweeping bottom to top. A horizontal parabola-like curve y = x² covers every x-value (domain all reals) but only produces y-values at or above 0 (range y ≥ 0), since squaring never gives a negative result.

Real-world application: functions as models

Functions describe dependence in the real world. The cost of a taxi ride is a function of distance; the temperature of a cooling cup of coffee is a function of time; the area of a square is a function of its side length, A(s) = s². In each case the domain is limited by reality: distance and time cannot be negative, and a side length must be positive. Recognizing the sensible domain of a model keeps your answers physically meaningful. For instance, in A(s) = s² the natural domain is s > 0, even though the algebra would accept negative inputs.

Common mistakes

• Reading f(x) as multiplication. It means "f evaluated at x," never "f times x."
• Dropping parentheses when substituting a negative number, which turns (-2)² = 4 into an incorrect -4.
• Confusing domain and range. Domain is the inputs (x); range is the outputs (y).
• Forgetting to check both restrictions when a function has a square root over a denominator.
• Calling a relation with repeated outputs a non-function. Repeated outputs are allowed; only repeated inputs with different outputs break the rule.
• Excluding a value from the domain of a polynomial. Polynomials never divide or take roots of the variable, so their domain is all real numbers.

Recap

A function pairs each input with exactly one output; a relation has no such restriction. The vertical line test checks this visually. Function notation f(x) names the output of function f at input x, and you evaluate by substituting the input, in parentheses, everywhere x appears. The domain is the set of legal inputs, restricted by forbidden division by zero and forbidden square roots of negatives; the range is the set of resulting outputs. Polynomials have domain of all real numbers, while rational and radical functions require careful exclusions. Choosing a sensible domain matters when a function models a real situation.

Sources

• OpenStax, Algebra and Trigonometry 2e, Chapter 3 (Functions), sections on function notation, domain, and range. Available free at openstax.org.
• OpenStax, Intermediate Algebra 2e, Chapter 3 (Graphs and Functions). Available free at openstax.org.
• Khan Academy, "Functions" unit, including "Evaluating functions" and "Domain and range," khanacademy.org.
• Paul's Online Math Notes, Lamar University, "Algebra - Functions," tutorial.math.lamar.edu.

Lines are everywhere

A straight line is the simplest and most useful graph in mathematics. It models any situation with a constant rate of change: a car traveling at steady speed, a phone plan with a flat per-minute fee, a candle burning down at a fixed rate. Understanding lines means understanding slope (how steep the line is and how fast the quantity changes) and the various equations we use to describe a line. By the end of this lesson you will compute slope from two points, write a line's equation in several forms, and tell at a glance whether two lines are parallel or perpendicular.

What slope measures

Slope measures steepness as "rise over run" - the change in the vertical direction divided by the change in the horizontal direction. Given two points (x₁, y₁) and (x₂, y₂), the slope is

m = (y₂ - y₁)/(x₂ - x₁),

the difference in y-values over the difference in x-values. The letter m is the traditional symbol for slope. Slope is also a rate of change: it tells you how many units y changes for each one-unit increase in x.

The four kinds of slope

• Positive slope: the line rises as you move to the right, like walking uphill.
• Negative slope: the line falls as you move to the right, like walking downhill.
• Zero slope: a horizontal line, perfectly flat; y never changes.
• Undefined slope: a vertical line. The run is zero, and division by zero is undefined, so a vertical line has no slope value at all.

Worked example 1: slope from two points

Find the slope through (1, 2) and (4, 11). Label the points: (x₁, y₁) = (1, 2) and (x₂, y₂) = (4, 11). Then the rise is y₂ - y₁ = 11 - 2 = 9, and the run is x₂ - x₁ = 4 - 1 = 3. So m = 9/3 = 3. The line climbs 3 units for every 1 unit to the right.

A common worry is whether the order of the points matters. It does not, as long as you are consistent. Using the points in the opposite order gives m = (2 - 11)/(1 - 4) = (-9)/(-3) = 3, the same answer. What you must not do is subtract the y-values in one order and the x-values in the other order.

Worked example 2: a negative slope

Find the slope through (-2, 5) and (2, -3). Rise = -3 - 5 = -8, run = 2 - (-2) = 4, so m = -8/4 = -2. The negative sign confirms the line falls to the right.

Slope-intercept form

The most common way to write a line is slope-intercept form, y = mx + b. It is beloved because it reveals two features at a glance: m is the slope, and b is the y-intercept, the y-value where the line crosses the y-axis (the point (0, b)). To graph such a line, plot the y-intercept first, then use the slope to step off more points.

Worked example 3: graphing from slope-intercept form

Graph y = 2x - 3. The y-intercept is b = -3, so plot (0, -3). The slope is 2, which you can read as 2/1: rise 2, run 1. From (0, -3), move up 2 and right 1 to reach (1, -1); repeat to reach (2, 1). Connect the points with a straight line. For a negative slope like -2/3, you would go down 2 and right 3.

Point-slope form

When you know the slope and any one point on the line, the fastest tool is point-slope form: y - y₁ = m(x - x₁), where (x₁, y₁) is the known point. This form flows directly from the definition of slope and is usually the quickest route to an equation.

Worked example 4: point-slope from a slope and a point

Write the equation of the line with slope 4 through (2, 5). Substitute into point-slope: y - 5 = 4(x - 2). To convert to slope-intercept form, distribute and solve for y: y - 5 = 4x - 8, so y = 4x - 3. Both forms describe the same line; point-slope is convenient to build, slope-intercept is convenient to graph.

Worked example 5: a line through two points

Find the line through (1, 3) and (3, 7). First compute the slope: m = (7 - 3)/(3 - 1) = 4/2 = 2. Then use point-slope with either point; using (1, 3): y - 3 = 2(x - 1). Distribute: y - 3 = 2x - 2, so y = 2x + 1. To double-check, verify the second point: 2(3) + 1 = 7, which matches.

Standard form and intercepts

A line can also be written in standard form, Ax + By = C, with A, B, and C integers. This form is handy for finding both intercepts quickly. To find the x-intercept, set y = 0 and solve; to find the y-intercept, set x = 0 and solve. For 2x + 3y = 12: setting y = 0 gives 2x = 12, so x = 6 and the x-intercept is (6, 0); setting x = 0 gives 3y = 12, so y = 4 and the y-intercept is (0, 4).

Horizontal and vertical lines

A horizontal line has the form y = c: every point shares the same y-value, and the slope is 0. A vertical line has the form x = c: every point shares the same x-value, and the slope is undefined. These are worth memorizing because their equations look unusual - a vertical line's equation has no y in it at all.

Parallel and perpendicular lines

Slopes reveal the geometric relationship between two lines. Two nonvertical lines are parallel exactly when they have equal slopes; they run in the same direction and never meet. Two lines are perpendicular exactly when their slopes are negative reciprocals of each other, meaning the product of the two slopes is -1. To find a perpendicular slope, flip the fraction and change its sign.

Worked example 6: parallel and perpendicular slopes

A line has slope 2/3. A line parallel to it also has slope 2/3. A line perpendicular to it has slope -3/2, the negative reciprocal: flip 2/3 to get 3/2, then negate to get -3/2. Check: (2/3)(-3/2) = -6/6 = -1, confirming perpendicularity. As another case, a line with slope 4 has perpendicular slope -1/4, since 4 is the same as 4/1.

Real-world application: rate of change

Slope is the mathematical name for "rate," and rates are everywhere. If a savings account grows by 50 dollars a month, the graph of balance versus time is a line with slope 50. If a road drops 3 meters over every 100 meters of horizontal distance, its grade is a slope of -3/100. A cell plan costing 20 dollars plus 5 dollars per line has cost equation C = 5n + 20, where the slope 5 is the cost per line and the y-intercept 20 is the fixed base fee. Reading slope as "change in output per unit change in input" turns an abstract number into a concrete, useful quantity.

Common mistakes

• Mixing up the order of subtraction in the slope formula. Subtract the y-values and the x-values in the same order.
• Confusing "rise over run" with "run over rise." Slope is vertical change divided by horizontal change, not the reverse.
• Saying a horizontal line has undefined slope. Horizontal lines have slope 0; vertical lines have undefined slope. It is easy to swap these.
• Forgetting the negative when finding a perpendicular slope. The perpendicular slope must be both flipped and sign-changed.
• Reading b as the x-intercept. In y = mx + b, the value b is the y-intercept.
• Distributing incorrectly when converting point-slope to slope-intercept form, especially forgetting to distribute the slope across both terms.

Recap

Slope, m = (y₂ - y₁)/(x₂ - x₁), measures steepness and rate of change; it is positive, negative, zero, or undefined. Slope-intercept form y = mx + b shows the slope and y-intercept and is ideal for graphing. Point-slope form y - y₁ = m(x - x₁) is the fastest way to build an equation from a slope and a point, and standard form Ax + By = C makes intercepts easy. Parallel lines have equal slopes; perpendicular lines have negative-reciprocal slopes whose product is -1. Horizontal lines are y = c with slope 0, and vertical lines are x = c with undefined slope.

Sources

• OpenStax, Algebra and Trigonometry 2e, Chapter 2 (Equations and Inequalities), sections on linear equations and lines. Available free at openstax.org.
• OpenStax, Intermediate Algebra 2e, Chapter 3 (Graphs and Functions), sections on slope and equations of lines. Available free at openstax.org.
• Khan Academy, "Forms of linear equations" and "Two-variable linear equations" units, khanacademy.org.
• Paul's Online Math Notes, Lamar University, "Algebra - Graphing and Functions: Lines," tutorial.math.lamar.edu.

What a system is

A system of equations is two or more equations considered together. A solution is a point that satisfies every equation at once - graphically, the point where the lines cross.

Substitution

The substitution method solves one equation for a variable, then substitutes that expression into the other equation.

Worked example 1: substitution

Solve y = 2x + 1 and 3x + y = 11.

1. Substitute 2x + 1 for y in the second equation: 3x + (2x + 1) = 11.
2. Combine: 5x + 1 = 11, so 5x = 10 and x = 2.
3. Back-substitute: y = 2(2) + 1 = 5.

The solution is (2, 5). Check: 3(2) + 5 = 11. Correct.

Elimination

The elimination method adds or subtracts the equations (after scaling) so one variable cancels.

Worked example 2: elimination

Solve 2x + 3y = 12 and 2x - y = 4.

1. Subtract the second equation from the first: (2x + 3y) - (2x - y) = 12 - 4, giving 4y = 8, so y = 2.
2. Substitute into 2x - y = 4: 2x - 2 = 4, so 2x = 6 and x = 3.

The solution is (3, 2).

Worked example 3: scaling before elimination

Solve 3x + 2y = 16 and x - y = 2. As written, adding the equations cancels nothing, so scale first. Multiply the second equation by 2 to get 2x - 2y = 4. Now add it to the first equation: (3x + 2y) + (2x - 2y) = 16 + 4, so 5x = 20 and x = 4. Substitute into x - y = 2: 4 - y = 2, so y = 2. The solution is (4, 2). Check: 3(4) + 2(2) = 12 + 4 = 16. Correct. When neither variable cancels as written, multiply one or both equations by numbers chosen so that one variable's coefficients become opposites, then add.

Special cases

If the variables cancel and you get a false statement like 0 = 5, the lines are parallel and there is no solution. If you get a true statement like 0 = 0, the equations describe the same line and there are infinitely many solutions.

Choosing the best method

All three approaches, graphing, substitution, and elimination, locate the same intersection point, so the real skill is picking the fastest tool for the system in front of you. Graphing builds intuition and gives quick estimates, but it is slow and imprecise when the answer is not a pair of small whole numbers. Substitution shines when a variable is already isolated, or has a coefficient of 1 so it can be isolated in one step. Elimination shines when both equations are lined up in standard form and the coefficients of one variable already match or can be matched with a single multiplication. With practice you will glance at a system and know immediately which method finishes it in the fewest lines.

Worked example 4: a ticket-sales application

A school play sells adult tickets for 8 dollars and child tickets for 5 dollars. One evening, 30 tickets sell for a total of 180 dollars. How many of each kind were sold? Let a be the number of adult tickets and c the number of child tickets.

1. The count equation: a + c = 30.
2. The money equation: 8a + 5c = 180.
3. Solve the count equation for a: a = 30 - c.
4. Substitute into the money equation: 8(30 - c) + 5c = 180, so 240 - 8c + 5c = 180, which simplifies to 240 - 3c = 180.
5. Subtract 240 from both sides: -3c = -60, so c = 20, and then a = 30 - 20 = 10.

Check: 10 adult tickets bring in 80 dollars and 20 child tickets bring in 100 dollars, a total of 180 dollars, and the counts add to 30. Both conditions are satisfied.

Worked example 5: a mixture problem

A chemist needs 30 liters of a 30 percent acid solution but stocks only a 20 percent solution and a 50 percent solution. How much of each should be mixed? Let x be liters of the 20 percent solution and y be liters of the 50 percent solution.

1. Total volume: x + y = 30.
2. Total acid: 0.20x + 0.50y = 0.30(30) = 9.
3. From the volume equation, x = 30 - y. Substitute: 0.20(30 - y) + 0.50y = 9.
4. Distribute: 6 - 0.20y + 0.50y = 9, so 6 + 0.30y = 9.
5. Subtract 6: 0.30y = 3, so y = 10 liters of the 50 percent solution and x = 20 liters of the 20 percent solution.

Check: the acid in the mix is 0.20(20) + 0.50(10) = 4 + 5 = 9 liters, and 9 liters out of 30 is exactly 30 percent.

Real-world applications

Systems of two equations appear whenever two conditions must hold at the same time. A business finds its break-even point by setting its cost equation equal to its revenue equation. Economists locate a market equilibrium by intersecting a supply line with a demand line. Everyday comparisons work the same way: if Plan A costs 30 dollars plus 0.10 dollars per minute and Plan B costs a flat 45 dollars, the plans cost the same when 30 + 0.10m = 45, so 0.10m = 15 and m = 150 minutes. Below 150 minutes Plan A is cheaper; above it, Plan B wins. Translating a two-condition story into a two-equation system is one of the most useful modeling skills in algebra.

Common misconceptions

• Reporting only x. The solution of a system is an ordered pair; find both coordinates and present the point.
• Scaling only part of an equation. When you multiply an equation to set up elimination, every term must be multiplied, including the constant on the right side.
• Sign slips when subtracting equations. Subtracting means subtracting every term of the second equation. Many students avoid the risk by multiplying one equation by -1 and adding instead.
• Confusing the special cases. A false statement such as 0 = 5 means no solution (parallel lines); a true statement such as 0 = 0 means infinitely many solutions (the same line twice).
• Back-substituting into the equation you just used. Substituting a found value back into the same rearranged equation produces an unhelpful identity. Use the other original equation to find the second coordinate.

Recap

A system of two linear equations asks for the point that satisfies both equations at once, which is where the two lines cross. Substitution replaces a variable with an equivalent expression from the other equation; elimination scales and adds the equations so one variable cancels. If both variables vanish, a false statement signals parallel lines with no solution, while a true statement signals the same line with infinitely many solutions. Systems model real two-condition problems such as ticket sales, mixtures, and break-even comparisons, and every answer should be checked in both original equations.

Sources

1. OpenStax, Intermediate Algebra 2e, Chapter 4 (Systems of Linear Equations), sections 4.1 through 4.3. Available free at openstax.org.
2. OpenStax, Algebra and Trigonometry 2e, Chapter 11 (Systems of Equations and Inequalities), section on systems of linear equations in two variables. Available free at openstax.org.
3. Khan Academy, "Systems of equations" unit, khanacademy.org.
4. Paul's Online Math Notes, Lamar University, "Algebra - Linear Systems with Two Variables," tutorial.math.lamar.edu.

Three equations, three unknowns

A system in three variables looks like three equations in x, y, and z. Each equation is a plane in three-dimensional space, and a unique solution is the single point where all three planes meet.

The strategy: eliminate down to two variables

The reliable plan is to eliminate the same variable from two different pairs of equations, producing two equations in two variables. Solve that smaller system, then back-substitute to find the third variable.

Worked example: solve a 3-by-3 system

Solve the system:

• Equation A: x + y + z = 6
• Equation B: 2x - y + z = 3
• Equation C: x + 2y - z = 2

1. Add A and C to eliminate z: (x + y + z) + (x + 2y - z) = 6 + 2, giving 2x + 3y = 8. Call this D.
2. Add B and C to eliminate z: (2x - y + z) + (x + 2y - z) = 3 + 2, giving 3x + y = 5. Call this E.
3. Now solve D and E. From E, y = 5 - 3x. Substitute into D: 2x + 3(5 - 3x) = 8, so 2x + 15 - 9x = 8, giving -7x = -7 and x = 1.
4. Then y = 5 - 3(1) = 2.
5. Back-substitute into A: 1 + 2 + z = 6, so z = 3.

The solution is (1, 2, 3). Check in B: 2(1) - 2 + 3 = 3. Correct.

Keeping the work organized

The most common mistake is losing track of signs while eliminating. Label your intermediate equations (D and E above), and always eliminate the same variable in both new equations so they form a clean two-variable system.

Worked example 2: a system that needs scaling

Solve the system:

• Equation A: x + y + z = 2
• Equation B: 2x - y + 3z = 9
• Equation C: x + 2y - z = -3

1. Add A and B to eliminate y: (x + y + z) + (2x - y + 3z) = 2 + 9, giving 3x + 4z = 11. Call this D.
2. To eliminate y from B and C, first multiply B by 2: 4x - 2y + 6z = 18. Add it to C: (4x - 2y + 6z) + (x + 2y - z) = 18 + (-3), giving 5x + 5z = 15, which divides by 5 to x + z = 3. Call this E.
3. Solve D and E together. From E, x = 3 - z. Substitute into D: 3(3 - z) + 4z = 11, so 9 - 3z + 4z = 11, giving 9 + z = 11 and z = 2.
4. Then x = 3 - 2 = 1.
5. Back-substitute into A: 1 + y + 2 = 2, so y = -1.

The solution is (1, -1, 2). Check in C: 1 + 2(-1) - 2 = 1 - 2 - 2 = -3. Correct.

The geometry of the special cases

Just as two lines can be parallel or identical, three planes can fail to share a single point. If elimination ever produces a false statement such as 0 = 4, the system has no solution: the planes may include a parallel pair, or they may pairwise intersect in three parallel lines like the faces of a triangular prism. If elimination produces the true statement 0 = 0 and no contradiction elsewhere, the system has infinitely many solutions: the three planes share a whole line (or are all the same plane), and the solutions are usually described with a parameter. A unique solution, one exact point in space, happens only when the planes cut each other cleanly, like two walls and a floor meeting at a room's corner.

Worked example 3: a number puzzle

The sum of three numbers is 16. The second number is 1 more than the first, and the third is three times the first. Find the numbers. Let the numbers be x, y, and z. The equations are x + y + z = 16, y = x + 1, and z = 3x. Because the last two are already solved for y and z, substitution is fastest: x + (x + 1) + 3x = 16, so 5x + 1 = 16, giving 5x = 15 and x = 3. Then y = 4 and z = 9. Check: 3 + 4 + 9 = 16. Correct.

Worked example 4: fitting a parabola through three points

Three-variable systems power a beautiful application: finding the quadratic y = ax² + bx + c that passes through three given points. Fit a parabola through (1, 2), (2, 3), and (3, 6). Each point gives one equation:

1. From (1, 2): a + b + c = 2.
2. From (2, 3): 4a + 2b + c = 3.
3. From (3, 6): 9a + 3b + c = 6.
4. Subtract the first equation from the second: 3a + b = 1. Subtract the second from the third: 5a + b = 3.
5. Subtract those two results: 2a = 2, so a = 1. Then 3(1) + b = 1 gives b = -2, and 1 - 2 + c = 2 gives c = 3.

The parabola is y = x² - 2x + 3. Check at x = 3: 9 - 6 + 3 = 6. Correct.

Real-world applications

Any situation with three unknown quantities tied together by three facts produces one of these systems. Nutrition planners solve for servings of three foods that hit calorie, protein, and carbohydrate targets simultaneously. Electrical engineers apply Kirchhoff's laws to a circuit and get three equations for three unknown currents. Economists balance three linked markets, and computer graphics software fits smooth curves through data points exactly as in the parabola example above. Larger versions of the same idea, with hundreds or millions of variables, are solved by computers using the very elimination process you practiced here, systematized as Gaussian elimination.

Common misconceptions

• Eliminating different variables from each pair. If one combination removes z and the other removes y, the two results do not form a solvable two-variable system. Eliminate the same variable both times.
• Using the same pair of equations twice. The two eliminations must involve all three original equations, or information is lost.
• Sign errors when subtracting. Subtracting an equation subtracts every one of its terms; many students multiply by -1 and add instead to stay safe.
• Stopping after two variables. The answer is an ordered triple (x, y, z); remember the final back-substitution step.
• Assuming a solution always exists. A false statement along the way means no solution, and 0 = 0 signals infinitely many; not every system pins down one point.

Recap

A linear equation in three variables graphs as a plane, and solving a three-variable system means finding where three planes meet. The strategy is systematic: eliminate one variable from two different pairs of equations, solve the resulting two-variable system, then back-substitute twice. A false statement during elimination means no solution and a true statement 0 = 0 means infinitely many. These systems solve number puzzles, fit parabolas through three points, and model any scenario where three facts constrain three unknowns.

Sources

1. OpenStax, Intermediate Algebra 2e, Chapter 4 (Systems of Linear Equations), section on solving systems with three variables. Available free at openstax.org.
2. OpenStax, Algebra and Trigonometry 2e, Chapter 11 (Systems of Equations and Inequalities), section on systems of linear equations with three variables. Available free at openstax.org.
3. Khan Academy, "Solving systems with three variables" lessons in the Systems of equations unit, khanacademy.org.
4. Paul's Online Math Notes, Lamar University, "Algebra - Linear Systems with Three Variables," tutorial.math.lamar.edu.

What a matrix is

A matrix is a rectangular array of numbers arranged in rows and columns. Its dimensions are written rows-by-columns; a matrix with 2 rows and 3 columns is a 2-by-3 matrix. Matrices give us a compact way to store the coefficients of a system.

Adding and subtracting

You add or subtract matrices of the same dimensions by combining corresponding entries. For example, the sum of the entries in row 1, column 1 is the sum of the two top-left numbers, and so on for every position.

Scalar multiplication

To multiply a matrix by a single number (a scalar), multiply every entry by that number. Multiplying a matrix by 3 triples each entry.

Matrix multiplication

To multiply two matrices, the number of columns of the first must equal the number of rows of the second. Each entry of the product is the dot product of a row from the first matrix with a column from the second: multiply matching pairs and add.

Worked example 1: a row-by-column product

Multiply the row [2 3] by the column with entries 4 and 5. The result is a single number: 2(4) + 3(5) = 8 + 15 = 23. Every entry of a matrix product is formed exactly this way.

The determinant of a 2-by-2 matrix

For a 2-by-2 matrix with rows [a b] and [c d], the determinant is ad - bc. It is a single number that tells us, among other things, whether a system has a unique solution (nonzero determinant) or not (zero determinant).

Worked example 2: a determinant

For rows [3 1] and [2 4], the determinant is ad - bc = 3(4) - 1(2) = 12 - 2 = 10.

Cramer's rule

For the system ax + by = e and cx + dy = f, let D = ad - bc. Then x = (ed - bf)/D and y = (af - ec)/D, provided D is not zero. Each numerator is a determinant formed by replacing a column of coefficients with the constants.

Worked example 3: solve with Cramer's rule

Solve 2x + 3y = 8 and x + 2y = 5. Here D = 2(2) - 3(1) = 1. Then x = (8·2 - 3·5)/1 = (16 - 15)/1 = 1 and y = (2·5 - 8·1)/1 = (10 - 8)/1 = 2. The solution is (1, 2).

Worked example 4: adding and scaling matrices

Let matrix A have rows [2 -1] and [3 4], and let matrix B have rows [1 5] and [-2 0]. To form A + B, add matching positions: row 1 is [2 + 1, -1 + 5] = [3 4], and row 2 is [3 + (-2), 4 + 0] = [1 4]. To form the scalar multiple 3A, triple every entry: rows [6 -3] and [9 12]. Notice that addition requires identical dimensions; a 2-by-2 matrix cannot be added to a 2-by-3 matrix because some entries would have no partner.

Worked example 5: multiplying two 2-by-2 matrices

Let A have rows [1 2] and [3 4], and let B have rows [5 6] and [7 8]. Each entry of AB is a row of A dotted with a column of B:

1. Row 1 of A with column 1 of B: 1(5) + 2(7) = 5 + 14 = 19.
2. Row 1 with column 2: 1(6) + 2(8) = 6 + 16 = 22.
3. Row 2 with column 1: 3(5) + 4(7) = 15 + 28 = 43.
4. Row 2 with column 2: 3(6) + 4(8) = 18 + 32 = 50.

So AB has rows [19 22] and [43 50]. Computing BA instead gives rows [23 34] and [31 46], a completely different matrix. Matrix multiplication is not commutative: order matters, which is one of the biggest differences between matrix algebra and ordinary number algebra.

Worked example 6: Cramer's rule again

Solve 4x - y = 2 and 2x + 3y = 8. The coefficient determinant is D = 4(3) - (-1)(2) = 12 + 2 = 14. Then x = (2·3 - (-1)·8)/14 = (6 + 8)/14 = 14/14 = 1, and y = (4·8 - 2·2)/14 = (32 - 4)/14 = 28/14 = 2. The solution is (1, 2). Check: 4(1) - 2 = 2 and 2(1) + 3(2) = 8. Both hold.

When the determinant is zero

Try the matrix with rows [2 4] and [1 2]: the determinant is 2(2) - 4(1) = 0. A zero determinant means the two rows are proportional, so the corresponding lines are parallel or identical, and Cramer's rule breaks down because dividing by zero is undefined. This is the matrix version of the special cases you met with elimination: D ≠ 0 guarantees exactly one solution, while D = 0 signals either no solution or infinitely many, and you must investigate further with elimination.

The identity matrix

The 2-by-2 identity matrix I has rows [1 0] and [0 1], with 1s on the main diagonal and 0s elsewhere. It behaves like the number 1: multiplying any 2-by-2 matrix by I, on either side, leaves it unchanged. The identity also lets a whole system be written compactly as the single matrix equation AX = B, where A holds the coefficients, X the unknowns, and B the constants, the form computers use to solve enormous systems.

Real-world applications

Matrices are the workhorse data structure of applied mathematics. A spreadsheet of sales, rows for products and columns for months, is a matrix, and combining two stores' spreadsheets is matrix addition. Computer graphics rotate, stretch, and reflect every point of a 3D model by multiplying coordinate matrices, which is why graphics cards are built to multiply matrices at staggering speed. Economists use matrices to track how industries feed into one another, and search engines rank pages using giant matrix computations. The little 2-by-2 determinant you learned generalizes to a test for whether any of these giant systems has a unique solution.

Common misconceptions

• Multiplying entry by entry. Matrix multiplication is row-times-column dot products, not matching-position products.
• Assuming AB = BA. Matrix multiplication is not commutative; the two orders usually give different results, and one order may not even be defined.
• Mixing up the determinant formula. It is ad - bc, the main diagonal product minus the off-diagonal product, in that order.
• Using Cramer's rule when D = 0. A zero determinant means the rule does not apply; the system has no solution or infinitely many.
• Forgetting dimension rules. Addition needs identical dimensions, and multiplication needs the columns of the first to match the rows of the second.

Recap

A matrix is a rectangular array of numbers described by its rows-by-columns dimensions. Add or subtract matrices entry by entry (same dimensions only), scale them by multiplying every entry, and multiply matrices by dotting rows with columns, remembering that order matters. The determinant ad - bc of a 2-by-2 coefficient matrix tells whether a system has a unique solution, and Cramer's rule turns determinants into a direct formula for that solution when D ≠ 0. The identity matrix plays the role of the number 1, and the equation AX = B packages any linear system in matrix form.

Sources

1. OpenStax, Algebra and Trigonometry 2e, Chapter 11 (Systems of Equations and Inequalities), sections on matrices, matrix operations, and Cramer's rule. Available free at openstax.org.
2. OpenStax, Intermediate Algebra 2e, Chapter 4 (Systems of Linear Equations), sections on solving systems with matrices and with determinants. Available free at openstax.org.
3. Khan Academy, "Matrices" unit, including matrix multiplication and determinants, khanacademy.org.
4. Paul's Online Math Notes, Lamar University, "Algebra - Augmented Matrices," tutorial.math.lamar.edu.

The parabola

A quadratic function has the form f(x) = ax² + bx + c with a ≠ 0. Its graph is a U-shaped curve called a parabola. If a > 0 the parabola opens upward; if a < 0 it opens downward. The larger |a| is, the narrower the parabola.

Vertex and axis of symmetry

The turning point is the vertex, and the vertical line through it is the axis of symmetry. For f(x) = ax² + bx + c, the axis of symmetry is x = -b/(2a), and the vertex's y-coordinate is found by evaluating the function there.

Worked example 1: find the vertex

Find the vertex of f(x) = x² - 6x + 5. Here a = 1, b = -6. The axis of symmetry is x = -(-6)/(2·1) = 6/2 = 3. The y-coordinate is f(3) = 9 - 18 + 5 = -4. So the vertex is (3, -4), and since a > 0, it is a minimum.

Vertex form

Vertex form is f(x) = a(x - h)² + k, where (h, k) is the vertex directly. The function f(x) = 2(x - 3)² - 4 has vertex (3, -4) and opens upward.

Worked example 2: standard to vertex form

Write f(x) = x² - 6x + 5 in vertex form by completing the square. Take half of -6, which is -3, and square it to get 9: f(x) = (x² - 6x + 9) + 5 - 9 = (x - 3)² - 4. This confirms the vertex (3, -4) we found above.

Intercepts of a parabola

Two more landmarks make a sketch accurate. The y-intercept is where the graph crosses the y-axis; substituting x = 0 into f(x) = ax² + bx + c leaves just c, so the y-intercept is always (0, c). The x-intercepts, if any, are where f(x) = 0. For f(x) = x² - 6x + 5, factoring gives (x - 1)(x - 5) = 0, so the graph crosses the x-axis at x = 1 and x = 5. Notice something elegant: the axis of symmetry x = 3 sits exactly halfway between the two x-intercepts, since (1 + 5)/2 = 3. Symmetry is a powerful shortcut, because every point on one side of the axis has a mirror twin on the other side.

Worked example 3: sketch from features

Sketch f(x) = -x² + 4x - 3 by collecting its features.

1. Direction: a = -1 is negative, so the parabola opens downward and the vertex is a maximum.
2. Axis and vertex: x = -b/(2a) = -4/(2 · (-1)) = 2, and f(2) = -4 + 8 - 3 = 1, so the vertex is (2, 1).
3. y-intercept: f(0) = -3, the point (0, -3); by symmetry (4, -3) is also on the graph.
4. x-intercepts: solve -x² + 4x - 3 = 0. Multiply by -1: x² - 4x + 3 = 0, which factors as (x - 1)(x - 3) = 0, so the graph crosses at x = 1 and x = 3.

Plot the vertex, the intercepts, and the mirror point, then draw a smooth downward-opening curve through them.

Worked example 4: vertex form when a is not 1

Write f(x) = 2x² + 8x + 3 in vertex form. Factor the leading coefficient out of the x-terms only: f(x) = 2(x² + 4x) + 3. Half of 4 is 2, and 2² = 4, so add and subtract 4 inside the parentheses, remembering that the added 4 is really 2 · 4 = 8 once the factor of 2 distributes: f(x) = 2(x² + 4x + 4) + 3 - 8 = 2(x + 2)² - 5. The vertex is (-2, -5), and the parabola opens upward, stretched twice as steep as y = x². Check: f(-2) = 2(4) - 16 + 3 = 8 - 16 + 3 = -5. Correct.

Maximum and minimum problems

Because the vertex is the highest or lowest point on the graph, quadratics answer optimization questions: what is the greatest height, the largest area, the maximum revenue? The recipe is always the same: build the quadratic model, find the vertex, and interpret its coordinates. The x-coordinate says where the extreme happens; the y-coordinate says what the extreme value is.

Worked example 5: a thrown ball

A ball is thrown upward so its height in feet after t seconds is h(t) = -16t² + 64t + 5. When is it highest, and how high does it get? The vertex is at t = -b/(2a) = -64/(2 · (-16)) = -64/(-32) = 2 seconds. The height there is h(2) = -16(4) + 64(2) + 5 = -64 + 128 + 5 = 69 feet. So the ball peaks 2 seconds after the throw at a height of 69 feet, and by symmetry it spends equal time rising and falling around that moment.

Real-world applications

Parabolas are everywhere in science and design. Projectiles, from basketballs to water in a fountain, trace parabolic arcs because gravity pulls with constant acceleration. Satellite dishes and car headlights use parabolic reflectors, since a parabola bounces incoming parallel rays through a single focus point. Businesses use quadratic revenue models: if a product priced at p dollars sells about 100 - 2p units, revenue is R(p) = p(100 - 2p) = 100p - 2p², which peaks at p = -100/(2 · (-2)) = 25 dollars, giving maximum revenue R(25) = 2500 - 1250 = 1250 dollars. Finding a vertex is the algebra behind all of these decisions.

Common misconceptions

• Dropping the sign in x = -b/(2a). The formula begins with negative b; when b is already negative, the axis of symmetry comes out positive.
• Reading vertex form with the wrong sign of h. In a(x - h)² + k, the vertex x-coordinate is the value that makes the parenthesis zero, so (x + 2)² means h = -2, not 2.
• Confusing the vertex y-coordinate with the maximum location. The x-coordinate tells where the extreme occurs; the y-coordinate is the extreme value itself.
• Forgetting to balance when completing the square with a leading coefficient. Adding 4 inside 2( ... ) really adds 8, so 8 must be subtracted outside.
• Assuming every parabola has x-intercepts. A parabola whose vertex sits above the x-axis and opens upward never crosses it; the quadratic has no real zeros.

Recap

A quadratic function f(x) = ax² + bx + c graphs as a parabola that opens up when a is positive and down when a is negative, with larger |a| making it narrower. The vertex lies on the axis of symmetry x = -b/(2a), and vertex form a(x - h)² + k displays the vertex directly; completing the square converts standard form into vertex form. The y-intercept is (0, c), the x-intercepts solve f(x) = 0, and the axis sits midway between them. Because the vertex is the extreme point, quadratics solve real maximum and minimum problems from projectile heights to pricing.

Sources

1. OpenStax, Algebra and Trigonometry 2e, Chapter 5 (Polynomial and Rational Functions), section on quadratic functions. Available free at openstax.org.
2. OpenStax, Intermediate Algebra 2e, Chapter 9 (Quadratic Equations and Functions), sections on graphing quadratic functions. Available free at openstax.org.
3. Khan Academy, "Quadratic functions and equations" unit, including vertex form and graphing parabolas, khanacademy.org.
4. Paul's Online Math Notes, Lamar University, "Algebra - Parabolas," tutorial.math.lamar.edu.

Four tools

A quadratic equation in standard form is ax² + bx + c = 0. You have several ways to solve one; pick the fastest that fits.

Factoring and the zero-product property

If the quadratic factors, use the zero-product property: if a product is zero, at least one factor is zero. Solve x² - 5x + 6 = 0: factor to (x - 2)(x - 3) = 0, so x = 2 or x = 3.

The square root property

If a quadratic has no middle term, isolate the square and take the square root of both sides, remembering the plus-or-minus. Solve x² = 49: x = ±7. Solve (x - 1)² = 16: x - 1 = ±4, so x = 5 or x = -3.

Completing the square

To complete the square for x² + bx, add (b/2)². Solve x² + 6x - 7 = 0. Move the constant: x² + 6x = 7. Half of 6 is 3, and 3² = 9, so add 9 to both sides: x² + 6x + 9 = 16, which is (x + 3)² = 16. Then x + 3 = ±4, giving x = 1 or x = -7.

The quadratic formula

The quadratic formula solves any quadratic: x = (-b ± √(b² - 4ac))/(2a).

Worked example: the quadratic formula

Solve 2x² - 4x - 3 = 0 with a = 2, b = -4, c = -3.

1. Discriminant: b² - 4ac = (-4)² - 4(2)(-3) = 16 + 24 = 40.
2. Square root: √40 = 2√10.
3. Apply: x = (4 ± 2√10)/4 = (2 ± √10)/2.

The two irrational solutions are (2 + √10)/2 and (2 - √10)/2.

The discriminant

The discriminant b² - 4ac tells the number and type of solutions: positive gives two real solutions, zero gives one real (repeated) solution, and negative gives two complex solutions (the topic of the next lesson).

Worked example 2: factoring with a leading coefficient

Solve 2x² + 5x - 3 = 0. Use the ac method: a · c = 2 · (-3) = -6, and the pair 6 and -1 multiplies to -6 while adding to 5. Split the middle term and factor by grouping:

1. 2x² + 6x - x - 3 = 0.
2. Group: 2x(x + 3) - 1(x + 3) = 0, so (2x - 1)(x + 3) = 0.
3. Zero-product property: 2x - 1 = 0 gives x = 1/2, and x + 3 = 0 gives x = -3.

Check x = 1/2: 2(1/4) + 5(1/2) - 3 = 1/2 + 5/2 - 3 = 3 - 3 = 0. Correct.

Worked example 3: completing the square with a leading coefficient

Solve 2x² - 8x + 6 = 0. When a is not 1, divide every term by a first: x² - 4x + 3 = 0. Move the constant: x² - 4x = -3. Half of -4 is -2, and (-2)² = 4, so add 4 to both sides: x² - 4x + 4 = 1, which is (x - 2)² = 1. Take square roots with the plus-or-minus: x - 2 = ±1, so x = 3 or x = 1. Both check in the original equation: 2(9) - 24 + 6 = 0 and 2(1) - 8 + 6 = 0.

Worked example 4: a negative discriminant

Solve x² + 2x + 5 = 0. The discriminant is 2² - 4(1)(5) = 4 - 20 = -16, which is negative, so there are no real solutions; the parabola never touches the x-axis. The quadratic formula still delivers the two complex solutions: x = (-2 ± √(-16))/2 = (-2 ± 4i)/2 = -1 ± 2i. The next lesson explains the number i; for now, notice how the discriminant predicted this outcome before any solving.

Real-world applications

Quadratic equations answer "when" and "what size" questions throughout science and daily life. A ball thrown from a rooftop with height h(t) = -16t² + 32t + 48 feet hits the ground when h(t) = 0: dividing by -16 gives t² - 2t - 3 = 0, which factors as (t - 3)(t + 1) = 0, so t = 3 or t = -1. Time cannot be negative, so the ball lands after 3 seconds; rejecting a solution that has no physical meaning is part of modeling. Similarly, a rectangular garden whose length is 3 meters more than its width and whose area is 40 square meters satisfies w(w + 3) = 40, so w² + 3w - 40 = 0, which factors as (w + 8)(w - 5) = 0. The width is 5 meters (rejecting -8) and the length is 8 meters.

Choosing a method wisely

All four tools solve the same problems, so pick by shape. If the quadratic factors quickly, factoring is fastest. If there is no middle term, or the equation is already a squared expression equal to a number, use the square root property. Completing the square is best when the equation has a = 1 and an even middle coefficient, and it is also the method that derives vertex form. The quadratic formula always works, making it the reliable fallback for messy coefficients, and its discriminant tells you in advance what kind of answers to expect.

Common misconceptions

• Forgetting the plus-or-minus. The equation x² = 49 has two solutions, 7 and -7; taking only the positive root loses half the answer.
• Dividing both sides by x. Solving x² = 5x by dividing by x discards the solution x = 0. Instead write x² - 5x = 0 and factor x(x - 5) = 0.
• Using the zero-product property when the product is not zero. From (x - 1)(x - 2) = 6 you may NOT conclude x - 1 = 6 or x - 2 = 6; first expand and rearrange so one side is 0.
• Sign errors in the formula. With b = -4, the formula's -b becomes +4, and b² is 16, always positive.
• Halving the wrong coefficient when completing the square. You halve b only after the equation is in the form x² + bx = k with leading coefficient 1.

Recap

To solve a quadratic, first put it in a helpful shape. Factoring plus the zero-product property is fastest when factors are easy to spot; the square root property handles equations like (x - 1)² = 16 directly; completing the square rewrites x² + bx by adding (b/2)²; and the quadratic formula x = (-b ± √(b² - 4ac))/(2a) solves anything. The discriminant b² - 4ac predicts two real solutions when positive, one repeated real solution when zero, and two complex solutions when negative. In applications, build the equation, solve it, and keep only the solutions that make sense in context.

Sources

1. OpenStax, Algebra and Trigonometry 2e, Chapter 2 (Equations and Inequalities), section on quadratic equations. Available free at openstax.org.
2. OpenStax, Intermediate Algebra 2e, Chapter 9 (Quadratic Equations and Functions), sections on completing the square and the quadratic formula. Available free at openstax.org.
3. Khan Academy, "Quadratic functions and equations" unit, including the quadratic formula and the discriminant, khanacademy.org.
4. Paul's Online Math Notes, Lamar University, "Algebra - Quadratic Equations, Part I and Part II," tutorial.math.lamar.edu.

The imaginary unit

There is no real number whose square is negative, so mathematicians defined the imaginary unit i with the property i² = -1, equivalently i = √(-1). With it we can take the square root of any negative number: √(-9) = √9 · √(-1) = 3i.

Complex numbers

A complex number has the form a + bi, where a is the real part and b is the imaginary part. Every real number is complex with b = 0.

Adding and subtracting

Combine real parts with real parts and imaginary parts with imaginary parts, just like combining like terms. For example, (3 + 2i) + (1 - 5i) = (3 + 1) + (2 - 5)i = 4 - 3i.

Multiplying

Multiply as you would binomials (FOIL), then replace i² with -1.

Worked example 1: multiply

Compute (2 + 3i)(1 - 4i).

1. FOIL: 2(1) + 2(-4i) + 3i(1) + 3i(-4i) = 2 - 8i + 3i - 12i².
2. Replace i² = -1: -12i² = -12(-1) = 12.
3. Combine: (2 + 12) + (-8 + 3)i = 14 - 5i.

The complex conjugate

The conjugate of a + bi is a - bi. Multiplying a complex number by its conjugate gives a real number: (a + bi)(a - bi) = a² + b². This is the key to division.

Worked example 2: divide

Compute (5 + i)/(2 - i). Multiply the top and bottom by the conjugate of the denominator, 2 + i:

1. Denominator: (2 - i)(2 + i) = 4 + 1 = 5.
2. Numerator: (5 + i)(2 + i) = 10 + 5i + 2i + i² = 10 + 7i - 1 = 9 + 7i.
3. Result: (9 + 7i)/5 = 9/5 + (7/5)i.

Complex solutions of quadratics

When the discriminant is negative, the quadratic formula produces complex solutions. Solve x² + 4 = 0: x² = -4, so x = ±2i. These come in conjugate pairs, always a + bi together with a - bi.

Worked example 3: a full quadratic with complex solutions

Solve x² - 4x + 13 = 0. The discriminant is (-4)² - 4(1)(13) = 16 - 52 = -36, so expect a conjugate pair. Apply the quadratic formula:

1. x = (4 ± √(-36))/2.
2. √(-36) = 6i, so x = (4 ± 6i)/2.
3. Divide each part by 2: x = 2 ± 3i.

The two solutions are 2 + 3i and 2 - 3i, a conjugate pair as promised. Check the first: (2 + 3i)² = 4 + 12i + 9i² = 4 + 12i - 9 = -5 + 12i; then (-5 + 12i) - 4(2 + 3i) + 13 = -5 + 12i - 8 - 12i + 13 = 0. Correct.

Powers of i

The powers of i repeat in a cycle of four: i¹ = i, i² = -1, i³ = -i, and i⁴ = 1, after which the pattern restarts. To simplify a large power, divide the exponent by 4 and use the remainder. For example, i²⁵: since 25 leaves remainder 1 when divided by 4, i²⁵ = i. Likewise 38 leaves remainder 2, so i³⁸ = i² = -1. This cycle is your first glimpse of the deep connection between i and rotation.

A pitfall: simplify before you multiply

The familiar rule √a · √b = √(ab) holds only when a and b are not both negative. Compute √(-4) · √(-9) correctly by converting to i-form first: (2i)(3i) = 6i² = -6. Multiplying the radicands first would give √36 = 6, which is wrong. Whenever a negative sits under a square root, pull out the i before doing anything else.

The complex plane and absolute value

Complex numbers have a geography. The complex plane plots a + bi at the point (a, b): the horizontal axis holds the real part and the vertical axis holds the imaginary part. The absolute value (or modulus) of a complex number is its distance from the origin, computed with the Pythagorean theorem: |a + bi| = √(a² + b²). For example, |3 + 4i| = √(9 + 16) = √25 = 5. Notice how this generalizes the absolute value of a real number, which is also just distance from zero.

Real-world applications

Complex numbers began as a trick for solving equations, but they are now essential working tools. Electrical engineers describe alternating-current circuits with complex impedance, where the real part is resistance and the imaginary part tracks how capacitors and inductors shift the current's timing. Signal processing, the mathematics inside music software, phone calls, and image compression, represents waves as complex numbers to make them easy to combine. Physicists write quantum mechanics in complex arithmetic, and the swirling Mandelbrot fractal is generated by repeatedly squaring complex numbers. Far from imaginary, these numbers quietly run much of modern technology.

Common misconceptions

• Treating i like a variable that equals some real number. No real number squares to -1; i is a new kind of number, defined by that property.
• Multiplying roots of negatives before converting to i-form. Convert first: the product of the square roots of -4 and -9 is -6, not 6.
• Forgetting that i² = -1 while simplifying. Every FOIL of complex numbers ends by replacing i² with -1 and combining; skipping that step leaves the answer unfinished.
• Conjugating both sign and real part. The conjugate of 3 - 7i is 3 + 7i; only the imaginary part flips sign.
• Expecting a lone complex root of a real quadratic. For polynomials with real coefficients, complex zeros always arrive in conjugate pairs.

Recap

The imaginary unit i satisfies i² = -1, which lets every negative number have square roots. A complex number a + bi adds and subtracts like combining like terms, multiplies by FOIL followed by replacing i² with -1, and divides by multiplying top and bottom by the conjugate of the denominator, since (a + bi)(a - bi) = a² + b² is real. Powers of i cycle with period four, complex numbers plot as points in the complex plane with modulus √(a² + b²), and quadratics with negative discriminants have conjugate-pair solutions. Always convert roots of negatives to i-form before computing.

Sources

1. OpenStax, Algebra and Trigonometry 2e, Chapter 2 (Equations and Inequalities), section on complex numbers. Available free at openstax.org.
2. OpenStax, Intermediate Algebra 2e, Chapter 8 (Roots and Radicals), section on the complex number system. Available free at openstax.org.
3. Khan Academy, "Complex numbers" unit, khanacademy.org.
4. Paul's Online Math Notes, Lamar University, "Algebra - Complex Numbers," tutorial.math.lamar.edu.

Anatomy of a polynomial

A polynomial is a sum of terms of the form (number)(variable raised to a whole-number power). The degree is the highest exponent. We name polynomials by degree (linear = 1, quadratic = 2, cubic = 3) and by number of terms (monomial = 1, binomial = 2, trinomial = 3).

Laws of exponents

These rules power all polynomial multiplication:

• Product rule: to multiply powers of the same base, add the exponents: xm · xn = xm+n. For example, x³ · x⁴ = x⁷.
• Power of a power: (xm)n = xmn; multiply the exponents. For example, (x²)³ = x⁶.
• Power of a product: (xy)ⁿ raises each factor. For example, (2x)³ = 8x³.

Adding and subtracting

Combine like terms - those with the same variable and exponent. To subtract, distribute the negative sign first. For example, (3x² + 2x) - (x² - 5x) = 3x² + 2x - x² + 5x = 2x² + 7x.

Multiplying

Use the distributive property so that every term of one polynomial multiplies every term of the other.

Worked example 1: multiply two binomials

Multiply (2x + 3)(x - 4) using FOIL: 2x·x + 2x·(-4) + 3·x + 3·(-4) = 2x² - 8x + 3x - 12 = 2x² - 5x - 12.

Worked example 2: a binomial times a trinomial

Multiply (x + 2)(x² - 3x + 1). Distribute each term of the first over the second:

1. x(x² - 3x + 1) = x³ - 3x² + x.
2. 2(x² - 3x + 1) = 2x² - 6x + 2.
3. Add: x³ - 3x² + 2x² + x - 6x + 2 = x³ - x² - 5x + 2.

Special products

Two patterns are worth memorizing: the square of a binomial (a + b)² = a² + 2ab + b², and the difference of squares (a + b)(a - b) = a² - b².

Worked example 3: squaring a binomial with coefficients

Expand (3x - 2)² using the pattern (a - b)² = a² - 2ab + b² with a = 3x and b = 2:

1. a² = (3x)² = 9x².
2. 2ab = 2(3x)(2) = 12x, taken with a minus sign.
3. b² = 2² = 4.

So (3x - 2)² = 9x² - 12x + 4. The most common wrong answer, 9x² + 4, forgets the middle term entirely; squaring a binomial always produces three terms.

Worked example 4: a trinomial times a binomial

Multiply (x² + 2x - 1)(x - 3). Distribute each term of the trinomial across the binomial:

1. x²(x - 3) = x³ - 3x².
2. 2x(x - 3) = 2x² - 6x.
3. -1(x - 3) = -x + 3.
4. Combine like terms: x³ + (-3 + 2)x² + (-6 - 1)x + 3 = x³ - x² - 7x + 3.

A quick sanity check: the leading term must be x² · x = x³ and the constant must be (-1)(-3) = 3. Both match.

Worked example 5: subtracting polynomials carefully

Simplify (4x³ - 2x + 7) - (x³ + 5x² - 3). The subtraction sign belongs to every term of the second polynomial, so distribute it first: 4x³ - 2x + 7 - x³ - 5x² + 3. Now gather like terms by degree: (4x³ - x³) - 5x² - 2x + (7 + 3) = 3x³ - 5x² - 2x + 10. Writing the answer in descending powers, called standard form, makes it easy to read the degree and leading coefficient at a glance.

Degrees under multiplication

When you multiply polynomials, the leading terms multiply, so the degrees add: a degree-3 polynomial times a degree-2 polynomial has degree 5. This gives a fast prediction of an answer's size and a built-in error check. Addition behaves differently: the sum of two degree-3 polynomials has degree at most 3, and can even drop lower if the leading terms cancel.

Real-world applications

Polynomials are the language of measurement formulas. An open-top box made by cutting squares of side x from a 10 by 8 sheet and folding up the sides has volume V = x(10 - 2x)(8 - 2x). Expanding shows the polynomial at work: (10 - 2x)(8 - 2x) = 80 - 20x - 16x + 4x² = 4x² - 36x + 80, so V = 4x³ - 36x² + 80x, a cubic in x. Businesses multiply a linear price expression by a linear quantity expression to get a quadratic revenue polynomial, and physics formulas for position under constant acceleration are quadratics in time. Being fluent at expanding and combining is what lets you build these models.

Common misconceptions

• Writing (a + b)² = a² + b². The square of a sum has a middle term: a² + 2ab + b².
• Adding exponents when adding terms. x³ + x³ = 2x³; exponents add only when you multiply powers of the same base.
• Forgetting to distribute a minus sign to every term of the polynomial being subtracted.
• Combining unlike terms, such as merging 3x² and 2x. Like terms must match in both variable and exponent.
• Multiplying only the first terms of each polynomial. Every term of one factor must multiply every term of the other; FOIL is just this rule specialized to two binomials.

Recap

A polynomial is a sum of terms with whole-number exponents, classified by degree and by its number of terms. The exponent laws, add exponents for products, multiply for powers of powers, and raise every factor for powers of products, drive all polynomial multiplication. Add and subtract by combining like terms, distributing any minus sign fully; multiply by distributing every term and then combining. The special patterns for the square of a binomial and the difference of squares save time and prevent errors, and degrees add under multiplication. These operations are the toolkit for building volume, revenue, and motion formulas.

Sources

1. OpenStax, Intermediate Algebra 2e, Chapter 5 (Polynomials and Polynomial Functions), sections on adding, subtracting, and multiplying polynomials. Available free at openstax.org.
2. OpenStax, Algebra and Trigonometry 2e, Chapter 1 (Prerequisites), sections on exponents and polynomials. Available free at openstax.org.
3. Khan Academy, "Polynomial arithmetic" unit, khanacademy.org.
4. Paul's Online Math Notes, Lamar University, "Algebra - Polynomials," tutorial.math.lamar.edu.

Always start with the GCF

The first step in any factoring problem is to pull out the greatest common factor, the largest monomial that divides every term. For 6x³ + 9x², the GCF is 3x², giving 3x²(2x + 3).

Factoring simple trinomials

To factor x² + bx + c, find two numbers that multiply to c and add to b. Factor x² + 7x + 12: the numbers 3 and 4 multiply to 12 and add to 7, so it factors as (x + 3)(x + 4).

Factoring when the leading coefficient is not 1

For ax² + bx + c, use the ac method: find two numbers that multiply to a·c and add to b, split the middle term, and factor by grouping.

Worked example 1: the ac method

Factor 2x² + 7x + 3. Here a·c = 2·3 = 6. Two numbers that multiply to 6 and add to 7 are 6 and 1. Split the middle term: 2x² + 6x + x + 3. Group: 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3).

Factoring by grouping

A four-term polynomial can often be grouped in pairs. Factor x³ + 2x² + 3x + 6: group as (x³ + 2x²) + (3x + 6) = x²(x + 2) + 3(x + 2) = (x² + 3)(x + 2).

Special forms

• Difference of squares: a² - b² = (a - b)(a + b). So x² - 25 = (x - 5)(x + 5).
• Sum of cubes: a³ + b³ = (a + b)(a² - ab + b²).
• Difference of cubes: a³ - b³ = (a - b)(a² + ab + b²).

Worked example 2: difference of cubes

Factor x³ - 8. Since 8 = 2³, use a = x, b = 2: x³ - 2³ = (x - 2)(x² + 2x + 4). The quadratic factor does not factor further over the real numbers.

Worked example 3: GCF first, then a pattern

Factor 2x³ - 8x completely.

1. The GCF of the two terms is 2x: 2x(x² - 4).
2. The leftover x² - 4 is a difference of squares: (x - 2)(x + 2).
3. Assemble: 2x³ - 8x = 2x(x - 2)(x + 2).

Skipping the GCF step makes the problem harder and often leads to an incomplete answer. Always ask "what divides every term?" before anything else.

Worked example 4: perfect square trinomials

Some trinomials are secretly squares. The patterns are a² + 2ab + b² = (a + b)² and a² - 2ab + b² = (a - b)². Factor 4x² - 12x + 9: the first term is (2x)² and the last is 3², so test the middle: 2(2x)(3) = 12x, which matches the middle term's size. Therefore 4x² - 12x + 9 = (2x - 3)². Check by expanding: 4x² - 6x - 6x + 9 = 4x² - 12x + 9. Correct.

Worked example 5: sum of cubes with coefficients

Factor 27x³ + 8. Recognize both terms as cubes: 27x³ = (3x)³ and 8 = 2³. Apply the sum-of-cubes formula with a = 3x and b = 2: (a + b)(a² - ab + b²) = (3x + 2)(9x² - 6x + 4). A memory aid for the signs in both cube formulas is SOAP: the first sign is the Same as the original, the second is Opposite, and the last is Always Positive.

A complete factoring strategy

Faced with any polynomial, run this checklist in order:

1. GCF: factor out the greatest common factor.
2. Count the terms. Two terms: look for a difference of squares, or a sum or difference of cubes. Three terms: try the simple trinomial method or the ac method. Four terms: try grouping.
3. Repeat. Examine every factor produced and factor it further if possible, as in x⁴ - 16 = (x² - 4)(x² + 4) = (x - 2)(x + 2)(x² + 4).
4. Check by multiplying the factors back together.

Factoring to solve equations

Factoring is the engine of the zero-product property. Solve x³ - 4x = 0: factor completely to x(x - 2)(x + 2) = 0, so x = 0, x = 2, or x = -2. A cubic can have three solutions, and complete factoring finds them all at once. This link between factors and solutions deepens in the next lesson as the factor theorem.

Real-world applications

Factoring converts messy expressions into revealing ones. Engineers factor polynomials to find the operating speeds at which a structure resonates. The projectile and area equations of the previous lessons were solved by factoring. In computer algebra and cryptography, the entire security of common encryption schemes rests on how hard it is to factor enormous numbers, a reminder that factoring is genuinely powerful mathematics, not just a classroom exercise. Even simplifying the rational expressions of Module 5 begins with factoring the numerator and denominator.

Common misconceptions

• Skipping the GCF. Factoring the GCF first makes every later step smaller and is required for a complete answer.
• Stopping too early. After one round, inspect each factor; differences of squares often hide inside, as in x⁴ - 16.
• Trying to factor a sum of squares. Over the real numbers, x² + 9 does not factor; only differences of squares split this way.
• Scrambling the cube-formula signs. Use SOAP: Same, Opposite, Always Positive.
• Confusing (x + a)(x + b) sign logic. For x² + bx + c, the two numbers multiply to c and add to b; when c is positive the signs match, and when c is negative they differ.

Recap

Factoring rewrites a polynomial as a product. Start with the GCF, then use the term count to choose a tool: difference of squares or the cube formulas for two terms, the simple or ac trinomial methods for three, grouping for four. Perfect square trinomials collapse to a squared binomial, and every result should be inspected for further factoring and verified by multiplying back. Factoring completely is what powers equation solving through the zero-product property and simplification of rational expressions later in the course.

Sources

1. OpenStax, Intermediate Algebra 2e, Chapter 6 (Factoring), sections on GCF, trinomials, and special products. Available free at openstax.org.
2. OpenStax, Algebra and Trigonometry 2e, Chapter 1 (Prerequisites), section on factoring polynomials. Available free at openstax.org.
3. Khan Academy, "Polynomial factorization" unit, khanacademy.org.
4. Paul's Online Math Notes, Lamar University, "Algebra - Factoring Polynomials," tutorial.math.lamar.edu.

Zeros of a polynomial

A zero (or root) of a polynomial f is a value of x that makes f(x) = 0. On a graph, the real zeros are the x-intercepts. A polynomial of degree n has exactly n zeros counting complex ones and multiplicities.

Synthetic division

Synthetic division is a quick shortcut for dividing a polynomial by a linear factor x - c. You bring down the leading coefficient, multiply by c, add to the next coefficient, and repeat.

Worked example 1: synthetic division

Divide x³ - 2x² - 5x + 6 by x - 3 (so c = 3). The coefficients are 1, -2, -5, 6.

1. Bring down 1.
2. Multiply 1·3 = 3; add to -2 to get 1.
3. Multiply 1·3 = 3; add to -5 to get -2.
4. Multiply -2·3 = -6; add to 6 to get 0.

The bottom row 1, 1, -2, 0 means the quotient is x² + x - 2 with remainder 0. Since the remainder is 0, x - 3 is a factor.

The remainder and factor theorems

The remainder theorem says that dividing f(x) by x - c leaves a remainder equal to f(c). The factor theorem follows: x - c is a factor exactly when f(c) = 0. This links dividing, factoring, and finding zeros.

Worked example 2: find all zeros

Find the zeros of f(x) = x³ - 2x² - 5x + 6. From above, x - 3 is a factor and the quotient is x² + x - 2. Factor the quotient: (x + 2)(x - 1). So f(x) = (x - 3)(x + 2)(x - 1), and the zeros are x = 3, -2, 1.

The rational root idea

To find a first zero to test, the rational root theorem says any rational zero is a factor of the constant term divided by a factor of the leading coefficient. For the polynomial above, the candidates come from factors of 6, and testing them with synthetic division finds the roots quickly.

Worked example 3: synthetic division with a remainder

Divide x³ + 2x² - 4x + 7 by x - 2, so c = 2 and the coefficients are 1, 2, -4, 7.

1. Bring down the 1.
2. Multiply 1 · 2 = 2; add to 2 to get 4.
3. Multiply 4 · 2 = 8; add to -4 to get 4.
4. Multiply 4 · 2 = 8; add to 7 to get 15.

The bottom row 1, 4, 4, 15 means the quotient is x² + 4x + 4 with remainder 15. The remainder theorem confirms it instantly: f(2) = 8 + 8 - 8 + 7 = 15. Because the remainder is not zero, x - 2 is not a factor. One caution: if the polynomial skips a power, insert a 0 for the missing coefficient. To divide x³ - 5x + 2, the coefficient row must be 1, 0, -5, 2.

Worked example 4: the rational root theorem, start to finish

Find all zeros of f(x) = x³ - 6x² + 11x - 6.

1. List candidates. The constant is -6 and the leading coefficient is 1, so any rational zero is among ±1, ±2, ±3, ±6.
2. Test one. f(1) = 1 - 6 + 11 - 6 = 0, so x = 1 is a zero and x - 1 is a factor.
3. Divide it out. Synthetic division with c = 1 on 1, -6, 11, -6: bring down 1; 1 · 1 = 1, -6 + 1 = -5; -5 · 1 = -5, 11 - 5 = 6; 6 · 1 = 6, -6 + 6 = 0. The quotient is x² - 5x + 6.
4. Finish with factoring. x² - 5x + 6 = (x - 2)(x - 3).

So f(x) = (x - 1)(x - 2)(x - 3) and the zeros are 1, 2, and 3. Each zero is an x-intercept of the graph.

Multiplicity: how a graph meets its zeros

A factor can repeat, and the repeat count, its multiplicity, controls the graph's behavior at that intercept. With odd multiplicity (such as a plain single factor) the graph crosses the x-axis; with even multiplicity the graph touches the axis and turns back. For f(x) = (x + 1)(x - 2)², the graph crosses at x = -1 but only bounces off the axis at x = 2. Counted with multiplicity, a degree-n polynomial has exactly n zeros among the complex numbers, a fact known as the fundamental theorem of algebra.

End behavior

Far to the left and right, a polynomial's graph follows its leading term. If the degree is even, both ends point the same way, up when the leading coefficient is positive, down when it is negative. If the degree is odd, the ends point opposite ways; for a positive leading coefficient the graph falls on the left and rises on the right, like y = x³. Combining end behavior with the zeros and their multiplicities lets you sketch a polynomial's overall shape without plotting dozens of points.

Real-world applications

Finding where a polynomial equals zero is one of the most common computational tasks in science and engineering. Designers solve cubic volume equations to size boxes, tanks, and shipping containers. Economists find break-even production levels where a cubic profit function crosses zero. Flight simulators and video games locate collisions by finding roots of polynomials that describe moving objects. The remainder theorem itself gives software a fast way to evaluate polynomials, and synthetic division is the hand version of the algorithms computers still use.

Common misconceptions

• Using the wrong sign for c. Dividing by x + 3 means c = -3, because x + 3 = x - (-3).
• Forgetting placeholder zeros for missing powers when setting up synthetic division.
• Believing every candidate from the rational root theorem is a zero. The theorem only lists possibilities; each must be tested.
• Confusing the remainder with the quotient. The final number in the synthetic row is the remainder, equal to f(c); the earlier numbers are the quotient's coefficients.
• Expecting the graph to cross at every zero. At zeros of even multiplicity the graph touches and turns instead of crossing.

Recap

Synthetic division divides a polynomial by x - c using only its coefficients, with the last number produced being the remainder. The remainder theorem says that remainder equals f(c), and the factor theorem upgrades a zero remainder into a factor. To find all zeros of a cubic, use the rational root theorem to list candidates, test until one works, divide it out, and factor the remaining quadratic. Multiplicity tells whether the graph crosses or touches at each intercept, and the leading term dictates end behavior. Together these tools turn a formula into an accurate sketch and a solved equation.

Sources

1. OpenStax, Algebra and Trigonometry 2e, Chapter 5 (Polynomial and Rational Functions), sections on dividing polynomials and zeros of polynomial functions. Available free at openstax.org.
2. OpenStax, College Algebra 2e, Chapter 5, sections on the remainder and factor theorems. Available free at openstax.org.
3. Khan Academy, "Polynomial division" and "Zeros of polynomials" lessons, khanacademy.org.
4. Paul's Online Math Notes, Lamar University, "Algebra - Dividing Polynomials" and "Zeroes/Roots of Polynomials," tutorial.math.lamar.edu.

What a rational expression is

A rational expression is a fraction whose numerator and denominator are polynomials, such as (x² - 1)/(x + 1). It is undefined wherever the denominator is zero, so we note those excluded values.

Simplifying

To simplify, factor the numerator and denominator completely and cancel common factors. Simplify (x² - 1)/(x + 1): the top factors as (x - 1)(x + 1), so the expression becomes (x - 1)(x + 1)/(x + 1) = x - 1, with x ≠ -1 still excluded.

Multiplying and dividing

Multiply by multiplying numerators and denominators, canceling common factors first. To divide, multiply by the reciprocal of the second expression.

Worked example 1: multiply

Multiply (x + 2)/(x - 3) · (x² - 9)/(x + 2). Factor x² - 9 = (x - 3)(x + 3). Cancel x + 2 and one x - 3: the result is x + 3.

Worked example 2: divide

Divide (x/6) ÷ (x²/3). Multiply by the reciprocal: (x/6)(3/x²) = 3x/(6x²) = 1/(2x).

Adding and subtracting

With a common denominator, combine numerators and keep the denominator. Without one, first rewrite each fraction over the least common denominator.

Worked example 3: add with a common denominator

Add 3/(x + 1) + 2/(x + 1). The denominators match, so add numerators: 5/(x + 1).

Worked example 4: unlike denominators

Add 1/x + 1/(2x). The LCD is 2x. Rewrite the first fraction: 2/(2x). Now add: 2/(2x) + 1/(2x) = 3/(2x).

Worked example 5: unlike polynomial denominators

Add 3/(x + 2) + 4/(x - 1). The denominators share no factor, so the LCD is their product (x + 2)(x - 1).

1. Scale each fraction: 3(x - 1)/((x + 2)(x - 1)) + 4(x + 2)/((x + 2)(x - 1)).
2. Expand the numerators: 3x - 3 and 4x + 8.
3. Add them: (3x - 3) + (4x + 8) = 7x + 5.

The sum is (7x + 5)/((x + 2)(x - 1)), with x ≠ -2 and x ≠ 1 excluded.

Worked example 6: subtracting with factoring first

Simplify 2x/(x² - 9) - 1/(x - 3). Factor the first denominator: x² - 9 = (x - 3)(x + 3), so the LCD is (x - 3)(x + 3). Rewrite the second fraction over the LCD: (x + 3)/((x - 3)(x + 3)). Now subtract the numerators, distributing the minus sign to both terms: 2x - (x + 3) = x - 3. The expression becomes (x - 3)/((x - 3)(x + 3)) = 1/(x + 3), with x ≠ 3 and x ≠ -3 still excluded. Factoring first revealed a dramatic simplification.

Complex fractions

A complex fraction has fractions inside a fraction, such as (1 + 1/x)/(1 - 1/x). The fastest cleanup is to multiply the top and bottom by the LCD of all the small fractions, here x: the numerator becomes x + 1 and the denominator becomes x - 1, so the whole expression is (x + 1)/(x - 1), for x ≠ 0 and x ≠ 1. One multiplication dissolves all the inner fractions at once.

Solving simple rational equations

When an equation contains rational expressions, multiply every term by the LCD to clear the fractions, solve, and then reject any candidate that makes a denominator zero. Solve 3/x + 1/2 = 5/(2x): the LCD is 2x. Multiplying every term gives 6 + x = 5, so x = -1. Since x = -1 makes no denominator zero, it is valid. Check: 3/(-1) + 1/2 = -3 + 1/2 = -5/2, and 5/(2(-1)) = -5/2. Correct. Had the answer been x = 0, it would be rejected as extraneous.

Real-world applications

Rational expressions model "per" quantities and shared work. If one painter finishes a room in 6 hours and another in 3 hours, they complete 1/6 + 1/3 = 1/2 of the room per hour together, so the job takes 2 hours. The same arithmetic computes electrical resistance: two parallel resistors of 6 ohms and 3 ohms satisfy 1/R = 1/6 + 1/3 = 1/2, so R = 2 ohms. Average cost in business is total cost divided by quantity, a rational function whose behavior for large quantities guides pricing. Lens equations in optics and concentration formulas in chemistry take the same form.

Common misconceptions

• Canceling terms instead of factors. In (x + 2)/x, the x cannot be canceled because the numerator's x is glued into a sum. Only whole multiplied factors cancel.
• Adding denominators. a/b + c/d is not (a + c)/(b + d); fractions must share a denominator before the numerators add.
• Losing excluded values after simplifying. Even though (x² - 1)/(x + 1) simplifies to x - 1, the value x = -1 remains excluded from the original expression.
• Flipping the wrong fraction when dividing. Keep the first expression, flip the second, then multiply.
• Forgetting to distribute a minus sign across a multi-term numerator when subtracting fractions.

Recap

A rational expression is a ratio of polynomials, undefined at its excluded values. Simplify by factoring completely and canceling common factors; multiply straight across after canceling; divide by multiplying by the reciprocal. To add or subtract, build each fraction up to the least common denominator, combine numerators carefully, and simplify the result. Clear complex fractions by multiplying top and bottom by the inner LCD, and solve rational equations by clearing denominators and rejecting extraneous candidates. These skills power work-rate, resistance, and average-cost problems.

Sources

1. OpenStax, Intermediate Algebra 2e, Chapter 7 (Rational Expressions and Functions). Available free at openstax.org.
2. OpenStax, Algebra and Trigonometry 2e, Chapter 1 (Prerequisites), section on rational expressions. Available free at openstax.org.
3. Khan Academy, "Rational expressions" lessons, khanacademy.org.
4. Paul's Online Math Notes, Lamar University, "Algebra - Rational Expressions," tutorial.math.lamar.edu.

Simplifying square roots

A radical like √n asks for a number that squares to n. To simplify, factor out the largest perfect square. Simplify √72: since 72 = 36 · 2, we get √36 · √2 = 6√2.

Adding and subtracting radicals

You can combine radicals only when they are like radicals (same index and same radicand). For example, 3√2 + 5√2 = 8√2, but √2 + √3 cannot be combined.

Multiplying radicals

Multiply the numbers outside and the radicands inside: √a · √b = √(ab). So √3 · √6 = √18 = 3√2.

Rationalizing the denominator

We prefer not to leave a radical in a denominator. To rationalize, multiply the top and bottom by a radical that clears it.

Worked example 1: rationalize

Simplify 3/√5. Multiply top and bottom by √5: (3√5)/(√5 · √5) = 3√5/5.

Rational exponents

A rational exponent is another way to write a radical. The rule is x raised to the power 1/n equals the n-th root of x, and x raised to m/n equals the n-th root of xⁿ. In words, the denominator is the root and the numerator is the power.

Worked example 2: evaluate a rational exponent

Evaluate 8 raised to the power 2/3. The denominator 3 says take the cube root of 8, which is 2; the numerator 2 says square it: 2² = 4. So the value is 4.

Worked example 3: convert notation

Rewrite √(x³) with a rational exponent. The square root is the 1/2 power and the cube is the 3rd power, so it is x raised to the power 3/2. Rational exponents let you apply the ordinary laws of exponents to radicals.

Simplifying radicals that contain variables

Variables under a radical follow the same perfect-square hunt as numbers. Since (x³)² = x⁶, we have √(x⁶) = x³ for nonnegative x. For a mixed radicand, factor out every perfect square: √(50x⁴y³) = √(25x⁴y² · 2y) = 5x²y√(2y), assuming the variables are nonnegative. The habit is always the same: split the radicand into (biggest perfect square) times (leftover), take the root of the square, and leave the leftover inside.

Worked example 4: rationalizing a two-term denominator

Simplify 4/(3 + √5). A one-term multiplier will not clear this denominator; instead multiply top and bottom by the conjugate 3 - √5, so the difference-of-squares pattern eliminates the radical:

1. Denominator: (3 + √5)(3 - √5) = 9 - 5 = 4.
2. Numerator: 4(3 - √5).
3. Divide: 4(3 - √5)/4 = 3 - √5.

The same conjugate trick you used for complex numbers works for radicals, and for the same structural reason.

Worked example 5: multiplying binomials with radicals

Radical binomials multiply by FOIL. First, (2 + √3)(2 - √3) = 4 - 3 = 1, a difference of squares. Second, (1 + √2)² = 1 + 2√2 + 2 = 3 + 2√2; note the middle term 2√2, which careless squaring omits.

Higher roots

The index of a radical tells which root to take. Cube roots undo cubing, so the cube root of 64 is 4, and unlike square roots they accept negative inputs: the cube root of -27 is -3, because (-3)³ = -27. Simplify higher radicals by hunting perfect powers of the index: the cube root of 54 equals the cube root of 27 · 2, which is 3 times the cube root of 2. The fourth root of 81 is 3, since 3⁴ = 81. Even-index roots of negative numbers are not real; odd-index roots are fine.

Worked example 6: computing with rational exponents

Rational exponents obey all the usual exponent laws, which makes many radical calculations mechanical.

1. 16 to the power 3/4: the fourth root of 16 is 2, then 2³ = 8.
2. 25 to the power -1/2: the negative exponent means reciprocal, so it equals 1/(25 to the 1/2) = 1/5.
3. x to the 1/2 times x to the 1/3: add exponents with a common denominator, 1/2 + 1/3 = 5/6, giving x to the 5/6.

Real-world applications

Radical formulas appear wherever quantities grow with squares or cubes. Crash investigators estimate a car's speed in miles per hour from skid marks with s = √(30fd), where d is the skid length in feet and f the road's friction factor; for f = 0.8 and d = 60, the speed is √(30 · 0.8 · 60) = √1440 = 12√10 ≈ 37.9 miles per hour. The period of a pendulum involves the square root of its length, which is why grandfather clocks are tall. In astronomy, Kepler's third law relates a planet's orbital period to the 3/2 power of its distance from the sun, a rational exponent doing serious scientific work.

Common misconceptions

• Distributing a root over addition. √(9 + 16) = √25 = 5, not 3 + 4 = 7. Roots distribute over products and quotients only.
• Combining unlike radicals. √2 + √3 cannot be condensed; only like radicals add.
• Reading x to the 1/2 as half of x. The exponent 1/2 means square root, an entirely different operation.
• Treating a negative exponent as a negative number. A negative exponent means reciprocal; 25 to the -1/2 is the positive number 1/5.
• Forgetting the middle term when squaring a radical binomial. (1 + √2)² is 3 + 2√2, not 1 + 2 = 3.

Recap

Simplify radicals by factoring out the largest perfect power of the index, combine only like radicals, and multiply radicals by multiplying radicands. Clear radicals from denominators by multiplying by a suitable radical, or by the conjugate when the denominator has two terms. Rational exponents translate radicals into exponent language, denominator as root and numerator as power, so all the exponent laws apply, including negative exponents as reciprocals. Odd-index roots accept negative radicands while even-index roots do not, and radical formulas govern real phenomena from skid marks to planetary orbits.

Sources

1. OpenStax, Intermediate Algebra 2e, Chapter 8 (Roots and Radicals). Available free at openstax.org.
2. OpenStax, Algebra and Trigonometry 2e, Chapter 1 (Prerequisites), sections on radicals and rational exponents. Available free at openstax.org.
3. Khan Academy, "Rational exponents and radicals" unit, khanacademy.org.
4. Paul's Online Math Notes, Lamar University, "Algebra - Radicals" and "Rational Exponents," tutorial.math.lamar.edu.

What makes a function exponential

An exponential function has the form f(x) = a · bx, where the variable lives in the exponent. The number a is the starting value, the output when x = 0, and b > 0 is the base (with b ≠ 1). The defining behavior is multiplicative: a linear function adds a fixed amount at every step, while an exponential function multiplies by a fixed factor at every step. That one difference produces dramatically different long-run behavior, which is why exponential models describe populations, investments, radioactivity, and viral spread, things that snowball rather than plod.

Growth versus decay

• If b > 1, the function shows exponential growth; the graph rises ever more steeply as x increases.
• If 0 < b < 1, the function shows exponential decay; the graph falls toward zero without ever reaching it.

In both cases (for positive a) the graph passes through (0, a), stays above the x-axis, and has the x-axis as a horizontal asymptote: a line the graph approaches but never touches. The domain is all real numbers and the range is all positive numbers.

Worked example 1: evaluate and read the pattern

For f(x) = 3 · 2x, find f(0), f(1), f(3), and f(-1).

1. f(0) = 3 · 20 = 3 · 1 = 3, the starting value.
2. f(1) = 3 · 2 = 6, and f(3) = 3 · 8 = 24. Each rise of 1 in x doubles the output.
3. f(-1) = 3 · 2-1 = 3 · (1/2) = 3/2. Negative inputs divide by the base, which is why the left side of the graph hugs the x-axis.

Linear versus exponential change

Compare two job offers starting at 40000 dollars: Offer A adds a 2000 dollar raise every year; Offer B grants a 4 percent raise every year. Offer A is linear, climbing by the same amount forever. Offer B is exponential with base 1.04, so its raises themselves grow: 1600 dollars the first year, but more each year after. Around year six the exponential offer overtakes the linear one, and the gap then widens permanently. Whenever change is proportional to the current amount, "4 percent of whatever you have now," the model is exponential, not linear.

Percent growth and decay factors

Real problems usually state a percent rate r, and the base is built from it: growth uses b = 1 + r and decay uses b = 1 - r. A 3 percent growth rate means multiplying by 1.03 each period; losing 15 percent means keeping 85 percent, multiplying by 0.85. Confusing the rate (0.03) with the factor (1.03) is the single most common modeling error.

Worked example 2: population growth

A town of 10000 people grows 3 percent per year, so P(t) = 10000(1.03)t. Predict the population in 5 years. Compute 1.03⁵ step by step: 1.03² = 1.0609, 1.03³ ≈ 1.0927, 1.03⁴ ≈ 1.1255, and 1.03⁵ ≈ 1.1593. So P(5) ≈ 10000 · 1.1593 = 11593 people, roughly 1600 more than the start, even though "3 percent" sounds small.

Worked example 3: a decay model

A car worth 20000 dollars loses 15 percent of its value per year, so it keeps 85 percent: V(t) = 20000(0.85)t. After 2 years, V(2) = 20000(0.85)² = 20000(0.7225) = 14450 dollars. Notice the value lost each year shrinks: 3000 dollars the first year but only 2550 the second, because 15 percent is taken of a smaller and smaller amount.

Worked example 4: doubling and half-life

Repeated doubling is exponential growth with base 2. A bacteria culture of 200 cells doubles every hour: N(t) = 200 · 2t, so after 6 hours N(6) = 200 · 64 = 12800 cells. Radioactive decay runs the same idea in reverse with base 1/2. A sample of 80 mg with a half-life of 3 days follows A(t) = 80(1/2) raised to the power t/3; after 9 days, three half-lives have passed, so the amount is 80 · (1/2)³ = 80/8 = 10 mg. Half-life thinking is central to radiocarbon dating and to how medicines wash out of the body.

Compound interest

Money at an annual rate r compounded once a year grows as A = P(1 + r)t, where P is the principal. Invest 500 dollars at 4 percent for 3 years: A = 500(1.04)³, and since 1.04³ = 1.124864, the balance is A ≈ 562.43 dollars. If interest compounds n times per year, each period applies the rate r/n, giving A = P(1 + r/n) raised to the power nt. For 1000 dollars at 6 percent compounded monthly for 2 years, A = 1000(1.005)²⁴ ≈ 1127.16 dollars, slightly more than yearly compounding would give, because interest starts earning interest sooner. As n grows without bound, the balance approaches Pe raised to the power rt, where e ≈ 2.71828 is the natural base you will meet with logarithms.

Real-world applications

Exponential models are everywhere once you look. Epidemiologists track outbreaks by their doubling time; a disease that doubles weekly can go from 100 cases to over 100000 in ten weeks. Computer engineers lived through decades of Moore's law, transistor counts doubling roughly every two years. Archaeologists date artifacts by the decayed fraction of carbon-14, and pharmacologists schedule doses around a drug's half-life. In every case the key question is the same: what is the constant factor, and how many times does it apply?

Common misconceptions

• Confusing 2^x with 2x. The first doubles repeatedly; the second just adds 2 each step. They agree at small values and then diverge spectacularly.
• Using the rate as the base. A 5 percent growth rate gives base 1.05, not 0.05; a 15 percent loss gives base 0.85, not 0.15.
• Thinking the graph reaches zero. Decay approaches the asymptote y = 0 but never lands on it; outputs stay positive.
• Assuming steady amounts of change. Exponential change is a steady percent, so the raw amount added or lost changes every period.
• Forgetting that negative exponents mean division. 2-3 = 1/8, a small positive number, never a negative one.

Recap

An exponential function f(x) = a · bx starts at a and multiplies by the base b for every unit step in x: growth when b is greater than 1, decay when b is between 0 and 1, always with the x-axis as horizontal asymptote. Build the base from a percent rate as 1 + r for growth or 1 - r for decay, use base 2 for doubling and base 1/2 for half-life, and model money with A = P(1 + r/n) to the power nt. Exponential change eventually outruns any linear change, which is exactly why these functions matter.

Sources

1. OpenStax, Algebra and Trigonometry 2e, Chapter 6 (Exponential and Logarithmic Functions), sections on exponential functions and their graphs. Available free at openstax.org.
2. OpenStax, Intermediate Algebra 2e, Chapter 10 (Exponential and Logarithmic Functions), section on evaluating and graphing exponential functions. Available free at openstax.org.
3. Khan Academy, "Exponential growth and decay" unit, khanacademy.org.
4. Paul's Online Math Notes, Lamar University, "Algebra - Exponential Functions," tutorial.math.lamar.edu.

A logarithm is an exponent

A logarithm answers the question "what exponent do I need?" The statement logb(y) = x means exactly bx = y. In words: the logarithm base b of y is the exponent you must put on b to produce y. A logarithm is therefore the inverse of an exponential function; where the exponential takes an exponent and returns a value, the logarithm takes a value and returns the exponent. Keeping the sentence "a logarithm IS an exponent" in mind demystifies everything in this lesson.

Converting between forms

The exponential and logarithmic forms carry identical information, and fluently translating between them is the core skill:

• 2³ = 8 in exponential form becomes log2(8) = 3 in logarithmic form.
• 10² = 100 becomes log10(100) = 2.
• Going the other way, log5(125) = 3 unpacks to 5³ = 125.

The common logarithm uses base 10 and is written log with no base shown. The natural logarithm uses the special base e ≈ 2.718 and is written ln. Scientific calculators carry buttons for both.

Worked example 1: evaluate by asking the question

Every evaluation is the same question in disguise: "the base to what power gives this number?"

1. log3(81): 3 to what power is 81? Since 3⁴ = 81, the answer is 4.
2. log2(1/8): since 2-3 = 1/8, the answer is -3. Logarithms of numbers between 0 and 1 are negative.
3. log7(7) = 1, because 7¹ = 7, and log7(1) = 0, because 7⁰ = 1. Every base gives log of 1 equal to 0.
4. log(0.01): since 10-2 = 0.01, the common log is -2.

The domain of a logarithm

Because a positive base raised to any real power is always positive, only positive numbers have logarithms. Expressions like log2(-4) and log(0) are undefined. Graphically, y = logb(x) lives entirely to the right of the y-axis, passes through (1, 0), and has the y-axis as a vertical asymptote. Its domain is x greater than 0 and its range is all real numbers, exactly the reverse of the exponential it inverts; in fact the two graphs are mirror images across the line y = x.

Properties of logarithms

Because logarithms are exponents, they inherit the exponent laws. Multiplying powers adds exponents, so logs turn multiplication into addition:

• Product: logb(MN) = logb(M) + logb(N).
• Quotient: logb(M/N) = logb(M) - logb(N).
• Power: logb(Mp) = p · logb(M).

These three rules let you expand a complicated logarithm into simple pieces or condense a sum of logarithms into one.

Worked example 2: expand

Expand log(x²y). Use the product rule, then the power rule: log(x²) + log(y) = 2 log(x) + log(y). A bigger one: log2(8x⁵/y) splits into log2(8) + 5 log2(x) - log2(y) = 3 + 5 log2(x) - log2(y), since 2³ = 8.

Worked example 3: condense

Write 3 log(x) - log(y) as a single logarithm. The power rule turns the coefficient into an exponent: log(x³) - log(y). The quotient rule combines them: log(x³/y). Similarly, 2 log(x) + 3 log(y) - log(z) condenses to log(x²y³/z).

Worked example 4: compute with known logs

Given log 2 ≈ 0.301 and log 3 ≈ 0.477, the properties unlock many other values without a calculator:

1. log 6 = log(2 · 3) = 0.301 + 0.477 = 0.778.
2. log 8 = log(2³) = 3(0.301) = 0.903.
3. log(3/2) = 0.477 - 0.301 = 0.176.

Before electronic calculators, tables of logarithms plus these rules were how scientists and navigators multiplied large numbers quickly; the slide rule is the same idea built from wood.

Change of base

Calculators only offer base 10 and base e, but any logarithm converts: logb(x) = log(x)/log(b). For example, log2(10) = log(10)/log(2) = 1/0.301 ≈ 3.32, meaning 2 raised to about 3.32 gives 10. The same formula works with ln in place of log.

Real-world applications

Logarithmic scales compress huge ranges into friendly numbers. Chemistry's pH scale is pH = -log[H+]: a solution with hydrogen ion concentration 10-4 has pH 4, and each single pH step means a tenfold change in acidity, so pH 3 is 100 times more acidic than pH 5. Earthquake magnitudes and sound decibels work the same way, one step up meaning ten times the energy or intensity. Computer scientists measure algorithm speed in logarithms, because each doubling of data adds only one more step to a binary search.

Common misconceptions

• Splitting a log across addition. log(a + b) is NOT log(a) + log(b); the product rule concerns multiplication inside the log.
• Multiplying logs. log(a) · log(b) is not log(ab); addition of logs does that job.
• Taking logs of zero or negatives. The argument must be positive; such expressions are undefined.
• Misreading division of logs. log(a)/log(b) is the change-of-base expression for logb(a), not log(a/b).
• Forgetting the anchors. For any base, log of 1 is 0 and log of the base itself is 1; these two facts catch many arithmetic slips.

Recap

A logarithm is an exponent: logb(y) = x means bx = y. Common logs use base 10 and natural logs use base e. Only positive numbers have logarithms, the graph of a log function passes through (1, 0) with a vertical asymptote at the y-axis, and the product, quotient, and power properties translate multiplication, division, and powers into addition, subtraction, and multiplication. Change of base converts any logarithm to calculator-friendly form, and logarithmic scales like pH, decibels, and earthquake magnitude tame enormous ranges of real-world quantities.

Sources

1. OpenStax, Algebra and Trigonometry 2e, Chapter 6 (Exponential and Logarithmic Functions), sections on logarithmic functions and properties of logarithms. Available free at openstax.org.
2. OpenStax, Intermediate Algebra 2e, Chapter 10 (Exponential and Logarithmic Functions), sections on evaluating logarithms and using their properties. Available free at openstax.org.
3. Khan Academy, "Logarithms" unit, including properties of logarithms and change of base, khanacademy.org.
4. Paul's Online Math Notes, Lamar University, "Algebra - Logarithm Functions," tutorial.math.lamar.edu.

Two families of equations

This lesson solves equations where the unknown is trapped in an exponent, like 3x = 20, or inside a logarithm, like log2(x) = 5. The strategies mirror each other because exponentials and logarithms are inverses: logarithms free a variable from an exponent, and rewriting in exponential form frees a variable from a logarithm. Three methods handle nearly everything you will meet.

Method 1: matching bases

If you can write both sides of an exponential equation as powers of the same base, the exponents must be equal, because exponential functions never repeat an output. Solve 2x = 32: write 32 = 2⁵, so 2x = 2⁵, which means x = 5.

Worked example 1: match bases with a fractional answer

Solve 9x = 27. Write both sides as powers of 3: 9 = 3² and 27 = 3³, so (3²)x = 3³, giving 32x = 3³. Equate exponents: 2x = 3, so x = 3/2. Check: 9 to the power 3/2 is the square root of 9, cubed: 3³ = 27. Correct.

Worked example 2: match bases with exponents on both sides

Solve 4x+1 = 8x. Both 4 and 8 are powers of 2: 4 = 2² and 8 = 2³. Rewrite: (2²)x+1 = (2³)x, so 22x+2 = 23x. Equate exponents: 2x + 2 = 3x, giving x = 2. Check: 4³ = 64 and 8² = 64. Correct.

Method 2: taking logarithms

When the bases refuse to match, take a logarithm of both sides and use the power property to pull the exponent down in front.

Worked example 3: logs on both sides

Solve 3x = 20. Take the common log of both sides: log(3x) = log(20), so x log(3) = log(20), giving the exact answer x = log(20)/log(3), which is about 2.727. Note carefully: this quotient of logs is not log(20/3); it is the change-of-base form of log3(20), exactly the exponent the question asks for.

Worked example 4: isolate the exponential first

Solve 5 · 2x = 40. Do not take logs yet; first divide both sides by 5: 2x = 8. Now the bases match easily: 8 = 2³, so x = 3. Isolating the exponential piece before anything else keeps the algebra painless, just as you isolate a squared expression before taking square roots.

Method 3: converting logarithmic equations

To solve an equation containing a logarithm, isolate the logarithm and rewrite the statement in exponential form. Solve log2(x) = 5: this says exactly x = 2⁵ = 32.

Worked example 5: convert with an expression inside

Solve log3(2x - 1) = 2. Convert: 2x - 1 = 3² = 9. Then 2x = 10, so x = 5. Domain check: the argument is 2(5) - 1 = 9, which is positive, so the solution stands.

Worked example 6: combine then convert

Solve log2(x) + log2(x - 2) = 3. Use the product rule to combine: log2(x(x - 2)) = 3. Convert to exponential form: x(x - 2) = 2³ = 8. Expand: x² - 2x - 8 = 0, which factors as (x - 4)(x + 2) = 0, so x = 4 or x = -2.

Check the domain: extraneous solutions

A logarithm requires a positive argument. The candidate x = -2 would make log2(x) undefined, so it is extraneous and rejected; the only valid solution is x = 4. This check is not optional. Combining logarithms can silently enlarge the domain, so every candidate must be tested against the original equation's logarithms. A related shortcut: if an equation says one log equals another with the same base, as in log(x + 4) = log(2x - 1), the arguments must be equal, so x + 4 = 2x - 1 gives x = 5, and both arguments equal the positive number 9. Valid.

Real-world applications: doubling time

These equations answer "how long?" questions that exponential functions pose. How many years does money at 6 percent annual interest take to double? Solve (1.06)t = 2: taking logs gives t log(1.06) = log(2), so t = log(2)/log(1.06) ≈ 0.3010/0.0253 ≈ 11.9 years. (Bankers' "rule of 72" estimates 72/6 = 12 years, remarkably close.) The same algebra finds when a growing town reaches a target population, how long a radioactive sample needs to decay to a safe level, and how many half-lives old a carbon-dated artifact is.

Common misconceptions

• Taking the log of each term instead of each side. The logarithm applies to an entire side of the equation at once; log(a + b) cannot be split.
• Confusing log(20)/log(3) with log(20/3). The first is log base 3 of 20 by change of base; the second is a different number entirely.
• Skipping the domain check. Candidates that make any original logarithm's argument zero or negative are extraneous and must be discarded.
• Taking logs before isolating. In 5 · 2x = 40, divide by 5 first; logging both sides immediately works but creates avoidable mess.
• Equating exponents with different bases. From 2x = 3² you may NOT conclude x = 2; exponents equate only when the bases already match.

Recap

Solve exponential equations by writing both sides with a matching base and equating exponents, or, when bases will not match, by isolating the exponential and taking a logarithm of both sides so the power property brings the exponent down. Solve logarithmic equations by condensing to a single logarithm and converting to exponential form, or by equating arguments when both sides are single logs of the same base. Always finish by checking each candidate keeps every original argument positive, and use these tools for doubling-time and decay-time questions in the real world.

Sources

1. OpenStax, Algebra and Trigonometry 2e, Chapter 6 (Exponential and Logarithmic Functions), section on exponential and logarithmic equations. Available free at openstax.org.
2. OpenStax, Intermediate Algebra 2e, Chapter 10 (Exponential and Logarithmic Functions), section on solving exponential and logarithmic equations. Available free at openstax.org.
3. Khan Academy, "Exponential and logarithmic equations" lessons in the Algebra 2 course, khanacademy.org.
4. Paul's Online Math Notes, Lamar University, "Algebra - Solving Exponential Equations" and "Solving Logarithm Equations," tutorial.math.lamar.edu.

Sequences

A sequence is an ordered list of numbers, each called a term. A sequence is arithmetic when each term differs from the previous one by the same fixed amount, the common difference d. For 3, 7, 11, 15, ..., the common difference is 4.

The nth-term formula

The explicit formula for the nth term of an arithmetic sequence is aₙ = a₁ + (n - 1)d, where a₁ is the first term. It says: start at the first term and add d a total of n - 1 times.

Worked example 1: find a term

For the sequence 3, 7, 11, 15, ..., find the 20th term. Here a₁ = 3 and d = 4. So a₂₀ = 3 + (20 - 1)(4) = 3 + 76 = 79.

Arithmetic series

A series is the sum of the terms of a sequence. The sum of the first n terms of an arithmetic sequence is Sₙ = (n/2)(a₁ + aₙ) - the number of terms times the average of the first and last terms.

Worked example 2: sum a series

Find the sum of the first 20 terms of 3, 7, 11, ..., where we found a₂₀ = 79. Then S₂₀ = (20/2)(3 + 79) = 10(82) = 820.

Worked example 3: sum 1 to 100

Add the whole numbers from 1 to 100. Here a₁ = 1, a₁₀₀ = 100, and n = 100. So S = (100/2)(1 + 100) = 50(101) = 5050. This is the famous shortcut for adding a long run of evenly spaced numbers.

Recursive and explicit descriptions

There are two standard ways to define an arithmetic sequence. A recursive formula gives a starting term and a stepping rule: a₁ = 3, aₙ = aₙ₋₁ + 4. It is easy to write but slow to use, because reaching the 200th term means taking 199 steps. The explicit formula aₙ = a₁ + (n - 1)d jumps straight to any term in one calculation. Notice that the explicit formula is really a linear function of n: aₙ = dn + (a₁ - d), so arithmetic sequences are exactly the sequences whose graphs are evenly spaced points along a line of slope d.

Worked example 4: which term is it?

Is 399 a term of 3, 7, 11, 15, ..., and if so, which one? Set the explicit formula equal to 399 and solve for n. Start with 3 + (n - 1)(4) = 399. Subtract 3 from both sides: (n - 1)(4) = 396. Divide by 4: n - 1 = 99. Add 1: n = 100. Since n came out a positive whole number, 399 is indeed a term, the 100th. If n had come out fractional, the number would simply not appear in the sequence.

Worked example 5: rebuild a sequence from two terms

Suppose a₃ = 14 and a₈ = 29. Find the sequence. Moving from the 3rd term to the 8th term takes 5 steps of size d, so 5d = 29 - 14 = 15, giving d = 3. Walk backward from the third term to the first: a₁ = 14 - 2(3) = 8. The sequence is 8, 11, 14, 17, 20, 23, 26, 29, ... Check: a₈ = 8 + (8 - 1)(3) = 8 + 21 = 29. Correct.

Summation notation

Long sums are written compactly with the Greek letter sigma. The expression "the sum from k = 1 to 30 of (2k + 1)" means: substitute k = 1, 2, 3, and so on up to 30 into 2k + 1, then add all the results. The first few terms are 3, 5, 7, ..., which is arithmetic with a₁ = 3 and d = 2, so the arithmetic series formula applies directly to sums written in sigma notation.

Worked example 6: evaluate a sigma sum

Evaluate the sum from k = 1 to 30 of (2k + 1). The first term (k = 1) is 3 and the last term (k = 30) is 2(30) + 1 = 61, with n = 30 terms in total. Apply the sum formula: S₃₀ = (30/2)(3 + 61) = 15(64) = 960.

A second sum formula

Substituting aₙ = a₁ + (n - 1)d into Sₙ = (n/2)(a₁ + aₙ) gives the alternative form Sₙ = (n/2)(2a₁ + (n - 1)d), handy when you know d but not the last term. For the first 12 terms of 7, 12, 17, ... (where a₁ = 7 and d = 5): S₁₂ = (12/2)(2 · 7 + 11 · 5) = 6(14 + 55) = 6(69) = 414.

Real-world applications

Arithmetic sequences model anything that grows by a constant amount per step. A theater has 24 seats in the first row and each row adds 2 more seats; with 20 rows, the last row holds a₂₀ = 24 + 19(2) = 62 seats and the whole house seats S₂₀ = (20/2)(24 + 62) = 10(86) = 860 people. A job that starts at 42,000 dollars with a 1,500 dollar raise each year pays 42,000 + 9(1,500) = 55,500 dollars in year 10, and the total earned across the decade is (10/2)(42,000 + 55,500) = 5(97,500) = 487,500 dollars. Stacked pipes, amphitheater rows, staircase materials, and simple-interest balances all follow the same pattern.

Common misconceptions

• Multiplying d by n instead of n - 1. The first term uses zero steps, so the 20th term adds d only 19 times.
• Confusing a sequence with a series. A sequence is the ordered list; a series is the sum of that list.
• Assuming d must be positive. The sequence 10, 7, 4, 1, ... is arithmetic with d = -3; constant subtraction is still a common difference.
• Using the averaging shortcut on non-arithmetic sums. S = (n/2)(first + last) works because the terms are evenly spaced; it fails for geometric or irregular lists.
• Misreading sigma notation limits. The index does not have to start at 1; the number of terms is the top value minus the bottom value, plus 1.

Recap

An arithmetic sequence adds a fixed common difference d at every step. The explicit formula aₙ = a₁ + (n - 1)d finds any term directly, and it runs in reverse to identify which position a value occupies or to rebuild a sequence from two known terms. A finite arithmetic series is summed with Sₙ = (n/2)(a₁ + aₙ), or with (n/2)(2a₁ + (n - 1)d) when the last term is unknown, and sigma notation is just a compact instruction to form and add such terms. These tools turn long repetitive additions, like seat counts and salary totals, into one-line calculations.

Sources

1. OpenStax, Intermediate Algebra 2e, Chapter 12 (Sequences, Series and Binomial Theorem), section on arithmetic sequences. Available free at openstax.org.
2. OpenStax, Algebra and Trigonometry 2e, Chapter 13 (Sequences, Probability, and Counting Theory), sections on sequences, arithmetic sequences, and series. Available free at openstax.org.
3. OpenStax, College Algebra 2e, Chapter 9 (Sequences, Probability, and Counting Theory), section on arithmetic sequences. Available free at openstax.org.
4. Khan Academy, "Arithmetic sequences" and "Arithmetic series" lessons, khanacademy.org.

Multiplying instead of adding

A sequence is geometric when each term is the previous one multiplied by the same fixed number, the common ratio r. For 2, 6, 18, 54, ..., each term is 3 times the one before, so r = 3. You find r by dividing any term by the previous term.

The nth-term formula

The explicit formula is aₙ = a₁ · rⁿ⁻¹: start at the first term and multiply by r a total of n - 1 times.

Worked example 1: find a term

For 2, 6, 18, 54, ..., find the 6th term. Here a₁ = 2 and r = 3. So a₆ = 2 · 3⁵ = 2 · 243 = 486.

Finite geometric series

The sum of the first n terms of a geometric sequence (with r ≠ 1) is Sₙ = a₁(1 - rⁿ)/(1 - r).

Worked example 2: sum a finite series

Sum the first 4 terms of 2, 6, 18, 54. Using the formula with a₁ = 2, r = 3, n = 4: S₄ = 2(1 - 3⁴)/(1 - 3) = 2(1 - 81)/(-2) = 2(-80)/(-2) = 80. A direct check: 2 + 6 + 18 + 54 = 80. Correct.

Infinite geometric series

When the common ratio satisfies |r| < 1, the terms shrink toward zero and the infinite sum settles on a finite value: S = a₁/(1 - r). If |r| ≥ 1, the terms do not shrink and the infinite sum does not exist.

Worked example 3: an infinite sum

Sum 1 + 1/2 + 1/4 + 1/8 + .... Here a₁ = 1 and r = 1/2, and since |r| < 1 the sum is S = 1/(1 - 1/2) = 1/(1/2) = 2. Adding infinitely many pieces gives a finite total of 2.

Arithmetic or geometric? Test, do not guess

To classify a sequence, test both patterns. For 2, 6, 18, 54: the differences are 4, 12, 36 (not constant), but the ratios are 6/2 = 3, 18/6 = 3, and 54/18 = 3 (constant), so it is geometric. For 2, 6, 10, 14 the differences are all 4, so it is arithmetic. A sequence can also be neither, like the squares 1, 4, 9, 16. Always compute r by dividing a term by the term before it, never the other way around.

Worked example 4: a negative ratio

For 81, -27, 9, -3, ..., the ratio is r = -27/81 = -1/3. The terms alternate sign because the ratio is negative, and they shrink because its absolute value is less than 1. The 6th term is a₆ = 81 · (-1/3)⁵ = 81 · (-1/243) = -81/243 = -1/3. Listing the terms confirms it: 81, -27, 9, -3, 1, -1/3.

Worked example 5: find the ratio from two terms

Suppose a₂ = 6 and a₅ = 48. Moving from the 2nd term to the 5th multiplies by r three times, so 6r³ = 48, giving r³ = 8 and r = 2. Then a₁ = a₂/r = 6/2 = 3, and the sequence is 3, 6, 12, 24, 48, ... Check: a₅ = 3 · 2⁴ = 3 · 16 = 48. Correct.

Worked example 6: a longer finite sum

Sum the first 8 terms of 5, 10, 20, 40, ... Here a₁ = 5, r = 2, and n = 8. Then S₈ = 5(1 - 2⁸)/(1 - 2) = 5(1 - 256)/(-1) = 5(-255)/(-1) = 1275. Adding the terms directly, 5 + 10 + 20 + 40 + 80 + 160 + 320 + 640 = 1275, confirms the formula. Notice how the doubling makes the last term (640) contribute more than everything before it combined; that top-heavy growth is the signature of geometric series.

Repeating decimals are geometric series

The decimal 0.7777... unpacks as 7/10 + 7/100 + 7/1000 + ..., an infinite geometric series with a₁ = 7/10 and r = 1/10. Its sum is S = (7/10)/(1 - 1/10) = (7/10)/(9/10) = 7/9. Every repeating decimal hides a geometric series, which is why every repeating decimal equals a fraction.

Real-world applications

Geometric sequences model anything that grows or shrinks by a constant percent per step. Money at 5 percent yearly interest turns 1,000 dollars into the sequence 1,000, 1,050, 1,102.50, ... with r = 1.05; after 10 years the balance is 1,000(1.05)10 ≈ 1,628.89 dollars. A ball dropped from 10 meters that rebounds to 60 percent of each previous height has rebound heights 6, 3.6, 2.16, ..., and the total of all its rebounds is the infinite sum 6/(1 - 0.6) = 6/0.4 = 15 meters. Radioactive decay, medicine concentrations between doses, and populations changing by a fixed percentage all run on the same mathematics.

Common misconceptions

• Using rⁿ instead of rⁿ⁻¹. The first term has been multiplied by r zero times, so the 6th term uses the 5th power.
• Assuming every infinite geometric series has a sum. The formula a₁/(1 - r) applies only when the absolute value of r is less than 1; the series 2 + 6 + 18 + ... grows without bound.
• Dividing in the wrong order for r. The ratio is any term divided by the term before it; dividing the earlier term by the later one gives 1/r.
• Dropping the sign of a negative ratio. With r negative the terms must alternate sign; if your listed terms do not alternate, recheck r.
• Mixing up the sum formulas. Arithmetic sums average the first and last terms; geometric sums use a₁(1 - rⁿ)/(1 - r). Classify the sequence before reaching for either.

Recap

A geometric sequence multiplies by a fixed common ratio r at each step, so the nth term is aₙ = a₁ · rⁿ⁻¹. Finite sums come from Sₙ = a₁(1 - rⁿ)/(1 - r), and when the absolute value of r is less than 1 the infinite series settles to a₁/(1 - r). Ratios can be negative (alternating signs) or fractional (shrinking terms), and two known terms are enough to recover r and rebuild the whole sequence. Compound interest, bouncing balls, repeating decimals, and percent growth of every kind are geometric series in disguise.

Sources

1. OpenStax, Intermediate Algebra 2e, Chapter 12 (Sequences, Series and Binomial Theorem), section on geometric sequences and series. Available free at openstax.org.
2. OpenStax, Algebra and Trigonometry 2e, Chapter 13 (Sequences, Probability, and Counting Theory), sections on geometric sequences and series. Available free at openstax.org.
3. Khan Academy, "Geometric sequences" and "Geometric series" lessons, khanacademy.org.
4. Paul's Online Math Notes, Lamar University, "Calculus II - Special Series" (geometric series), tutorial.math.lamar.edu.

Ratios in a right triangle

In a right triangle, pick one acute angle. The side across from it is the opposite side, the side next to it (not the hypotenuse) is the adjacent side, and the longest side across from the right angle is the hypotenuse. The three basic trigonometric ratios compare these sides.

SOH-CAH-TOA

The mnemonic SOH-CAH-TOA records the three ratios:

• Sine = Opposite / Hypotenuse.
• Cosine = Adjacent / Hypotenuse.
• Tangent = Opposite / Adjacent.

Worked example 1: ratios in a 3-4-5 triangle

A right triangle has legs 3 (opposite) and 4 (adjacent) and hypotenuse 5. For the angle θ at the corner between the adjacent side and hypotenuse: sinθ = 3/5, cosθ = 4/5, and tanθ = 3/4.

Worked example 2: find a missing side

In a right triangle, an angle measures 30 degrees and the hypotenuse is 10. Find the side opposite the 30-degree angle. Use sine: sin 30° = opposite/10. Since sin 30° = 1/2, we get opposite = 10 · (1/2) = 5.

The Pythagorean theorem

The legs and hypotenuse always satisfy a² + b² = c², where c is the hypotenuse. If the legs are 6 and 8, then c² = 36 + 64 = 100, so c = 10. Solving a right triangle means finding all its unknown sides and angles, using these ratios together with the Pythagorean theorem.

Worked example 3: find a height with tangent

From a point 50 meters from the base of a tower, the angle of elevation to the top is 30 degrees. The tower's height is the side opposite the angle and the 50-meter distance is adjacent, so tangent is the right ratio: tan 30° = h/50. Multiply both sides by 50: h = 50 tan 30° = 50(√3/3) ≈ 50(0.577) ≈ 28.9 meters. Choosing the ratio that links one known side to the side you want is the whole game; the third side never needs to enter the calculation.

Finding angles: inverse trig ratios

When you know two sides and want the angle, run a ratio backward with an inverse trig function. If tanθ = 3/4 = 0.75, then θ = tan-1(0.75) ≈ 36.9°. Likewise sin^-1 and cos^-1 recover angles from sine and cosine ratios. In the 3-4-5 triangle the two acute angles are about 36.9 degrees and 53.1 degrees, and they add to 90 degrees, as the two acute angles of a right triangle always must.

Worked example 4: solve a complete right triangle

A right triangle has a 40-degree angle and hypotenuse 12. Solve the triangle. The other acute angle is 90 - 40 = 50 degrees. The side opposite the 40-degree angle uses sine: opp = 12 sin 40° ≈ 12(0.643) ≈ 7.7. The adjacent side uses cosine: adj = 12 cos 40° ≈ 12(0.766) ≈ 9.2. Check: 7.7² + 9.2² ≈ 59.3 + 84.6 = 143.9 ≈ 144 = 12². The Pythagorean theorem confirms the pair.

Special right triangles

Two triangles appear so often that their side ratios are worth memorizing. In a 45-45-90 triangle the legs are equal and the hypotenuse is a leg times √2, so legs of 7 give a hypotenuse of 7√2 ≈ 9.9. In a 30-60-90 triangle the sides are in the ratio x : x√3 : 2x: the shorter leg (opposite the 30-degree angle) is x, the longer leg (opposite the 60-degree angle) is x√3, and the hypotenuse is 2x.

Worked example 5: a 30-60-90 triangle

The shorter leg of a 30-60-90 triangle is 5. Then the hypotenuse is 2(5) = 10 and the longer leg is 5√3 ≈ 8.66. Check with the Pythagorean theorem: 5² + (5√3)² = 25 + 75 = 100 = 10². These exact ratios are where the exact values sin 30° = 1/2 and cos 30° = √3/2 come from.

Real-world applications: elevation and depression

The angle of elevation is measured upward from horizontal and the angle of depression downward, and surveyors, pilots, and builders live on them. An accessible wheelchair ramp rises 1 unit for every 12 units of run, so its angle is tan-1(1/12) ≈ 4.8°, comfortably gentle. An airliner on a standard 3-degree descent path at an altitude of 900 meters is about 900/tan 3° ≈ 900/0.0524 ≈ 17,200 meters, roughly 17.2 kilometers, from the runway threshold. The same triangle logic measures tree heights from shadows, cliff heights from boat sightings, and roof pitches from rise and run.

Common misconceptions

• Treating opposite and adjacent as fixed labels. They depend on which acute angle you pick; switch angles and the two legs trade names.
• Letting the hypotenuse play adjacent. The hypotenuse is always the side across from the right angle; adjacent means the other side touching your angle.
• Wrong calculator mode. Computing sin 30 in radian mode returns about -0.988 instead of 0.5; check the mode before trusting any decimal.
• Reading sinθ as multiplication. Sine is a function of the angle, not sin times theta; sin 60° is not twice sin 30°.
• Using SOH-CAH-TOA in non-right triangles. These ratios require a right angle; oblique triangles need the laws of sines and cosines instead.

Recap

Right-triangle trigonometry names the sides relative to a chosen acute angle and compares them with three ratios: sine (opposite over hypotenuse), cosine (adjacent over hypotenuse), and tangent (opposite over adjacent). One known angle and one known side unlock the remaining sides; two known sides unlock the angles through inverse trig functions; and the Pythagorean theorem ties all three sides together as a built-in check. The 45-45-90 and 30-60-90 patterns supply exact values, and angles of elevation and depression carry the whole toolkit into surveying, aviation, and construction.

Sources

1. OpenStax, Algebra and Trigonometry 2e, Chapter 7 (The Unit Circle: Sine and Cosine Functions), section on right triangle trigonometry. Available free at openstax.org.
2. Khan Academy, "Right triangles and trigonometry" unit, khanacademy.org.
3. Khan Academy, "Special right triangles" lessons, khanacademy.org.
4. Paul's Online Math Notes, Lamar University, "Algebra/Trig Review - Trig Function Evaluation," tutorial.math.lamar.edu.

Two ways to measure angles

Angles can be measured in degrees, where a full circle is 360 degrees, or in radians, where a full circle is 2π radians. The bridge between them is 180 degrees = π radians. To convert degrees to radians, multiply by π/180; to convert radians to degrees, multiply by 180/π.

Worked example 1: convert

Convert 60 degrees to radians: 60 · π/180 = π/3. Convert 3π/4 radians to degrees: (3π/4)(180/π) = 135 degrees.

The unit circle

The unit circle is the circle of radius 1 centered at the origin. For an angle θ in standard position (vertex at the origin, initial side along the positive x-axis), the point where the terminal side meets the circle has coordinates (cosθ, sinθ). That single fact extends sine and cosine to every angle, not just those in a triangle.

Special angles

Three first-quadrant angles appear constantly. Their exact values are:

Angle	sin	cos
30 degrees (pi/6)	1/2	the square root of 3, over 2
45 degrees (pi/4)	the square root of 2, over 2	the square root of 2, over 2
60 degrees (pi/3)	the square root of 3, over 2	1/2

Reference angles and signs

A reference angle is the acute angle between the terminal side and the x-axis. It lets you reduce any angle to a first-quadrant version, then attach the sign for the quadrant. For example, 150 degrees lies in Quadrant II with a reference angle of 30 degrees, so its sine matches sin 30 (positive, 1/2) while its cosine is the negative of cos 30. In Quadrant II sine is positive and cosine is negative.

Radians and arc length

Radians are not an arbitrary convention: an angle in radians is the ratio of arc length to radius, which makes circle formulas clean. An arc cut by a central angle θ (in radians) on a circle of radius r has length s = rθ. On a circle of radius 6 centimeters, a central angle of π/3 cuts an arc of s = 6(π/3) = 2π ≈ 6.28 centimeters. The same formula in degrees would need an extra factor of π/180, which is exactly why science and calculus prefer radians.

Worked example 2: conversions both ways

Convert 270 degrees to radians: 270 · π/180 = 3π/2. Convert -45 degrees: -45 · π/180 = -π/4. Convert 5π/6 radians to degrees: (5π/6)(180/π) = 150 degrees. A quick sanity check: since π radians is a half turn, 3π/2 should be three quarters of a turn, and 270 degrees is exactly that.

Coterminal angles

Adding or subtracting full turns of 360 degrees (or 2π radians) leaves the terminal side unchanged. Such angles are coterminal and share the same sine and cosine values. So 400 degrees is coterminal with 40 degrees, and -30 degrees is coterminal with 330 degrees. To reduce 850 degrees, strip out two full turns: 850 - 720 = 130 degrees.

Quadrantal angles

Angles that land on an axis are read straight from the circle's coordinates: at 0 degrees the point is (1, 0); at 90 degrees, (0, 1); at 180 degrees, (-1, 0); at 270 degrees, (0, -1). Therefore cos 180° = -1, sin 270° = -1, and sin 180° = 0. No reference angle is needed for these; the coordinates are the values.

Worked example 3: evaluate with reference angles

Find sin 225°. The angle lies in Quadrant III, where its reference angle is 225 - 180 = 45 degrees. Sine is negative in Quadrant III, so sin 225° = -√2/2. Now find cos 300°. The angle lies in Quadrant IV with reference angle 360 - 300 = 60 degrees, and cosine is positive there, so cos 300° = cos 60° = 1/2.

The sign pattern by quadrant

The mnemonic All Students Take Calculus records which functions are positive, moving counterclockwise from Quadrant I: All three in Quadrant I, Sine only in Quadrant II, Tangent only in Quadrant III, and Cosine only in Quadrant IV. Combined with reference angles, it reduces any angle to a first-quadrant lookup plus a sign.

Worked example 4: a radian-measured angle

Evaluate sin(7π/6) and cos(7π/6). Converting, 7π/6 = 210 degrees, which sits in Quadrant III with a reference angle of 30 degrees. Both sine and cosine are negative there, so sin(7π/6) = -1/2 and cos(7π/6) = -√3/2.

Real-world applications

Arc length powers real measurement. Each degree of latitude on Earth (radius about 6,371 km) spans s = 6,371 · π/180 ≈ 111 kilometers, which is how navigators first sized the planet. A 40-meter wind-turbine blade sweeping a quarter turn moves its tip through 40(π/2) = 20π ≈ 62.8 meters. And because the unit circle extends sine and cosine to all angles, it is the mathematical engine behind rotating machinery, alternating current, and the seasonal swing of daylight hours that the next lesson graphs.

Common misconceptions

• Thinking π radians is a full circle. A full circle is 2π; the number π corresponds to a half turn of 180 degrees.
• Writing the point as (sin, cos). The x-coordinate comes first and x is cosine: the point is (cosθ, sinθ).
• Measuring reference angles from the y-axis. A reference angle is always the acute angle to the x-axis; for 150 degrees it is 30, not 60.
• Ignoring calculator mode. Evaluating sin(30) in radian mode or sin(π/6) in degree mode both give wrong answers; match the mode to the units.
• Skipping the sign step. The reference angle gives the size of the value; the quadrant gives its sign, and dropping either half loses the answer.

Recap

Degrees and radians are two rulers for the same angles, linked by 180 degrees = π radians, and radians earn their keep through the arc-length formula s = rθ. The unit circle defines cosine and sine as the x and y coordinates of the terminal-side point, extending both functions to every angle, with coterminal angles sharing the same values. Quadrantal angles read straight off the axes, and every other angle reduces to the special-angle table by finding its reference angle and then attaching the quadrant's sign using All Students Take Calculus.

Sources

1. OpenStax, Algebra and Trigonometry 2e, Chapter 7 (The Unit Circle: Sine and Cosine Functions), sections on angles and the unit circle. Available free at openstax.org.
2. Khan Academy, "Unit circle introduction" lessons in the Trigonometry course, khanacademy.org.
3. Khan Academy, "Radians" lessons in the Trigonometry course, khanacademy.org.
4. Paul's Online Math Notes, Lamar University, "Algebra/Trig Review - Trig Function Evaluation," tutorial.math.lamar.edu.

Two waves

The graphs of sine and cosine are smooth repeating waves. The sine curve starts at 0, rises to 1, returns to 0, falls to -1, and comes back to 0 over one cycle. The cosine curve has the same shape but starts at its maximum of 1. Because they repeat, they are periodic, and their natural period is 2π.

Amplitude

The amplitude is the height from the middle line to a peak. For y = a sin(x) or y = a cos(x), the amplitude is |a|. The function y = 3 sin(x) rises to 3 and falls to -3.

Midline

The midline is the horizontal line halfway between the maximum and minimum. Adding a constant shifts it: y = sin(x) + 2 has midline y = 2, oscillating between 1 and 3.

Period

The coefficient of x changes how fast the wave repeats. For y = sin(bx), the period is 2π/|b|. A larger b squeezes more cycles into the same width.

Worked example 1: read a graph's features

For y = 4 cos(2x), the amplitude is |4| = 4, so the curve runs between 4 and -4. The period is 2π/2 = π, so one full cycle finishes in a horizontal span of π instead of 2π.

Worked example 2: build an equation

Write a cosine function with amplitude 5 and period π. The amplitude gives a = 5. For the period, solve 2π/b = π, which gives b = 2. So one answer is y = 5 cos(2x). Understanding amplitude, midline, and period lets you both read and construct these waves.

Five key points per cycle

Every sine or cosine cycle is anchored by five evenly spaced points: the start, quarter, half, three-quarter, and end marks of one period. For y = sin(x) they are (0, 0), (π/2, 1), (π, 0), (3π/2, -1), and (2π, 0): midline, maximum, midline, minimum, midline. Cosine hits the same five x-values in the order maximum, midline, minimum, midline, maximum. Plot the five anchors, join them with a smooth wave, and the sketch is done.

Worked example 3: sketch y = 3 sin(x)

The amplitude is 3 and the period is 2π, so divide one period into quarters at x = 0, π/2, π, 3π/2, 2π. Scale the sine pattern by 3: the anchor heights are 0, 3, 0, -3, 0. The curve rises from the origin to a peak of 3 at π/2, returns to the midline at π, dips to -3 at 3π/2, and closes the cycle at 2π. Only the heights changed; the timing of the wave is untouched by a.

The full transformation form

The general sinusoid is y = a sin(b(x - c)) + d, or the same with cosine. Each letter controls one feature: |a| is the amplitude, 2π/|b| is the period, c is the phase shift (right if positive, left if negative), and d raises the midline to y = d. The maximum is d + |a| and the minimum is d - |a|. If a is negative, the wave is reflected across its midline and starts by falling instead of rising.

Phase shift

Replacing x with x - π/3 slides the whole wave right by π/3: the graph of y = sin(x - π/3) crosses its midline heading up at x = π/3 instead of at 0. Watch for equations like y = sin(2x - π): factor the inside as 2(x - π/2) first, so the shift is π/2, not π. Phase shift also explains a family resemblance: shifting sine left by π/2 lands exactly on cosine, since sin(x + π/2) = cos(x); the two curves are one shape viewed from different starting posts.

Worked example 4: read every feature

Analyze y = 2 sin(3(x - π/6)) + 1. Amplitude: |2| = 2. Period: 2π/3. Phase shift: π/6 to the right. Midline: y = 1. So the wave oscillates between 1 + 2 = 3 and 1 - 2 = -1, completing a full cycle every 2π/3 units, with its upward midline crossing at x = π/6.

Worked example 5: write an equation from a description

A wave oscillates between a maximum of 8 and a minimum of 2, has period 4π, and starts at its midline heading upward, so sine is the natural choice. Midline: d = (8 + 2)/2 = 5. Amplitude: a = (8 - 2)/2 = 3. Period: solve 2π/b = 4π to get b = 1/2. The equation is y = 3 sin(x/2) + 5. Check: maximum 5 + 3 = 8, minimum 5 - 3 = 2, period 2π/(1/2) = 4π. All three match.

Real-world applications

Sinusoids are the mathematics of anything that cycles. Tides: a harbor's depth h(t) = 1.5 sin(πt/6) + 4 swings between 2.5 and 5.5 meters around a 4-meter midline with period 2π/(π/6) = 12 hours, roughly the twice-daily tide. A Ferris wheel of radius 20 meters whose hub sits 25 meters up gives riders the height h(t) = 25 - 20 cos(πt/20): a boarding height of 5 meters at t = 0, a top height of 45 meters, and a 40-second revolution. Household electricity in the United States is V(t) = 170 sin(120πt), a sinusoid with period 2π/(120π) = 1/60 second, which is exactly 60 cycles per second. Daylight length across the year, sound waves, and heartbeats on an EKG all wear the same shape.

Common misconceptions

• Reading the period as b. The period is 2π/|b|; y = sin(2x) has period π and repeats faster, not slower.
• Calling the amplitude negative. Amplitude is the distance |a|; a negative a reflects the wave, but the amplitude stays positive.
• Taking the phase shift straight from an unfactored formula. In y = sin(2x - π), factor to 2(x - π/2); the shift is π/2, not π.
• Confusing maximum with amplitude. The maximum is d + |a|; with a raised midline, an amplitude-3 wave can peak at 8.
• Expecting a or b to move the midline. Changes to a and b stretch the wave but never lift it; only the constant d added outside shifts it vertically.

Recap

Sine and cosine graph as endless smooth waves with natural period 2π, and the transformed form y = a sin(b(x - c)) + d dials in every feature: |a| sets the amplitude, 2π/|b| the period, c the phase shift, and d the midline, with a negative a flipping the wave. Five anchor points per cycle make sketching fast, and the same four numbers can be read off a graph to reconstruct its equation. Tides, Ferris wheels, alternating current, and every other steady cycle in nature are described by exactly these curves.

Sources

1. OpenStax, Algebra and Trigonometry 2e, Chapter 8 (Periodic Functions), section on graphs of the sine and cosine functions. Available free at openstax.org.
2. Khan Academy, "Graphs of sin(x), cos(x), and tan(x)" lessons, khanacademy.org.
3. Khan Academy, "Amplitude, midline, and period" lessons, khanacademy.org.
4. Paul's Online Math Notes, Lamar University, "Algebra/Trig Review - Graphs of Trig Functions," tutorial.math.lamar.edu.