Precalculus

What a function is

A function is a rule that assigns to each input exactly one output. Think of it as a machine: you feed in a number, and the machine returns a single number. The set of allowed inputs is the domain, and the set of outputs that actually come out is the range. The key word is "exactly one" - an input can never lead to two different outputs.

We name functions with letters and write function notation like f(x), read "f of x." This does not mean f times x; it means "the output of f when the input is x." If f(x) = 2x + 1, then f(3) = 2(3) + 1 = 7. The symbol f(3) is just the height of the graph above x = 3.

The vertical line test

A graph represents a function exactly when no vertical line crosses it more than once. This is the vertical line test. A vertical line fixes one x-value; if it hits the graph twice, that single input has two outputs, which a function is not allowed to do. A circle fails the test, but a parabola opening upward passes it.

Finding the domain

When a function is given by a formula and no domain is stated, the domain is every real number for which the formula makes sense. Two operations create restrictions:

• Division by zero is forbidden. Any x that makes a denominator zero must be excluded.
• Even roots of negatives are not real. The expression under a square root must be greater than or equal to zero.

Worked example 1: domain of a rational function

Find the domain of f(x) = 1 / (x - 4). The denominator is zero when x - 4 = 0, that is x = 4. Every other real number is allowed, so the domain is all real numbers except 4.

Worked example 2: domain of a square root

Find the domain of g(x) = √(x - 2). We need x - 2 ≥ 0, so x ≥ 2. The domain is all real numbers greater than or equal to 2. Because a square root is never negative, the range here is all outputs greater than or equal to 0.

Worked example 3: evaluating and reading a graph

Let f(x) = x² - 1. Then f(-3) = (-3)² - 1 = 9 - 1 = 8, and f(0) = -1. The smallest output is -1 (at x = 0) and the outputs grow without bound, so the range is all real numbers greater than or equal to -1, while the domain is all real numbers.

Reading a graph, the domain is the set of x-values the curve covers (its shadow on the x-axis) and the range is the set of y-values it covers (its shadow on the y-axis). Getting comfortable moving between the formula, the graph, and these two sets is the foundation for everything that follows.

Combining restrictions

Many formulas carry more than one restriction at the same time. The rule is simple: apply every restriction separately, then keep only the x-values that satisfy all of them at once.

Worked example 4: two restrictions at once

Find the domain of h(x) = √(x + 6) / (x - 2).

1. The square root needs x + 6 ≥ 0, which means x ≥ -6.
2. The denominator cannot equal zero, so x ≠ 2.

Both conditions must hold together, so the domain is every x greater than or equal to -6 except x = 2. Checking a few values makes it concrete: x = -6 works (the numerator is √0 = 0 and the denominator is -8), x = -7 fails the square root, and x = 2 fails the denominator.

Piecewise-defined functions

A piecewise function uses different formulas on different pieces of its domain. Real life is full of them: postage rates, tax brackets, and parking fees all switch rules at certain break points. To evaluate a piecewise function, first decide which piece the input belongs to, then use only that formula.

Worked example 5: evaluate a piecewise function

Let f be defined by f(x) = x + 1 when x < 0 and f(x) = x² when x ≥ 0.

1. f(-2): since -2 < 0, use the first piece: f(-2) = -2 + 1 = -1.
2. f(0): since 0 satisfies x ≥ 0, use the second piece: f(0) = 0² = 0.
3. f(3): since 3 ≥ 0, use the second piece: f(3) = 3² = 9.

The single most common error with piecewise functions is plugging the input into the wrong piece, so always check the condition before you compute.

Average rate of change

The average rate of change of a function between x = a and x = b is the slope of the line through those two points on the graph: (f(b) - f(a)) / (b - a). It answers the question "on average, how fast did the output change per unit of input?" and it is the precalculus ancestor of the derivative you will meet in calculus.

Worked example 6: average rate of change

Find the average rate of change of f(x) = x² from x = 1 to x = 3.

1. Compute the outputs: f(3) = 3² = 9 and f(1) = 1² = 1.
2. Divide the change in output by the change in input: (9 - 1) / (3 - 1) = 8 / 2 = 4.

On average the function climbs 4 output units per input unit across that interval, even though it climbs slowly near x = 1 and faster near x = 3.

Real-world applications

Domain thinking is everywhere once you look for it. If C(n) is the cost of printing n posters, the practical domain is whole numbers with n ≥ 0, whatever the formula would allow. If h(t) gives the height of a ball t seconds after it is thrown, the model only makes sense from t = 0 until the ball lands; for h(t) = -16t² + 32t, setting h(t) = 0 gives -16t(t - 2) = 0, so the sensible domain is 0 ≤ t ≤ 2 seconds. A phone plan that charges 30 dollars for the first 5 GB and 10 dollars per extra GB is a piecewise function of data used. In every case, the algebra of the formula and the common sense of the situation both shape the domain.

Common misconceptions

• "f(x) means f times x." It does not. It is the output of the function f at the input x.
• "The range is the same as the domain." They usually differ: the domain lists inputs, the range lists the outputs actually produced.
• "Every formula has all real numbers as its domain." Denominators and even roots create exclusions you must hunt down.
• "A graph that fails the horizontal line test is not a function." The vertical line test decides whether a graph is a function; the horizontal test only checks whether it is one-to-one.
• "f(0) is always 0." f(0) is the y-intercept and can be any value; for f(x) = x² - 1 it is -1.

Recap

A function assigns exactly one output to each input. Its domain is the set of legal inputs, restricted by denominators (never zero) and even roots (never negative inside), and its range is the set of outputs it actually produces. Function notation f(x) names the output at x, the vertical line test identifies function graphs, piecewise functions switch formulas at break points, and the average rate of change (f(b) - f(a)) / (b - a) measures the slope between two points on the curve. These are the working tools for every chapter that follows.

Sources

1. OpenStax, Precalculus 2e, Chapter 1: Functions (Sections 1.1 to 1.3). OpenStax, Rice University. Free at openstax.org/details/books/precalculus-2e.
2. OpenStax, Algebra and Trigonometry 2e, Chapter 3: Functions. Free at openstax.org/details/books/algebra-and-trigonometry-2e.
3. Khan Academy, Precalculus: functions review and composite functions unit. khanacademy.org/math/precalculus.
4. Paul Dawkins, Paul's Online Math Notes, Algebra: Graphing and Functions. tutorial.math.lamar.edu.

Parent functions

A parent function is the simplest form of a family, such as y = x² for parabolas or y = |x| for absolute value. Once you know a parent's shape, you can produce many relatives by shifting, reflecting, and stretching it. These predictable moves are called transformations.

Vertical and horizontal shifts

Starting from y = f(x):

• f(x) + k shifts the graph up by k (or down if k is negative).
• f(x) - k shifts the graph down by k.
• f(x - h) shifts the graph right by h. This one surprises students: the minus sign moves it right.
• f(x + h) shifts the graph left by h.

Vertical shifts do what you expect; horizontal shifts go the opposite way of the sign, because the change is happening inside the function to the input.

Reflections

• -f(x) reflects the graph across the x-axis (flips it upside down).
• f(-x) reflects the graph across the y-axis (flips it left to right).

Stretches and compressions

Multiplying the whole function by a constant a scales it vertically. If a > 1, the graph is a vertical stretch (taller and steeper); if 0 < a < 1, it is a vertical compression (flatter). For example, y = 3x² is three times as tall as y = x² at every x.

Worked example: read a transformed function

Describe the graph of g(x) = -2(x - 3)² + 4 compared to the parent y = x². Work from the inside out:

1. (x - 3): shift right 3.
2. The factor 2: vertical stretch by a factor of 2.
3. The minus sign in front: reflect across the x-axis, so the parabola now opens downward.
4. + 4: shift up 4.

The vertex of the parent at (0, 0) moves to (3, 4), and the parabola opens down. A general rule of thumb: handle the horizontal shift and any horizontal effects first, then stretches and reflections, then the vertical shift last. Checking one point confirms it: at x = 3, g(3) = -2(0)² + 4 = 4, matching the vertex.

Even and odd functions

Symmetry is a special case of reflection. A function is even if f(-x) = f(x), meaning its graph is symmetric across the y-axis (like x²). It is odd if f(-x) = -f(x), meaning it has rotational symmetry through the origin (like x³).

Worked example: test even or odd algebraically

Classify f(x) = x³ - 4x. Replace x with -x and simplify carefully:

1. f(-x) = (-x)³ - 4(-x) = -x³ + 4x.
2. Factor out -1: -x³ + 4x = -(x³ - 4x) = -f(x).

Since f(-x) = -f(x), the function is odd. By contrast, g(x) = x² + 1 gives g(-x) = (-x)² + 1 = x² + 1 = g(x), so g is even. And h(x) = x² + x gives h(-x) = x² - x, which is neither h(x) nor -h(x), so h is neither even nor odd. Most functions are neither; the labels describe special symmetry, not a complete classification.

Horizontal stretches and compressions

Multiplying the input scales the graph horizontally, and just like horizontal shifts, the effect runs opposite to intuition. The graph of f(bx) with b > 1 is a horizontal compression by a factor of 1/b (the wave of activity happens b times as fast), while 0 < b < 1 stretches the graph horizontally. For example, y = √(2x) reaches height 2 at x = 2, whereas the parent y = √x needs x = 4 to reach height 2: everything happens in half the horizontal room.

Worked example: track a single point

Transformations become fully concrete when you follow one point. Suppose (4, 2) lies on y = f(x). Find the matching point on y = 2 f(x - 3) + 1.

1. The inside change x - 3 shifts the point right 3: the input 4 becomes 4 + 3 = 7.
2. The outside factor 2 doubles the output: 2 becomes 2 · 2 = 4.
3. The outside + 1 raises the output: 4 becomes 4 + 1 = 5.

The new point is (7, 5). Notice inside changes act on x and outside changes act on y; that single sentence organizes the entire topic.

Worked example: build a formula from a description

Write the equation for the parent y = √x after it is shifted left 1, reflected across the x-axis, and shifted up 3.

1. Shift left 1: replace x with x + 1 to get y = √(x + 1).
2. Reflect across the x-axis: multiply by -1 to get y = -√(x + 1).
3. Shift up 3: add 3 to get y = -√(x + 1) + 3.

Check with a point: at x = 0, y = -√1 + 3 = -1 + 3 = 2. The parent point (1, 1) moved left to (0, 1), flipped to (0, -1), and rose to (0, 2). The check matches, so the formula is right.

Real-world applications

Transformations are the language of modeling. A company whose daily profit follows P(t) but opens two hours later shifts to P(t - 2). Doubling every price scales a revenue curve vertically by 2. In physics, a sound wave played at double speed is a horizontal compression of the original wave, and raising the volume is a vertical stretch. Reading a formula as "the same underlying shape, moved and scaled" is often the fastest way to understand a new model.

Common misconceptions

• "f(x - 3) shifts left because of the minus." It shifts right 3. Inside changes run opposite to their sign.
• "f(x) + k and f(x + k) do the same thing." The first moves the graph vertically, the second horizontally.
• "-f(x) and f(-x) are the same reflection." The first flips across the x-axis, the second across the y-axis.
• "Order of transformations never matters." It does when a stretch or reflection combines with a shift in the same direction; apply stretches and reflections before the vertical shift.
• "Any graph is either even or odd." Most functions are neither; symmetry is special.

Recap

Every transformed function is a parent shape that has been shifted, reflected, or scaled. Outside changes act on outputs: f(x) + k shifts vertically, a f(x) stretches vertically, -f(x) flips across the x-axis. Inside changes act on inputs and run opposite to their sign: f(x - h) shifts right h, f(bx) compresses horizontally for b > 1, and f(-x) flips across the y-axis. Track a known point through each step to verify your work, and use f(-x) to test for even (matches f(x)) or odd (matches -f(x)) symmetry.

Sources

1. OpenStax, Precalculus 2e, Section 1.5: Transformation of Functions. OpenStax, Rice University. Free at openstax.org/details/books/precalculus-2e.
2. OpenStax, Algebra and Trigonometry 2e, Section 3.5: Transformation of Functions. Free at openstax.org/details/books/algebra-and-trigonometry-2e.
3. Khan Academy, Precalculus and Algebra 2: transformations of functions. khanacademy.org/math/precalculus.
4. Paul Dawkins, Paul's Online Math Notes, Algebra: Transformations. tutorial.math.lamar.edu.

Composing functions

The composition of f and g, written (f ∘ g)(x) or f(g(x)), means "do g first, then feed the result into f." The output of the inside function becomes the input of the outside function. Order matters: f(g(x)) is usually different from g(f(x)).

Worked example 1: evaluate a composition

Let f(x) = x² + 1 and g(x) = 3x - 2. Find f(g(4)).

1. Inside first: g(4) = 3(4) - 2 = 10.
2. Outside next: f(10) = 10² + 1 = 101.

To build the general formula f(g(x)), substitute all of g into f: f(g(x)) = (3x - 2)² + 1.

Inverse functions

An inverse function, written f⁻¹, undoes what f does. If f turns 3 into 7, then f⁻¹ turns 7 back into 3. Formally, f⁻¹(f(x)) = x and f(f⁻¹(x)) = x. Only a one-to-one function has an inverse: each output must come from just one input, which you can check with the horizontal line test.

Worked example 2: find an inverse

Find the inverse of f(x) = 2x + 6. A clean recipe: replace f(x) with y, swap x and y, then solve for y.

1. Write y = 2x + 6.
2. Swap: x = 2y + 6.
3. Solve: subtract 6 to get x - 6 = 2y, then divide by 2: y = (x - 6)/2.

So f⁻¹(x) = (x - 6)/2. The swap step reflects the reasoning perfectly, because the graph of an inverse is the mirror image of the original across the line y = x.

Worked example 3: verify with composition

Check that the answer above is truly the inverse by composing. Compute f(f⁻¹(x)) = 2 · ((x - 6)/2) + 6 = (x - 6) + 6 = x. Because the composition returns x, the two functions undo each other, confirming the inverse.

Why inverses matter

Inverses let us reverse a process to recover the original input. When you solve an equation, you are really applying inverse operations one step at a time. Later in this course, the logarithm will appear precisely as the inverse of the exponential function, and the inverse trigonometric functions will let you recover an angle from a ratio.

Worked example 4: compose in both orders

Order genuinely matters in composition. With f(x) = x² + 1 and g(x) = 3x - 2:

1. f(g(x)) = (3x - 2)² + 1 = 9x² - 12x + 4 + 1 = 9x² - 12x + 5.
2. g(f(x)) = 3(x² + 1) - 2 = 3x² + 3 - 2 = 3x² + 1.

These are different functions. Test them at x = 1: f(g(1)) = 9 - 12 + 5 = 2, while g(f(1)) = 3 + 1 = 4. Never assume the two orders agree.

The domain of a composition

For f(g(x)), an input x must first be legal for g, and then the output g(x) must be legal for f. Example: let g(x) = √x and f(x) = 1/(x - 3). Then f(g(x)) = 1/(√x - 3). We need x ≥ 0 for the square root, and we need √x ≠ 3, which excludes x = 9. The domain is all x ≥ 0 except x = 9.

Decomposing a function

Running composition backward is just as useful: seeing a complicated function as simple layers. The function h(x) = √(3x + 1) is f(g(x)) where the inside is g(x) = 3x + 1 and the outside is f(x) = √x. Spotting the inside-outside structure of a function is exactly the skill the chain rule will demand in calculus, so practice it now.

Worked example 5: inverse of a rational function

Find the inverse of f(x) = (2x + 1)/(x - 3). Use the swap-and-solve recipe:

1. Write y = (2x + 1)/(x - 3), then swap: x = (2y + 1)/(y - 3).
2. Multiply both sides by (y - 3): x(y - 3) = 2y + 1, so xy - 3x = 2y + 1.
3. Collect the y terms on one side: xy - 2y = 3x + 1.
4. Factor out y: y(x - 2) = 3x + 1, so y = (3x + 1)/(x - 2).

Therefore f⁻¹(x) = (3x + 1)/(x - 2). Check with a number: f(4) = (8 + 1)/(4 - 3) = 9, and f⁻¹(9) = (27 + 1)/(9 - 2) = 28/7 = 4. The inverse sent 9 back to 4, exactly as it should.

Restricting a domain to create an inverse

The parabola f(x) = x² fails the horizontal line test (both 2 and -2 map to 4), so it has no inverse on its full domain. But if we restrict it to x ≥ 0, it becomes one-to-one, and its inverse is √x. This restriction trick is standard: it is exactly how the inverse trigonometric functions will be defined later in the course.

Real-world applications

Composition chains processes: if a currency converter gives euros from dollars and a second function gives yen from euros, their composition converts dollars straight to yen. A 20 percent discount followed by 8 percent sales tax is the composition t(d(p)) = 1.08(0.80p) = 0.864p. Inverses reverse processes: the function F(c) = (9/5)c + 32 converts Celsius to Fahrenheit, and solving backward gives c = (5/9)(F - 32), the inverse that converts Fahrenheit to Celsius.

Common misconceptions

• "f⁻¹(x) means 1/f(x)." The superscript -1 on a function names the inverse function, not a reciprocal.
• "f(g(x)) and g(f(x)) are the same." They usually differ; only special pairs (such as a function and its inverse) commute.
• "Every function has an inverse." Only one-to-one functions do; others need a domain restriction first.
• "To verify an inverse, checking one composition is enough." Strictly, both f(f⁻¹(x)) = x and f⁻¹(f(x)) = x should hold on the appropriate domains.
• "The inverse graph looks like the original flipped upside down." It is the mirror image across the diagonal line y = x, not across the x-axis.

Recap

Composition f(g(x)) feeds the output of g into f; work inside out, and expect the two orders to differ. The domain of a composition honors both layers. An inverse function undoes f, exists only when f is one-to-one (horizontal line test), and is found by swapping x and y and solving. Its graph is the reflection of f across y = x, and compositions of a function with its inverse return x. Decomposing functions into layers prepares you directly for the chain rule in calculus.

Sources

1. OpenStax, Precalculus 2e, Section 1.4: Composition of Functions and Section 1.7: Inverse Functions. Free at openstax.org/details/books/precalculus-2e.
2. OpenStax, Algebra and Trigonometry 2e, Chapter 3: Functions. Free at openstax.org/details/books/algebra-and-trigonometry-2e.
3. Khan Academy, Precalculus: Composite and inverse functions unit. khanacademy.org/math/precalculus.
4. Paul Dawkins, Paul's Online Math Notes, Algebra: Combining Functions and Inverse Functions. tutorial.math.lamar.edu.

Anatomy of a polynomial

A polynomial function is a sum of terms of the form (number) times x raised to a whole-number power, such as P(x) = 2x³ - 5x² + x - 7. The largest exponent is the degree, and the coefficient of that highest-degree term is the leading coefficient. Here the degree is 3 and the leading coefficient is 2.

End behavior

End behavior describes where the graph heads as x goes far to the right or far to the left. It is controlled entirely by the leading term:

Degree	Leading coefficient	Left end	Right end
even	positive	up	up
even	negative	down	down
odd	positive	down	up
odd	negative	up	down

Even-degree graphs have both ends pointing the same way, like a parabola. Odd-degree graphs have ends pointing in opposite directions, like a line or a cubic.

Zeros and x-intercepts

A zero (or root) of a polynomial is a value of x that makes P(x) = 0. Each real zero is an x-intercept of the graph. If a factor appears more than once, the zero has a multiplicity equal to how many times the factor repeats, and multiplicity controls the graph's behavior there:

• Odd multiplicity: the graph crosses the x-axis at that zero.
• Even multiplicity: the graph touches the x-axis and turns back without crossing.

Worked example: sketch the key features

Analyze P(x) = (x - 2)²(x + 1).

1. Degree: multiplying gives a leading term of x³, so degree 3, leading coefficient positive. End behavior: down on the left, up on the right.
2. Zeros: x = 2 with multiplicity 2 (even, so the graph touches and turns), and x = -1 with multiplicity 1 (odd, so the graph crosses).
3. y-intercept: P(0) = (0 - 2)²(0 + 1) = 4 · 1 = 4, giving the point (0, 4).

From these facts you can draw the graph: it rises from the lower left, crosses at x = -1, passes through (0, 4), dips to touch the axis at x = 2, and rises to the upper right. Notice a degree-3 polynomial has at most 2 turning points, which is exactly what appears.

Turning points and smoothness

Two facts shape every polynomial graph. First, polynomials are smooth and continuous: no breaks, no holes, no sharp corners, ever. Second, a polynomial of degree n has at most n - 1 turning points (places where the graph switches between rising and falling). A degree-5 polynomial can turn at most 4 times; if a graph turns 5 times, its degree is at least 6. Counting turns is a quick way to estimate degree from a picture.

Worked example 2: end behavior practice

Describe the end behavior of P(x) = -2x⁴ + 7x³ - 1. Only the leading term -2x⁴ matters. The degree 4 is even, so both ends point the same way, and the leading coefficient -2 is negative, so both ends point down. Far to the left and far to the right the graph falls without bound, whatever the middle terms do.

Worked example 3: build a polynomial from its zeros

Write a degree-3 polynomial with leading coefficient 1 whose zeros are -3 (multiplicity 1) and 1 (multiplicity 2). Each zero c contributes a factor (x - c) raised to its multiplicity:

1. Set up the factors: P(x) = (x + 3)(x - 1)².
2. Expand the square: (x - 1)² = x² - 2x + 1.
3. Multiply by (x + 3): x³ - 2x² + x + 3x² - 6x + 3 = x³ + x² - 5x + 3.

Verify both zeros: P(1) = 1 + 1 - 5 + 3 = 0 and P(-3) = -27 + 9 + 15 + 3 = 0. Both check, so the expansion is correct. The graph crosses at x = -3 (odd multiplicity) and touches at x = 1 (even multiplicity).

Locating zeros by sign change

Because polynomials are continuous, if P(a) is negative and P(b) is positive, the graph must cross the x-axis somewhere between a and b. This is the intermediate value idea. Example: for P(x) = x³ - x - 2, compute P(1) = 1 - 1 - 2 = -2 (below the axis) and P(2) = 8 - 2 - 2 = 4 (above the axis). A real zero therefore lies between 1 and 2. Squeezing the interval tighter (testing 1.5, then 1.52, and so on) approximates the zero as precisely as you like, which is exactly how calculators locate roots.

Real-world applications

Polynomials model quantities built from products of lengths. An open-top box made from a 12 by 10 inch sheet by cutting x-inch squares from the corners has volume V(x) = x(12 - 2x)(10 - 2x), a cubic whose practical domain is 0 < x < 5. Engineers use polynomial curves for roller coaster segments and road transitions because they are smooth, and economists fit revenue and cost data with low-degree polynomials to find maximum-profit production levels near turning points.

Common misconceptions

• "The degree equals the number of turning points." Degree n allows at most n - 1 turns; the graph may turn fewer times.
• "The graph crosses the axis at every zero." At even-multiplicity zeros it touches and turns back without crossing.
• "End behavior depends on the whole formula." Only the leading term controls the far ends.
• "A degree-n polynomial always has n real zeros." It has at most n real zeros; some may be repeated or not real at all.
• "A missing x-intercept means a mistake." Polynomials like x² + 1 never touch the x-axis; that is perfectly legal.

Recap

A polynomial is a sum of whole-number-power terms; its degree and leading coefficient control end behavior (even degree: matching ends; odd: opposite ends; the sign points them up or down). Real zeros are x-intercepts, with odd multiplicity crossing and even multiplicity touching. Graphs are smooth and continuous with at most n - 1 turning points, sign changes locate zeros between test points, and factored form lets you build or read a polynomial zero by zero.

Sources

1. OpenStax, Precalculus 2e, Section 3.3: Power Functions and Polynomial Functions and Section 3.4: Graphs of Polynomial Functions. Free at openstax.org/details/books/precalculus-2e.
2. OpenStax, Algebra and Trigonometry 2e, Chapter 5: Polynomial and Rational Functions. Free at openstax.org/details/books/algebra-and-trigonometry-2e.
3. Khan Academy, Precalculus and Algebra 2: polynomial graphs and end behavior. khanacademy.org/math/precalculus.
4. Paul Dawkins, Paul's Online Math Notes, Algebra: Graphing Polynomials. tutorial.math.lamar.edu.

The remainder and factor theorems

The remainder theorem says that when a polynomial P(x) is divided by (x - c), the remainder equals P(c). So you can find a remainder just by plugging in c, with no long division. The closely related factor theorem says (x - c) is a factor of P(x) exactly when P(c) = 0. In words: c is a zero if and only if (x - c) divides evenly.

Worked example 1: test a zero

Is x = 2 a zero of P(x) = x³ - 3x² + 4? Evaluate P(2) = 8 - 12 + 4 = 0. Since the value is 0, yes, x = 2 is a zero and (x - 2) is a factor.

Dividing to reduce the degree

Once you know one factor, dividing by it produces a lower-degree quotient you can often factor by hand. Continuing the example, divide x³ - 3x² + 0x + 4 by (x - 2). Using synthetic division with c = 2 and coefficients 1, -3, 0, 4:

• Bring down 1. Multiply by 2: 2. Add to -3: -1.
• Multiply -1 by 2: -2. Add to 0: -2.
• Multiply -2 by 2: -4. Add to 4: 0 (the remainder, confirming the factor).

The quotient is x² - x - 2, so P(x) = (x - 2)(x² - x - 2).

Worked example 2: finish the factoring

Factor the quotient x² - x - 2 = (x - 2)(x + 1). Therefore P(x) = (x - 2)(x - 2)(x + 1) = (x - 2)²(x + 1). The zeros are x = 2 (multiplicity 2) and x = -1.

The rational zero test

When you have no obvious first zero, the rational zero test lists the candidates to try. Any rational zero of a polynomial with integer coefficients has the form (factor of the constant term) divided by (factor of the leading coefficient). For 2x² - 3x - 2, the constant is -2 (factors 1, 2) and the leading coefficient is 2 (factors 1, 2), so the possible rational zeros are plus or minus 1, 2, 1/2. Testing x = 2: 2(4) - 3(2) - 2 = 0, a zero. The other factor gives x = -1/2. This test turns a search into a short, finite checklist.

Worked example 3: a complete solve, start to finish

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

1. List candidates. Factors of the constant -6 over factors of the leading 1: possible rational zeros are ±1, ±2, ±3, ±6.
2. Test one. P(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 coefficients 1, -6, 11, -6: bring down 1; multiply by 1 and add to -6 to get -5; multiply by 1 and add to 11 to get 6; multiply by 1 and add to -6 to get 0. The quotient is x² - 5x + 6.
4. Factor the quotient. x² - 5x + 6 = (x - 2)(x - 3).

So P(x) = (x - 1)(x - 2)(x - 3) and the zeros are 1, 2, and 3. Quick check of another zero: P(3) = 27 - 54 + 33 - 6 = 0. The full pipeline is always the same: candidates, test, divide, factor what remains.

Synthetic division details that save you

Two habits prevent most errors. First, write the coefficients in descending order of degree and insert a 0 for every missing power: dividing x³ - 3x² + 4 requires the row 1, -3, 0, 4. Second, remember synthetic division only works for linear divisors of the form (x - c); dividing by (x + 3) means using c = -3, because x + 3 = x - (-3). For divisors of higher degree you fall back on polynomial long division, which works like numeric long division: divide leading terms, multiply back, subtract, repeat.

The fundamental theorem of algebra

How many zeros should you expect? The fundamental theorem of algebra answers: a degree-n polynomial has exactly n zeros if you count multiplicity and allow complex numbers. A quadratic like x² + 9 = 0 has no real solutions, but it has two complex ones, x = 3i and x = -3i, where i² = -1. For polynomials with real coefficients, complex zeros always arrive in conjugate pairs: if 3 + 2i is a zero, then 3 - 2i is too. So a cubic with real coefficients always has at least one real zero, because complex zeros pair up and three is odd.

Real-world applications

Zero-finding is how models get solved. If the box volume from the previous lesson must equal a target, you set the cubic equal to that value and hunt zeros of the difference. Break-even analysis sets profit P(x) = 0 and factors. Engineers locate the natural frequencies of structures as roots of characteristic polynomials, using exactly this candidates-test-divide workflow.

Common misconceptions

• "Dividing by (x + 3) means plugging in 3." The factor form is (x - c), so (x + 3) corresponds to c = -3.
• "Synthetic division works for any divisor." It only handles linear divisors (x - c); use long division otherwise.
• "Forgetting placeholder zeros is harmless." Omitting the 0 coefficient for a missing power scrambles every later column.
• "No rational zero means no zeros." The rational zero test only lists rational candidates; irrational and complex zeros can still exist (x² - 2 has zeros ±√2).
• "A remainder of 0 is a failure." The opposite: remainder 0 confirms the divisor is a factor.

Recap

The remainder theorem evaluates a division instantly: dividing P(x) by (x - c) leaves remainder P(c). The factor theorem turns zeros into factors and back. The standard solve is: list rational candidates (constant factors over leading factors), test until one gives zero, divide it out synthetically, and factor the lower-degree quotient. The fundamental theorem of algebra promises exactly n zeros counting multiplicity over the complex numbers, with complex zeros of real polynomials arriving in conjugate pairs.

Sources

1. OpenStax, Precalculus 2e, Section 3.5: Dividing Polynomials and Section 3.6: Zeros of Polynomial Functions. Free at openstax.org/details/books/precalculus-2e.
2. OpenStax, Algebra and Trigonometry 2e, Chapter 5: Polynomial and Rational Functions. Free at openstax.org/details/books/algebra-and-trigonometry-2e.
3. Khan Academy, Algebra 2 and Precalculus: polynomial division and zeros. khanacademy.org/math/precalculus.
4. Paul Dawkins, Paul's Online Math Notes, Algebra: Dividing Polynomials and Finding Zeroes of Polynomials. tutorial.math.lamar.edu.

What a rational function is

A rational function is a ratio of two polynomials, R(x) = P(x) / Q(x). Its interesting features come from where the denominator is zero and from how the top and bottom degrees compare.

Vertical asymptotes and holes

Set the denominator equal to zero to find candidate breaks. After canceling any common factors of top and bottom:

• A factor that remains only in the denominator gives a vertical asymptote, a vertical line the graph shoots up or down along but never touches.
• A factor that cancels from both top and bottom gives a hole, a single missing point in an otherwise continuous curve.

Worked example 1: find the vertical features

Analyze R(x) = (x - 1) / ((x - 1)(x + 3)). The factor (x - 1) cancels, leaving 1 / (x + 3). So there is a hole at x = 1 and a vertical asymptote at x = -3 (from the surviving denominator factor).

Horizontal asymptotes by degree

The horizontal asymptote describes the far-left and far-right height of the graph. Compare the degree of the numerator (call it n) with the degree of the denominator (call it d):

Comparison	Horizontal asymptote
n < d	y = 0 (the x-axis)
n = d	y = ratio of leading coefficients
n > d	none (there is a slant or higher asymptote instead)

Worked example 2: full analysis

Analyze R(x) = (2x² + 1) / (x² - 4).

1. Vertical asymptotes: denominator x² - 4 = (x - 2)(x + 2) = 0 gives lines at x = 2 and x = -2 (neither factor cancels).
2. Horizontal asymptote: both degrees are 2, so y = the ratio of leading coefficients = 2/1 = 2.
3. y-intercept: R(0) = 1 / (-4) = -1/4.

The graph approaches the horizontal line y = 2 at the far edges and breaks at the two vertical lines. Plotting a couple of test points in each region between the asymptotes reveals whether each piece is above or below y = 2, letting you sketch the whole curve.

Slant asymptotes

When the numerator degree is exactly one more than the denominator degree, the graph approaches a tilted line called a slant asymptote (or oblique asymptote). To find it, divide the polynomials; the quotient (ignoring the remainder) is the line.

Worked example 3: find a slant asymptote

Analyze R(x) = (x² + 1)/(x - 1). Long division: x² + 1 divided by x - 1 gives quotient x + 1 with remainder 2, because (x - 1)(x + 1) + 2 = x² - 1 + 2 = x² + 1. So R(x) = x + 1 + 2/(x - 1). As x grows huge, the fraction 2/(x - 1) shrinks to nothing and the graph hugs the line y = x + 1. There is also a vertical asymptote at x = 1.

Worked example 4: the exact location of a hole

Find the hole of R(x) = (x² - 9)/(x - 3). Factor and cancel: (x - 3)(x + 3)/(x - 3) = x + 3 for every x ≠ 3. The graph is the line y = x + 3 with one missing point. To find its height, substitute x = 3 into the simplified form: 3 + 3 = 6. The hole sits at (3, 6). A hole has coordinates, not just an x-value, and the y-coordinate always comes from the simplified expression.

Can a graph cross an asymptote?

A graph never touches a vertical asymptote, because the function is undefined there. But a graph can cross a horizontal or slant asymptote in the middle region; the asymptote only promises what happens at the far ends. For example, R(x) = x/(x² + 1) has horizontal asymptote y = 0 yet passes straight through (0, 0). Treating horizontal asymptotes as untouchable fences is one of the most common errors in the course.

A sketching checklist

1. Factor the numerator and denominator completely.
2. Cancel common factors and mark each cancellation as a hole.
3. Set the remaining denominator to zero for vertical asymptotes.
4. Compare degrees for the horizontal or slant asymptote.
5. Plot the intercepts: x-intercepts from the remaining numerator, y-intercept from R(0).
6. Test one point in each region between vertical asymptotes to place the branches.

Real-world applications

Rational functions describe averages and shared resources. If a batch of parts costs 500 dollars in setup plus 3 dollars per part, the average cost per part is C(x) = (500 + 3x)/x = 500/x + 3. Its horizontal asymptote y = 3 is the long-run cost per part: the setup cost spreads ever thinner but never fully disappears. Drug concentration after a dose, lens equations in optics, and predator-prey saturation models all lean on the same asymptotic reading.

Common misconceptions

• "Every denominator zero is a vertical asymptote." If the factor cancels, it is a hole instead; factor before you conclude.
• "Graphs never touch any asymptote." Only vertical asymptotes are forbidden; horizontal and slant asymptotes can be crossed in the middle.
• "The horizontal asymptote describes the whole graph." It describes only the far left and far right ends.
• "A hole is visible as a gap." It is a single missing point; plotting software often cannot show it, so you must find it algebraically.
• "When degrees are equal the asymptote is y = 1." It is the ratio of the leading coefficients, which is 1 only when they match.

Recap

A rational function is a ratio of polynomials. Factor and cancel first: canceled factors give holes (with y-coordinates from the simplified form), surviving denominator zeros give vertical asymptotes. Compare degrees for far-end behavior: smaller numerator degree gives y = 0, equal degrees give the leading-coefficient ratio, and one-more gives a slant asymptote found by dividing. Intercepts and one test point per region complete an accurate sketch.

Sources

1. OpenStax, Precalculus 2e, Section 3.7: Rational Functions. Free at openstax.org/details/books/precalculus-2e.
2. OpenStax, Algebra and Trigonometry 2e, Chapter 5: Polynomial and Rational Functions. Free at openstax.org/details/books/algebra-and-trigonometry-2e.
3. Khan Academy, Precalculus: Rational functions unit. khanacademy.org/math/precalculus.
4. Paul Dawkins, Paul's Online Math Notes, Algebra: Rational Functions. tutorial.math.lamar.edu.

The exponential form

An exponential function has the variable in the exponent: f(x) = a · bₓ, where the base b is positive and not 1, and a is the starting value (the y-intercept, since f(0) = a). This is fundamentally different from a power function like x², where the variable is in the base.

Growth versus decay

• If b > 1, the function shows exponential growth: it increases, faster and faster, as x increases. Example base: 2.
• If 0 < b < 1, the function shows exponential decay: it decreases toward zero. Example base: 1/2.

Every basic exponential graph passes through (0, a) and has the x-axis as a horizontal asymptote: the curve gets arbitrarily close to y = 0 but never reaches it on the decay side.

Worked example 1: evaluate

Let f(x) = 3 · 2ₓ. Then f(0) = 3 · 1 = 3, f(2) = 3 · 4 = 12, and f(-1) = 3 · (1/2) = 1.5. Each step to the right multiplies the output by the base 2; each step left divides by 2.

The natural base e

One base is so useful it has its own name: e, approximately 2.718. It arises naturally in continuous growth, such as continuously compounded interest, and it is the base that makes the calculus of exponentials simplest. Functions written f(x) = eₓ behave like any other growth exponential, just with this special base.

Worked example 2: a growth model

A colony of 200 bacteria doubles every hour. After t hours the population is P(t) = 200 · 2ₓ. After 3 hours: P(3) = 200 · 2³ = 200 · 8 = 1600 bacteria. Doubling is exponential growth because the multiplier (times 2) is constant per hour rather than a constant amount added.

Worked example 3: a decay model

A medicine starts at 80 mg and 25 percent leaves the body each hour, so 75 percent remains. The amount is A(t) = 80 · (0.75)ₓ. After 2 hours: A(2) = 80 · 0.5625 = 45 mg. Because the base 0.75 is between 0 and 1, the graph decays toward zero without ever reaching it.

Percent change and the base

Real problems usually state a percent rate, and the base packages it. Growing by rate r per step means multiplying by b = 1 + r; shrinking by rate r means b = 1 - r. A 6 percent annual raise gives base 1.06; losing 15 percent of value each year gives base 0.85. Translating percent language into the base is the first move in nearly every application.

Worked example 4: compound interest

Money compounded n times per year follows A = P(1 + r/n)nt, where P is the starting amount, r the annual rate, and t the years. Invest 1000 dollars at 8 percent compounded quarterly for one year:

1. The quarterly rate is 0.08/4 = 0.02, and there are 4 · 1 = 4 compoundings.
2. A = 1000(1.02)⁴. Compute in stages: 1.02² = 1.0404, then 1.0404² = 1.08243216.
3. A ≈ 1082.43 dollars.

Notice this beats simple 8 percent (1080 dollars) because interest earns interest. Compounding continuously (letting n grow without bound) gives the formula A = Pert; here 1000 · e0.08 ≈ 1083.29 dollars, only slightly more. This is exactly where the natural base e earns its keep.

Worked example 5: half-life

A half-life is the time for half of a decaying quantity to disappear. A 100 mg sample with a 6 hour half-life follows A(t) = 100 · (1/2)t/6. After 18 hours, the exponent is 18/6 = 3, so A(18) = 100 · (1/2)³ = 100/8 = 12.5 mg. Each 6 hour block cuts the amount in half: 100 to 50 to 25 to 12.5.

Exponential versus linear growth

Linear growth adds a fixed amount per step; exponential growth multiplies by a fixed factor. Compare y = 100x with y = 2^x: at x = 5 the line is far ahead (500 versus 32), at x = 10 they are close (1000 versus 1024), and by x = 20 the exponential has exploded (2000 versus 1,048,576). Any growth exponential eventually overtakes any line, no matter the slopes involved. That eventual takeover is what "exponential growth" really means, not merely "fast."

Transformations of exponentials

Everything from the transformations lesson applies. The graph of y = 2x + 3 is the parent shifted up 3, so its horizontal asymptote rises to y = 3. The graph of y = 2-x reflects across the y-axis, turning growth into decay, which is why 2-x = (1/2)x. Reading the asymptote from the vertical shift is a frequent exam task.

Worked example 6: depreciation

A 20,000 dollar car loses 15 percent of its value each year, so V(t) = 20000 · (0.85)ₓ. After 2 years: V(2) = 20000 · 0.85² = 20000 · 0.7225 = 14,450 dollars. The car does not lose the same dollar amount each year; it loses 15 percent of an ever-smaller value, which is exactly the multiplicative signature of exponential decay.

Common misconceptions

• "Exponential just means fast." Exponential decay shrinks, and growth can start slowly; the defining feature is a constant multiplier per step.
• "The base can be any number." The base must be positive and not 1; negative bases would oscillate wildly and are excluded.
• "x² and 2^x are basically alike." One is a power function (variable in the base), the other exponential (variable in the exponent); their long-run behaviors are utterly different.
• "Doubling adds the same amount each time." Doubling adds more each time; equal additions are linear, not exponential.
• "The graph eventually reaches the asymptote." It approaches y = 0 (or the shifted asymptote) forever without touching it.

Recap

An exponential function f(x) = a · bx starts at a and multiplies by b each step: growth when b > 1, decay when 0 < b < 1. Percent rates convert to bases via 1 + r or 1 - r, compound interest uses P(1 + r/n)nt with continuous compounding Pert, and half-life models use base 1/2 with the exponent counting half-lives. The graph passes through (0, a), hugs a horizontal asymptote, and eventually outruns every linear function.

Sources

1. OpenStax, Precalculus 2e, Section 4.1: Exponential Functions and Section 4.2: Graphs of Exponential Functions. Free at openstax.org/details/books/precalculus-2e.
2. OpenStax, Algebra and Trigonometry 2e, Chapter 6: Exponential and Logarithmic Functions. Free at openstax.org/details/books/algebra-and-trigonometry-2e.
3. Khan Academy, Algebra 2 and Precalculus: exponential growth and decay models. khanacademy.org/math/precalculus.
4. Paul Dawkins, Paul's Online Math Notes, Algebra: Exponential Functions. tutorial.math.lamar.edu.

The logarithm as an inverse

A logarithm answers the question "what exponent do I need?" The statement log_b(y) = x means exactly bₓ = y. In other words, the log is the inverse of the exponential: it takes the output y and returns the exponent x. Because they are inverses, log_b(bₓ) = x and b^(log_b y) = y.

Reading a log

log₂(8) = 3 because 2³ = 8. And log₁₀(1000) = 3 because 10³ = 1000. Two bases are so common they get shorthand: the common log log(x) means base 10, and the natural log ln(x) means base e.

Worked example 1: convert forms

• log₃(81) = 4 becomes 3⁴ = 81. Check: 3 to the 4th is 81.
• 5² = 25 becomes log₅(25) = 2.

The laws of logarithms

Because logs turn products into sums (their defining superpower), three laws follow:

1. Product law: log_b(MN) = log_b(M) + log_b(N).
2. Quotient law: log_b(M/N) = log_b(M) - log_b(N).
3. Power law: log_b(Mⁿ) = p · log_b(M).

Worked example 2: expand a logarithm

Expand log₂(8x³). Use the product law, then the power law:

1. Product law: log₂(8) + log₂(x³).
2. Evaluate and apply the power law: 3 + 3·log₂(x).

Worked example 3: condense a logarithm

Write 2·log(x) - log(y) as one logarithm. Power law turns the coefficient into an exponent: log(x²) - log(y). Then the quotient law combines them: log(x² / y).

The change-of-base formula

Calculators usually only have log (base 10) and ln (base e), so to compute a log in another base use the change-of-base formula: log_b(x) = ln(x) / ln(b). For instance, log₅(20) = ln(20)/ln(5), which is about 2.996/1.609 ≈ 1.86.

Special values worth memorizing

Two evaluations work in every base b: log_b(1) = 0, because b⁰ = 1, and log_b(b) = 1, because b¹ = b. Inverse pairs also collapse instantly: ln(e⁷) = 7 and 10^(log 5) = 5, since a function applied to its inverse returns the input.

Worked example 4: evaluate without a calculator

1. log₄(1/16): ask "4 to what power gives 1/16?" Since 4² = 16, we need 4⁻² = 1/16, so the answer is -2.
2. log(0.001): base 10, and 0.001 = 10⁻³, so the answer is -3.
3. ln(√e): the square root is the 1/2 power, so the answer is 1/2.

Fractional inputs give negative logs, and roots give fractional logs; the log can output any real number even though its input must be positive.

The graph and domain of a logarithm

Because y = log_b(x) is the inverse of y = bₓ, its graph is the exponential curve reflected across the line y = x. It passes through (1, 0) and (b, 1), climbs slowly forever, and has a vertical asymptote at x = 0: you can only take the log of a positive number. Shifts move that wall. For y = log₂(x - 3), the argument must satisfy x - 3 > 0, so the domain is x > 3 and the vertical asymptote stands at x = 3.

Worked example 5: expand and condense with all three laws

Expand log_b(x²y / z):

1. Quotient law: log_b(x²y) - log_b(z).
2. Product law: log_b(x²) + log_b(y) - log_b(z).
3. Power law: 2 log_b(x) + log_b(y) - log_b(z).

Condensing runs the same steps in reverse. For example, 3 log(x) + log(y) becomes log(x³) + log(y) = log(x³y). Keep coefficients as exponents before combining; that ordering avoids nearly all condensing errors.

Real-world applications: logarithmic scales

When quantities span many powers of 10, scientists measure the exponent instead of the number. The Richter-style magnitude of an earthquake is the log of its intensity ratio, so a magnitude 6 quake shakes 10² = 100 times harder than a magnitude 4. Decibels measure loudness as 10 times the log of a power ratio, and pH is the negative log of hydrogen ion concentration, so lemon juice at pH 2 is 100,000 times more acidic than water at pH 7. Logarithms turn multiplication of intensities into addition of scale points, which is exactly the product law at work in the real world.

Common misconceptions

• "log(a + b) = log(a) + log(b)." False. The product law says log(ab) = log(a) + log(b); there is no law for the log of a sum.
• "log(a)/log(b) = log(a) - log(b)." False. The quotient law applies to log(a/b), not to a quotient of two logs; log(a)/log(b) is change of base.
• "You can take the log of any number." The argument must be strictly positive; log(0) and log(-5) are undefined.
• "A negative log output means an error." Logs of numbers between 0 and 1 are negative by design.
• "ln means log base 10." ln is base e (about 2.718); log with no base written usually means base 10.

Recap

A logarithm is an exponent: log_b(y) = x means bₓ = y, making the log the inverse of the exponential. Memorize log_b(1) = 0 and log_b(b) = 1, expand and condense with the product, quotient, and power laws, and reach any base through change of base ln(x)/ln(b). The graph passes through (1, 0) with a vertical asymptote where its argument hits zero, and logarithmic scales (magnitude, decibels, pH) compress huge ranges by measuring exponents.

Sources

1. OpenStax, Precalculus 2e, Sections 4.3 to 4.5: Logarithmic Functions, Graphs of Logarithmic Functions, and Logarithmic Properties. Free at openstax.org/details/books/precalculus-2e.
2. OpenStax, Algebra and Trigonometry 2e, Chapter 6: Exponential and Logarithmic Functions. Free at openstax.org/details/books/algebra-and-trigonometry-2e.
3. Khan Academy, Algebra 2 and Precalculus: logarithms and properties of logarithms. khanacademy.org/math/precalculus.
4. Paul Dawkins, Paul's Online Math Notes, Algebra: Logarithm Functions. tutorial.math.lamar.edu.

Two solving strategies

To solve an exponential equation where the variable is in the exponent, the main tool is to take a logarithm of both sides, which brings the exponent down using the power law. For a logarithmic equation, the main tool is to rewrite it in exponential form to peel the log away.

Worked example 1: same base

Solve 2ₓ = 32. Write 32 as a power of 2: 32 = 2⁵. With equal bases, the exponents must match: x = 5. This shortcut works only when both sides share a base.

Worked example 2: take a logarithm

Solve 3ₓ = 20. Take the natural log of both sides and use the power law:

1. ln(3ₓ) = ln(20).
2. Power law: x · ln(3) = ln(20).
3. Divide: x = ln(20)/ln(3) ≈ 2.996/1.099 ≈ 2.73.

Worked example 3: a logarithmic equation

Solve log₂(x) = 5. Rewrite in exponential form: x = 2⁵ = 32. Since 32 is positive, it is a valid input to the log, so x = 32.

Watch for extraneous solutions

Logarithms only accept positive inputs, so after solving you must check that each answer keeps every log's argument positive. A value that fails is called an extraneous solution and is discarded.

Worked example 4: check the domain

Solve log(x) + log(x - 3) = 1. Condense the left side with the product law: log(x(x - 3)) = 1. Rewrite in exponential form (base 10): x(x - 3) = 10¹ = 10. Expand: x² - 3x - 10 = 0, which factors as (x - 5)(x + 2) = 0, so x = 5 or x = -2. Now check: x = 5 makes both logs positive and works, but x = -2 makes log(x) undefined, so it is extraneous. The only solution is x = 5.

Isolate the exponential first

When the exponential term is wrapped in other arithmetic, peel that away before touching logarithms. Solve 5 · 2ₓ = 40: divide both sides by 5 to get 2ₓ = 8, and since 8 = 2³, x = 3. Likewise for 7 + 3ₓ = 34: subtract 7 first to get 3ₓ = 27, so x = 3. Taking a log while the extra terms are still attached is the classic wrong first move, because there is no law for the log of a sum.

Worked example 5: logs on both sides

Solve log₃(2x - 1) = log₃(x + 4). When the same base appears on both sides, one-to-one-ness lets you equate the arguments directly:

1. 2x - 1 = x + 4.
2. Subtract x from both sides: x - 1 = 4, so x = 5.
3. Check the domains: 2(5) - 1 = 9 > 0 and 5 + 4 = 9 > 0. Both arguments are positive, so x = 5 stands.

Worked example 6: doubling time

How long does money take to double at 6 percent compounded annually? We need (1.06)ᵗ = 2.

1. Take the natural log: t · ln(1.06) = ln(2).
2. Divide: t = ln(2)/ln(1.06) ≈ 0.6931/0.0583 ≈ 11.9 years.

The banker's shortcut called the rule of 72 estimates doubling time as 72 divided by the percent rate: 72/6 = 12 years, agreeably close to the exact answer. The rule works precisely because of this logarithm computation.

Worked example 7: solving for time in a decay model

A 100 mg dose decays with an 8 hour half-life: A(t) = 100 · (0.5)t/8. When does 20 mg remain?

1. Set up: 100 · (0.5)t/8 = 20, so (0.5)t/8 = 0.2.
2. Take the natural log and use the power law: (t/8) · ln(0.5) = ln(0.2).
3. Divide: t/8 = ln(0.2)/ln(0.5) ≈ (-1.609)/(-0.693) ≈ 2.32.
4. Multiply by 8: t ≈ 18.6 hours.

Sanity check: two half-lives (16 hours) leave 25 mg and three (24 hours) leave 12.5 mg, so hitting 20 mg between 16 and 24 hours is exactly right.

Real-world applications

These equations answer "when" questions across science and finance: when a population reaches a threshold, when a radioactive tracer decays below a safe level, when an investment reaches a goal, and how old a fossil is (carbon dating solves a decay equation for t). Any time the unknown lives in an exponent, a logarithm is the tool that pulls it down.

Common misconceptions

• "Take the log of both sides immediately." Isolate the exponential term first; log(a + b) cannot be split.
• "ln(20)/ln(3) equals ln(20/3) or ln(20 - 3)." Neither; a quotient of logs is just a number (and also the change-of-base form).
• "All solutions of the algebra are solutions of the equation." Log equations demand a domain check; discard any value making an argument nonpositive.
• "Exponential equations always have nice answers." Most need logarithms and give decimals like ln(20)/ln(3); the same-base shortcut is the special case.
• "Dividing exponents solves b to the x = c." The exponent comes down only via logarithms (or matching bases), never by dividing.

Recap

To solve exponential equations: isolate the exponential, then either match bases or take a logarithm of both sides and use the power law. To solve logarithmic equations: condense to a single log, rewrite in exponential form (or equate arguments when both sides are logs of the same base), solve, and always check for extraneous solutions against the domain. The same recipes answer real "how long until" questions, from doubling money to drug decay.

Sources

1. OpenStax, Precalculus 2e, Section 4.6: Exponential and Logarithmic Equations and Section 4.7: Exponential and Logarithmic Models. Free at openstax.org/details/books/precalculus-2e.
2. OpenStax, Algebra and Trigonometry 2e, Chapter 6: Exponential and Logarithmic Functions. Free at openstax.org/details/books/algebra-and-trigonometry-2e.
3. Khan Academy, Algebra 2 and Precalculus: solving exponential and logarithmic equations. khanacademy.org/math/precalculus.
4. Paul Dawkins, Paul's Online Math Notes, Algebra: Solving Exponential Equations and Solving Logarithm Equations. tutorial.math.lamar.edu.

Two ways to measure an angle

Angles can be measured in degrees, where a full circle is 360 degrees, or in radians, where a full circle is 2π radians. A radian is the angle that wraps one radius-length of arc around the circle, which is why it ties so cleanly to circular motion. The key bridge is 180 degrees = π radians.

Converting

• Degrees to radians: multiply by π/180.
• 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.

Standard position and the unit circle

An angle is in standard position when its vertex is at the origin and its initial side lies along the positive x-axis. The unit circle is the circle of radius 1 centered at the origin. For an angle in standard position, the point where its terminal side meets the unit circle has coordinates (cosθ, sinθ). That single fact defines sine and cosine for every angle.

Reference angles

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 correct sign for the quadrant. For example, 150 degrees is in Quadrant II with a reference angle of 30 degrees, so its sine matches sin 30 (positive) and its cosine matches negative cos 30.

Special angle values

Memorizing a few first-quadrant values unlocks the whole circle:

Angle	cos	sin
0	1	0
30 deg = pi/6	(the square root of 3)/2	1/2
45 deg = pi/4	(the square root of 2)/2	(the square root of 2)/2
60 deg = pi/3	1/2	(the square root of 3)/2
90 deg = pi/2	0	1

Coterminal angles

Angles that differ by a full rotation share the same terminal side and are called coterminal. Add or subtract 360 degrees (or 2π radians) as many times as needed: 400 degrees is coterminal with 40 degrees, and -30 degrees is coterminal with 330 degrees. Coterminal angles have identical sine and cosine values, which is why trig functions repeat every revolution.

Worked example 2: evaluate outside the first quadrant

Find cos 210° and sin 210°.

1. Locate the quadrant: 210 degrees is 30 degrees past 180, in Quadrant III.
2. Reference angle: 210 - 180 = 30 degrees.
3. Attach signs: in Quadrant III both x and y are negative, so cosine and sine are both negative.
4. Answer: cos 210° = -√3/2 and sin 210° = -1/2.

Every angle on the circle reduces to a first-quadrant value plus a sign decision. That two-step habit (reference angle, then sign) is the fastest reliable method.

Arc length and sector area

Radians pay off immediately in measurement formulas. For a circle of radius r and a central angle θ in radians:

• Arc length: s = rθ.
• Sector area: A = (1/2) r²θ.

Worked example 3: arc length and sector area

A circle has radius 6 cm and a central angle of π/3 (that is, 60 degrees).

1. Arc length: s = 6 · π/3 = 2π ≈ 6.28 cm.
2. Sector area: A = (1/2)(6²)(π/3) = (1/2)(36)(π/3) = 6π ≈ 18.85 square cm.

Both formulas fail if you plug in degrees; convert to radians first. This is the practical reason radians are the default in calculus and physics.

Worked example 4: linear speed from rotation

A wheel of radius 0.3 m spins at 10 radians per second. A point on the rim travels v = rω = 0.3 · 10 = 3 meters per second. Radian measure makes rotational and straight-line speed interchangeable through nothing more than multiplication by the radius.

Real-world applications

Radians run the rotating world: gear trains and bike sprockets convert angular speed between wheels of different radii, satellite ground tracks use arc length along the Earth's surface (one degree of latitude is about 111 km because Earth's radius is about 6371 km), and a clock's minute hand of length 10 cm sweeps 2π radians per hour, so its tip moves about 62.8 cm each hour. Whenever something spins, s = rθ is nearby.

Common misconceptions

• "π equals 180." π radians equals 180 degrees; π itself is just the number 3.14159...
• "The calculator is always in the right mode." Degree mode versus radian mode silently changes every answer; check before evaluating.
• "Reference angles are measured to the y-axis." Always to the x-axis; that is what makes the table values reusable.
• "Coterminal angles are equal angles." They share a terminal side but represent different rotation amounts, such as 40 and 400 degrees.
• "Arc length works with degrees." s = rθ requires radians; using degrees inflates the answer by a factor of about 57.3.

Recap

Angles are measured in degrees (full circle 360) or radians (full circle 2π), linked by 180 degrees = π radians. On the unit circle, the terminal side of an angle in standard position lands at (cosθ, sinθ). Reference angles plus quadrant signs evaluate any angle from a small memorized table, coterminal angles repeat values every revolution, and radian measure powers the formulas s = rθ, A = (1/2)r²θ, and v = rω.

Sources

1. OpenStax, Precalculus 2e, Section 5.1: Angles and Section 5.2: Unit Circle: Sine and Cosine Functions. Free at openstax.org/details/books/precalculus-2e.
2. OpenStax, Algebra and Trigonometry 2e, Chapter 7: The Unit Circle: Sine and Cosine Functions. Free at openstax.org/details/books/algebra-and-trigonometry-2e.
3. Khan Academy, Precalculus and Trigonometry: radians, the unit circle, and reference angles. khanacademy.org/math/precalculus.
4. Paul Dawkins, Paul's Online Math Notes, Trig Review: Trig Functions and the unit circle. tutorial.math.lamar.edu.

From two functions to six

Sine and cosine come straight from the unit circle. The other four are built from them:

• Tangent: tanθ = sinθ / cosθ.
• Cosecant: cscθ = 1 / sinθ.
• Secant: secθ = 1 / cosθ.
• Cotangent: cotθ = cosθ / sinθ.

These are the reciprocal and quotient relationships. A function is undefined wherever its denominator is zero; for instance, tangent is undefined at 90 degrees because cosine is zero there.

Right-triangle definitions: SOH-CAH-TOA

For an acute angle in a right triangle, the classic mnemonic is SOH-CAH-TOA:

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

Worked example 1: right-triangle ratios

A right triangle has legs 3 and 4 and hypotenuse 5. For the angle whose opposite side is 3:

• sinθ = 3/5 = 0.6.
• cosθ = 4/5 = 0.8.
• tanθ = 3/4 = 0.75.

Signs by quadrant

The sign of each function depends on the signs of x and y in that quadrant. A common memory aid is "All Students Take Calculus," giving which functions are positive:

Quadrant	Positive functions
I	All
II	Sine (and cosecant)
III	Tangent (and cotangent)
IV	Cosine (and secant)

Worked example 2: use a quadrant sign

Find tanθ if sinθ = 3/5 and θ is in Quadrant II. In Quadrant II cosine is negative. Using the Pythagorean relationship, cosθ = -4/5. Then tanθ = sinθ/cosθ = (3/5)/(-4/5) = -3/4. The negative sign is exactly what the quadrant table predicts, since tangent is negative in Quadrant II.

Worked example 3: all six functions from a point

The terminal side of θ passes through the point (-3, 4). Find all six trig values. First compute the distance from the origin: r = √((-3)² + 4²) = √(9 + 16) = √25 = 5. For a general point, sinθ = y/r, cosθ = x/r, and tanθ = y/x:

• sinθ = 4/5 and its reciprocal cscθ = 5/4.
• cosθ = -3/5 and its reciprocal secθ = -5/3.
• tanθ = 4/(-3) = -4/3 and its reciprocal cotθ = -3/4.

The point is in Quadrant II (x negative, y positive), and indeed only sine and cosecant came out positive, matching "All Students Take Calculus."

Tangent values for special angles

Dividing the sine and cosine columns of the special-angle table produces tangent values worth knowing: tan 0 = 0, tan 30° = √3/3, tan 45° = 1, tan 60° = √3, and tan 90° is undefined because cos 90 = 0 puts a zero in the denominator. Tangent grows without bound as the angle approaches 90 degrees, which matches the geometry: a ramp of 89.9 degrees is nearly vertical, with a huge rise for a tiny run.

Worked example 4: angle of elevation

From a spot 50 meters from the base of a tower, the angle of elevation to the top is 35 degrees. How tall is the tower?

1. The 50 m distance is the side adjacent to the angle; the height h is opposite. Tangent relates them: tan 35° = h/50.
2. Solve: h = 50 · tan 35° ≈ 50 · 0.700 = 35.0 meters.

Choosing the right ratio is the whole game: we knew the adjacent side and wanted the opposite, and TOA points straight to tangent.

Cofunctions

In a right triangle the two acute angles add to 90 degrees, and the side opposite one is adjacent to the other. That symmetry produces the cofunction identities: sin(90° - θ) = cosθ, cos(90° - θ) = sinθ, and tan(90° - θ) = cotθ. This is literally why cosine is named "co-sine": it is the sine of the complementary angle. Check: sin 30 = 1/2 and cos 60 = 1/2.

Real-world applications

Right-triangle trigonometry is the surveyor's toolkit: heights of trees and buildings from angle-of-elevation sightings, aircraft glide slopes (a 3 degree descent path), wheelchair ramp regulations (a rise-to-run ratio is a tangent), and the grade of a road (a 6 percent grade means tan of about 3.4 degrees). GPS receivers and cell networks triangulate positions with these same ratios at scale.

Common misconceptions

• "sin⁻¹(x) means 1/sin(x)." On calculators, sin⁻¹ is the inverse sine (arcsine), which returns an angle; the reciprocal of sine is cosecant.
• "SOH-CAH-TOA works in any triangle." Only in right triangles; oblique triangles need the laws of sines and cosines.
• "The hypotenuse can be any side." It is always the side opposite the right angle and always the longest.
• "Trig values depend on the triangle's size." Similar triangles share ratios; trig values depend only on the angle.
• "Tangent is always defined." Tangent and secant blow up where cosine is 0 (90 and 270 degrees); cotangent and cosecant fail where sine is 0.

Recap

Sine and cosine come from the unit circle, and the other four functions are their quotients and reciprocals: tan = sin/cos, cot = cos/sin, sec = 1/cos, csc = 1/sin. In right triangles, SOH-CAH-TOA picks the ratio; from a point (x, y) at distance r, sin = y/r, cos = x/r, tan = y/x. Quadrant signs follow "All Students Take Calculus," cofunctions swap an angle with its complement, and tangent is undefined wherever cosine vanishes.

Sources

1. OpenStax, Precalculus 2e, Section 5.3: The Other Trigonometric Functions and Section 5.4: Right Triangle Trigonometry. Free at openstax.org/details/books/precalculus-2e.
2. OpenStax, Algebra and Trigonometry 2e, Chapter 7: The Unit Circle: Sine and Cosine Functions. Free at openstax.org/details/books/algebra-and-trigonometry-2e.
3. Khan Academy, Trigonometry: right triangles and the reciprocal trig functions. khanacademy.org/math/trigonometry.
4. Paul Dawkins, Paul's Online Math Notes, Trig Review. tutorial.math.lamar.edu.

The shape of a sinusoid

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

The four controls

A general sinusoid is y = A · sin(Bx - C) + D. Each constant changes the wave in a specific way:

• A - amplitude: half the distance from the highest point to the lowest; the wave rises A above and falls A below the midline. Use the absolute value of A.
• B - affects period: the period is 2π / |B|. A larger B squeezes the wave into a shorter period.
• C - phase shift: a horizontal shift equal to C/B.
• D - vertical shift (midline): raises or lowers the center line to y = D.

Worked example 1: amplitude and period

For y = 3 sin(2x): the amplitude is 3, so the wave reaches from -3 to 3. The period is 2π/2 = π, so one complete wave finishes in a horizontal distance of π instead of the usual 2π. The midline is y = 0.

Worked example 2: include a vertical shift

For y = 2 cos(x) + 1: amplitude 2 and midline y = 1. So the graph oscillates between 1 + 2 = 3 at the top and 1 - 2 = -1 at the bottom, with period 2π/1 = 2π. Because it is cosine, it starts at its maximum height of 3 when x = 0.

Worked example 3: read a graph backward

Suppose a wave has a maximum of 7, a minimum of 1, and completes a cycle every 4 units. The midline is the average, (7 + 1)/2 = 4, so D = 4. The amplitude is half the gap, (7 - 1)/2 = 3, so A = 3. The period 4 gives B = 2π/4 = π/2. A cosine model is therefore y = 3 cos((π/2)x) + 4. Reading these three numbers off a graph is the reverse of the plotting process and is a common exam task.

Phase shift in action

For y = 2 sin(2x - π/2), the phase shift is C/B = (π/2)/2 = π/4 to the right. A reliable habit is to factor the inside: 2x - π/2 = 2(x - π/4), which displays the shift directly. The amplitude is 2 and the period is 2π/2 = π, so this wave is a sine curve that starts its cycle at x = π/4 instead of at 0.

The five key points method

To graph one full cycle accurately, divide the period into four equal steps and plot five landmark points. For sine (no shifts): start on the midline heading up, reach the maximum at one quarter, return to the midline at half, hit the minimum at three quarters, and finish on the midline. For y = 3 sin(2x), the period is π, so the quarter step is π/4:

• (0, 0): midline, rising.
• (π/4, 3): maximum, since sin(π/2) = 1 and 3 · 1 = 3.
• (π/2, 0): midline again.
• (3π/4, -3): minimum.
• (π, 0): cycle complete.

Connect the five points with a smooth wave, then copy the cycle left and right as far as needed.

Reading maximum and minimum from the formula

The extremes are always midline plus and minus amplitude: maximum D + |A| and minimum D - |A|. For y = 3 sin(x) + 2: maximum 2 + 3 = 5, minimum 2 - 3 = -1. For y = 4 cos(x) - 1: maximum 3, minimum -5. No graphing needed.

Worked example 4: model a Ferris wheel

A Ferris wheel of radius 20 m has its center 25 m above the ground and completes one revolution every 30 seconds. A rider boards at the lowest point at t = 0. Model the height.

1. Midline and amplitude: the center height gives D = 25 and the radius gives amplitude 20.
2. Period: 30 seconds, so B = 2π/30 = π/15.
3. Starting position: the rider starts at the minimum. A negative cosine starts at its minimum, so use h(t) = 25 - 20 cos((π/15)t).

Check: h(0) = 25 - 20(1) = 5 m, the boarding height (25 - 20). At t = 15 (half a turn), cos(π) = -1 gives h = 45 m, the top. The model matches the physical wheel at every landmark.

A word on tangent graphs

Tangent behaves differently from sine and cosine: its natural period is π rather than 2π, it has no amplitude (it climbs without bound), and it has vertical asymptotes wherever cosine is zero, at x = π/2 + πk. Between consecutive asymptotes it sweeps from negative infinity up through 0 to positive infinity.

Real-world applications

Sinusoids model anything that cycles: hours of daylight across a year, tides in a harbor, alternating current in a wall outlet (60 cycles per second in North America), sound waves (amplitude is loudness, period sets pitch), and seasonal temperature curves. Fitting D, A, and B to data, exactly as in the worked examples, is how scientists turn periodic measurements into predictive formulas.

Common misconceptions

• "The period of sin(Bx) is B." The period is 2π/B; a bigger B means a shorter period, not a longer one.
• "Amplitude is the distance from top to bottom." It is half that distance, measured from the midline.
• "The maximum of A sin(x) + D is A." It is D + |A|; the vertical shift moves the extremes too.
• "In sin(Bx - C) the phase shift is C." It is C/B; factor the inside to see the true shift.
• "A negative amplitude makes heights negative." The negative sign reflects the wave across the midline; amplitude itself is |A|.

Recap

A sinusoid y = A sin(Bx - C) + D is controlled by four dials: amplitude |A| (height above the midline), period 2π/|B|, phase shift C/B, and midline y = D. Extremes sit at D plus or minus |A|. Graph one cycle with five key points a quarter period apart, and build models by reading midline, amplitude, and period from the situation, choosing sine or cosine (possibly reflected) to match the starting position.

Sources

1. OpenStax, Precalculus 2e, Section 6.1: Graphs of the Sine and Cosine Functions and Section 6.2: Graphs of the Other Trigonometric Functions. Free at openstax.org/details/books/precalculus-2e.
2. OpenStax, Algebra and Trigonometry 2e, Chapter 8: Periodic Functions. Free at openstax.org/details/books/algebra-and-trigonometry-2e.
3. Khan Academy, Trigonometry and Precalculus: graphing sinusoidal functions and modeling with periodic functions. khanacademy.org/math/trigonometry.
4. Paul Dawkins, Paul's Online Math Notes, Trig Review: Graphs of Trig Functions. tutorial.math.lamar.edu.

What an identity is

A trigonometric identity is an equation that is true for every angle where both sides are defined. Identities are the algebra of trigonometry: they let you rewrite a complicated expression as a simpler equivalent one.

The core identities

Reciprocal identities: csc = 1/sin, sec = 1/cos, cot = 1/tan.

Quotient identities: tan = sin/cos and cot = cos/sin.

Pythagorean identities, which come from the equation of the unit circle x² + y² = 1:

• sin²θ + cos²θ = 1 (the fundamental one).
• 1 + tan²θ = sec²θ.
• 1 + cot²θ = csc²θ.

The notation sin²θ means (sinθ)². The first Pythagorean identity is worth memorizing cold; the other two follow by dividing it through by cos² or sin².

Worked example 1: use the Pythagorean identity

If cosθ = 3/5 and θ is in Quadrant I, find sinθ. From sin²θ + cos²θ = 1: sin²θ = 1 - 9/25 = 16/25, so sinθ = 4/5 (positive in Quadrant I).

Worked example 2: simplify an expression

Simplify sinθ · cotθ. Replace cot with its quotient form: sinθ · (cosθ/sinθ). The sine cancels, leaving cosθ. So the whole expression is just cosθ.

Worked example 3: prove an identity

Prove that (1 - cos²θ) / sinθ = sinθ. Work on the left side only. By the Pythagorean identity, 1 - cos²θ = sin²θ. Substitute: sin²θ / sinθ. Cancel one factor of sine: sinθ, which equals the right side. The identity is proved.

The standard proof strategy is to pick the messier side, convert everything to sines and cosines, and simplify until it matches the other side. Never move terms across the equals sign as if you already knew it were true; transform one side into the other.

Even-odd identities

Negating an angle reflects its terminal side across the x-axis, which flips y but not x. Consequently sin(-θ) = -sinθ (sine is odd), cos(-θ) = cosθ (cosine is even), and tan(-θ) = -tanθ. Quick use: simplify sin(-x)/cos(-x) = -sin(x)/cos(x) = -tan(x).

Sum and difference identities

These unlock exact values beyond the special angles:

• sin(A ± B) = sinA cosB ± cosA sinB.
• cos(A ± B) = cosA cosB ∓ sinA sinB (note the sign flips for cosine).

Worked example 4: an exact value for cos 15 degrees

Write 15 as a difference of special angles: 15 = 45 - 30.

1. cos(45 - 30) = cos45 cos30 + sin45 sin30.
2. Substitute exact values: (√2/2)(√3/2) + (√2/2)(1/2).
3. Multiply: √6/4 + √2/4 = (√6 + √2)/4.

Decimal check: (2.449 + 1.414)/4 ≈ 0.966, and a calculator confirms cos 15° ≈ 0.966. The identity produced an exact form no table lookup could give.

Double-angle identities

Setting B = A in the sum identities gives the double-angle formulas:

• sin 2θ = 2 sinθ cosθ.
• cos 2θ = cos²θ - sin²θ = 2cos²θ - 1 = 1 - 2sin²θ (three equivalent forms).

Worked example 5: apply the double angle

Given sinθ = 3/5 with θ in Quadrant I, find sin 2θ and cos 2θ. From the Pythagorean identity, cosθ = 4/5. Then:

1. sin 2θ = 2(3/5)(4/5) = 24/25.
2. cos 2θ = (4/5)² - (3/5)² = 16/25 - 9/25 = 7/25.

Consistency check: (24/25)² + (7/25)² = (576 + 49)/625 = 625/625 = 1. The doubled angle still sits on the unit circle, as it must.

Worked example 6: a two-sided simplification proof

Prove tan x + cot x = sec x · csc x. Convert the left side to sines and cosines:

1. tan x + cot x = sin x/cos x + cos x/sin x.
2. Common denominator: (sin²x + cos²x)/(cos x sin x).
3. Pythagorean identity collapses the numerator to 1: 1/(cos x sin x).
4. Split the product: (1/cos x)(1/sin x) = sec x · csc x.

Every identity proof in this course uses the same three moves: convert to sine and cosine, combine fractions, and apply a Pythagorean identity.

Real-world applications

Identities are workhorses in physics and engineering. Adding two sound or radio waves uses the sum formulas, the double-angle form of a projectile's range formula R = (v² sin 2θ)/g explains why 45 degrees maximizes distance, and power calculations in alternating-current circuits rest on rewriting products of sinusoids with these same identities. Your calculator itself uses identity-based reductions to evaluate trig functions quickly.

Common misconceptions

• "sin(A + B) = sin A + sin B." False; the sum identity has the cross terms sinA cosB + cosA sinB.
• "sin²θ means sin(θ²)." It means (sinθ)², the square of the sine value.
• "You can cancel sin from both sides of an identity." Identities are proved by transforming one side, not by operating on both sides as if the statement were already known.
• "cos 2θ = 2 cosθ." Doubling the angle is not doubling the value; use one of the three double-angle forms.
• "There is one right proof." Any chain of valid identities that transforms one side into the other is a complete proof.

Recap

The identity toolkit: reciprocal (sec = 1/cos), quotient (tan = sin/cos), Pythagorean (sin² + cos² = 1 and its two divided forms), even-odd (cosine even, sine and tangent odd), sum and difference formulas for exact values like cos 15°, and double-angle formulas sin 2θ = 2 sinθ cosθ and cos 2θ in three forms. Prove identities by converting one side to sines and cosines and simplifying with these tools until it matches the other side.

Sources

1. OpenStax, Precalculus 2e, Chapter 7: Trigonometric Identities and Equations (sum, difference, and double-angle identities). Free at openstax.org/details/books/precalculus-2e.
2. OpenStax, Algebra and Trigonometry 2e, Chapter 9: Trigonometric Identities and Equations. Free at openstax.org/details/books/algebra-and-trigonometry-2e.
3. Khan Academy, Trigonometry and Precalculus: trigonometric identities. khanacademy.org/math/trigonometry.
4. Paul Dawkins, Paul's Online Math Notes, Trig Review: Trig Formulas and Identities. tutorial.math.lamar.edu.

What is being asked

A trigonometric equation asks for the angles that make a trig statement true. Because trig functions repeat, there are usually infinitely many solutions, so problems either restrict you to one revolution, often 0 ≤ θ < 2π, or ask for a general solution that lists them all.

The solving recipe

1. Isolate the trig function, just like isolating a variable.
2. Find the reference angle from the known value.
3. Place solutions in every quadrant where the function has the required sign.
4. If asked for all solutions, add the period to each.

Worked example 1: a sine equation

Solve 2 sinθ - 1 = 0 for 0 ≤ θ < 2π.

1. Isolate: sinθ = 1/2.
2. Reference angle: sine is 1/2 at π/6 (30 degrees).
3. Sine is positive in Quadrants I and II, so the solutions are θ = π/6 and θ = π - π/6 = 5π/6.

Worked example 2: a cosine equation

Solve cosθ = -1/2 for 0 ≤ θ < 2π. The reference angle for cosine 1/2 is π/3. Cosine is negative in Quadrants II and III, so θ = π - π/3 = 2π/3 and θ = π + π/3 = 4π/3.

Worked example 3: factor first

Solve 2 sin²θ - sinθ = 0 on 0 ≤ θ < 2π. Factor out sine: sinθ(2 sinθ - 1) = 0. Set each factor to zero.

• sinθ = 0 gives θ = 0 and θ = π.
• 2 sinθ - 1 = 0 gives sinθ = 1/2, so θ = π/6 and 5π/6.

All four values solve the equation. Treating trig equations like algebraic ones (isolate, factor, use the zero-product rule) is the recurring theme; the only new idea is that each value repeats every period.

Writing the general solution

To capture every solution, add whole periods to each solution in one revolution. Sine and cosine repeat every 2π, so sinθ = 1/2 has general solution θ = π/6 + 2πk or θ = 5π/6 + 2πk for any integer k. Tangent repeats every π, so tanθ = 1 has the single compact family θ = π/4 + πk, which covers both π/4 and 5π/4 in one formula.

Worked example 4: quadratic in form

Solve 2cos²θ + cosθ - 1 = 0 on 0 ≤ θ < 2π. Treat cosθ like a variable u, so the equation is 2u² + u - 1 = 0.

1. Factor: (2u - 1)(u + 1) = 0, that is (2cosθ - 1)(cosθ + 1) = 0. Check by expanding: 2cos²θ + 2cosθ - cosθ - 1 = 2cos²θ + cosθ - 1. Correct.
2. First factor: cosθ = 1/2, giving θ = π/3 and θ = 5π/3 (Quadrants I and IV).
3. Second factor: cosθ = -1, giving θ = π.

Three solutions in one revolution: π/3, π, 5π/3.

Worked example 5: multiple angles

Solve sin 2θ = 1 on 0 ≤ θ < 2π. Substitute u = 2θ; as θ runs from 0 to 2π, u runs from 0 to 4π, two full revolutions, so expect the solution list to double.

1. Solve for u: sine equals 1 at u = π/2, and adding a period gives u = π/2 + 2π = 5π/2 (both less than 4π).
2. Divide by 2: θ = π/4 and θ = 5π/4.

Both check: sin(π/2) = 1 and sin(5π/2) = 1. With multiple angles, always solve for the whole angle first and divide last.

Worked example 6: when the value is not special

Solve sinθ = 0.4 on 0 ≤ θ < 2π. No special angle has sine 0.4, so use the inverse function: θ = sin⁻¹(0.4) ≈ 0.4115 radians. That is the Quadrant I answer; sine is also positive in Quadrant II, so the second solution is π - 0.4115 ≈ 2.7301 radians. A calculator's inverse key always returns just one angle; the quadrant analysis supplies the rest.

The division trap

Solve sinθ cosθ = cosθ. Dividing both sides by cosθ looks tempting but silently discards every angle where cosθ = 0. Instead move everything to one side and factor: cosθ(sinθ - 1) = 0. Then cosθ = 0 gives θ = π/2, 3π/2, and sinθ = 1 gives θ = π/2 (already listed). Solutions: π/2 and 3π/2. Division by a possibly-zero factor is the single most common way to lose solutions.

Real-world applications

Trig equations answer timing questions in periodic systems: at what times is a tide 2 meters above mean level, when during the year does a city get exactly 12 hours of daylight, at what angles does a projectile achieve a given range, and when does an alternating current cross a threshold. In each case the model is a sinusoid and the question "when does it equal this value" is precisely a trigonometric equation.

Common misconceptions

• "One answer is enough." On one revolution most values occur twice (once per quadrant with the right sign); the general solution has infinitely many.
• "Divide both sides by a trig factor." Factoring is safe; dividing throws away the solutions where that factor is zero.
• "The calculator's inverse gives all solutions." It gives one; use reference angles and quadrant signs to find the others.
• "For sin 2 theta, divide the interval answers by 2 at the start." Solve for the doubled angle over the doubled interval first, then divide.
• "Squaring both sides is harmless." Squaring can create extraneous roots; check candidates in the original equation.

Recap

Solve trig equations like algebra problems: isolate the trig function or factor (including quadratic-in-form and zero-product patterns), find the reference angle, place solutions in the quadrants with the correct sign, and extend by the period for the general solution. For multiple angles, solve for the whole angle across the stretched interval before dividing. Use inverse functions for non-special values, and never divide away a factor that could be zero.

Sources

1. OpenStax, Precalculus 2e, Chapter 7: Trigonometric Identities and Equations (solving trigonometric equations). Free at openstax.org/details/books/precalculus-2e.
2. OpenStax, Algebra and Trigonometry 2e, Chapter 9: Trigonometric Identities and Equations. Free at openstax.org/details/books/algebra-and-trigonometry-2e.
3. Khan Academy, Trigonometry and Precalculus: solving sinusoidal equations. khanacademy.org/math/trigonometry.
4. Paul Dawkins, Paul's Online Math Notes, Calculus I Review: Solving Trig Equations. tutorial.math.lamar.edu.

Beyond right triangles

SOH-CAH-TOA only works in right triangles. For any triangle, two laws relate its sides (labeled a, b, c) to its opposite angles (A, B, C).

The law of sines

The law of sines states that each side is proportional to the sine of its opposite angle:

sin A / a = sin B / b = sin C / c

Use it when you know an angle and its opposite side, plus one more piece. That covers angle-angle-side (AAS), angle-side-angle (ASA), and the tricky side-side-angle (SSA) case.

Worked example 1: law of sines

In a triangle, A = 40 degrees, B = 60 degrees, and side a = 10. Find side b. Set up the proportion:

1. sin 40 / 10 = sin 60 / b.
2. Cross multiply: b · sin 40 = 10 · sin 60.
3. b = 10 · sin 60 / sin 40 ≈ 10 · 0.866 / 0.643 ≈ 13.5.

The law of cosines

The law of cosines generalizes the Pythagorean theorem to any triangle:

c² = a² + b² - 2ab · cos C

Use it when the law of sines cannot start, namely for side-angle-side (SAS) (two sides and the angle between them) or side-side-side (SSS) (all three sides, solving for an angle).

Worked example 2: law of cosines (SAS)

Two sides are a = 5 and b = 7 with the included angle C = 60 degrees. Find side c.

1. c² = 5² + 7² - 2(5)(7) cos 60.
2. c² = 25 + 49 - 70(0.5) = 74 - 35 = 39.
3. c = √39 ≈ 6.2.

Choosing the right law

You are given	Use
AAS or ASA (an angle-side opposite pair)	Law of sines
SSA (ambiguous case)	Law of sines, check for two triangles
SAS (two sides and included angle)	Law of cosines
SSS (three sides)	Law of cosines

A quick check: notice the law of cosines reduces to c² = a² + b² when C = 90 degrees, because cos 90 = 0. That recovers the Pythagorean theorem, a nice confirmation the formula is right.

Worked example 3: SSS, solving for an angle

A triangle has sides a = 7, b = 8, c = 9. Find angle C (opposite the side of length 9). Rearrange the law of cosines to solve for the cosine:

1. cos C = (a² + b² - c²) / (2ab).
2. Substitute: cos C = (49 + 64 - 81) / (2 · 7 · 8) = 32/112 = 2/7 ≈ 0.2857.
3. Invert: C = cos⁻¹(2/7) ≈ 73.4 degrees.

A sanity check: the largest side (9) should face the largest angle, and 73.4 degrees will indeed turn out to be the biggest angle of this triangle. A helpful bonus of solving for cosines: a negative cosine immediately announces an obtuse angle, with no quadrant ambiguity at all.

The ambiguous case, worked out

Given A = 30 degrees, a = 5, b = 8 (an SSA setup), find angle B. The law of sines gives sin B = b sin A / a = 8(0.5)/5 = 0.8. Two angles between 0 and 180 degrees have sine 0.8:

• B ≈ 53.1 degrees, leaving C = 180 - 30 - 53.1 = 96.9 degrees. Valid.
• B ≈ 180 - 53.1 = 126.9 degrees, leaving C = 180 - 30 - 126.9 = 23.1 degrees. Also valid.

Both angle sums stay under 180 degrees, so two different triangles match the given data. Always test the supplement in SSA problems: if the supplement still leaves a positive third angle, a second triangle exists; if not, only one (and if sin B computes to more than 1, none at all).

The area formula

Two sides and the included angle also give the area directly: Area = (1/2) a b sin C. For sides 5 and 7 meeting at 60 degrees: Area = (1/2)(5)(7) sin 60 ≈ 17.5 · 0.866 ≈ 15.2 square units. When C = 90 degrees this collapses to the familiar half base times height, since the sine of a right angle is 1. (For three known sides, Heron's formula computes area from the side lengths alone.)

Real-world applications

These laws are the mathematics of indirect measurement. Surveyors find the width of a lake by measuring two distances and the included angle from a single shore station (SAS, law of cosines). Sailors and pilots triangulate position from two bearings to known landmarks (ASA, law of sines). Cell networks locate phones, astronomers gauge stellar distances by parallax, and engineers resolve forces in truss bridges, all with the same two equations.

Common misconceptions

• "The law of sines works with any three pieces of data." It needs a known angle-opposite-side pair to anchor the proportion; otherwise start with the law of cosines.
• "The calculator's arcsin answer is the only angle." Sine is positive in two quadrants, so SSA problems must also test the supplement.
• "The law of cosines needs a right angle." It works in every triangle; the right-angle case merely reduces it to the Pythagorean theorem.
• "Any angle can pair with any two sides in SAS." The angle must be the included angle, between the two given sides.
• "Angles can be matched to sides arbitrarily." Each side faces its opposite angle, and the largest side always faces the largest angle; use this to catch errors.

Recap

The law of sines (sin A / a = sin B / b = sin C / c) solves triangles with a known angle-opposite-side pair (AAS, ASA, and the ambiguous SSA, which may have two solutions). The law of cosines (c² = a² + b² - 2ab cos C) handles SAS and SSS, and its rearranged form finds angles with no ambiguity. The area of any SAS triangle is (1/2)ab sin C. Together the two laws solve every solvable triangle.

Sources

1. OpenStax, Precalculus 2e, Section 8.1: Non-right Triangles: Law of Sines and Section 8.2: Non-right Triangles: Law of Cosines. Free at openstax.org/details/books/precalculus-2e.
2. OpenStax, Algebra and Trigonometry 2e, Chapter 10: Further Applications of Trigonometry. Free at openstax.org/details/books/algebra-and-trigonometry-2e.
3. Khan Academy, Trigonometry: solving general triangles with the laws of sines and cosines. khanacademy.org/math/trigonometry.

Sequences as ordered lists

A sequence is an ordered list of numbers called terms, written a₁, a₂, a₃, .... Two families are especially important because their patterns are so regular.

Arithmetic sequences

An arithmetic sequence adds a fixed number, the common difference d, to get from one term to the next. The list 3, 7, 11, 15, ... has d = 4. The explicit formula for the nth term is:

aₙ = a₁ + (n - 1)d

Worked example 1: arithmetic term

Find the 10th term of 3, 7, 11, 15, ... Here a₁ = 3 and d = 4, so a₁₀ = 3 + (10 - 1)(4) = 3 + 36 = 39.

Geometric sequences

A geometric sequence multiplies by a fixed number, the common ratio r, to advance. The list 2, 6, 18, 54, ... has r = 3. Its explicit formula is:

aₙ = a₁ · rⁿ⁻¹

Worked example 2: geometric term

Find the 5th term of 2, 6, 18, 54, ... With a₁ = 2 and r = 3: a₅ = 2 · 3⁴ = 2 · 81 = 162.

Telling them apart

To classify a sequence, test the pattern between consecutive terms. If the difference is constant, it is arithmetic; if the ratio is constant, it is geometric. The list 5, 10, 20, 40 is geometric (ratio 2), while 5, 10, 15, 20 is arithmetic (difference 5). Some sequences are neither, such as the squares 1, 4, 9, 16, whose differences 3, 5, 7 grow and whose ratios also change.

Worked example 3: build a formula

A theater has 20 seats in the first row and 4 more in each row after. How many seats in row 15? This is arithmetic with a₁ = 20 and d = 4: a₁₅ = 20 + (15 - 1)(4) = 20 + 56 = 76 seats. Recognizing "the same amount added each time" as arithmetic, or "multiplied by the same factor" as geometric, is the whole skill.

Recursive formulas

Besides the explicit formula, a sequence can be defined recursively: state the first term, then a rule for getting each term from the one before. Arithmetic: a₁ = 2 and aₙ = aₙ₋₁ + 7 generates 2, 9, 16, 23, ... Geometric: a₁ = 5 and aₙ = 2 · aₙ₋₁ generates 5, 10, 20, 40, ... Recursive rules match how spreadsheets and computer loops build lists, while explicit formulas jump straight to term 1000 without computing the first 999.

Worked example 4: recover the difference from two terms

An arithmetic sequence has a₃ = 11 and a₇ = 23. Find d and a₁.

1. From term 3 to term 7 is 4 steps, so a₇ - a₃ = 4d, giving 23 - 11 = 12 = 4d, so d = 3.
2. Walk back from a₃: a₁ = 11 - 2(3) = 5.
3. Check with the explicit formula: a₇ = 5 + (7 - 1)(3) = 5 + 18 = 23. Correct.

Worked example 5: recover the ratio from two terms

A geometric sequence has a₂ = 6 and a₅ = 48. From term 2 to term 5 is 3 multiplications, so r³ = 48/6 = 8, giving r = 2. Then a₁ = 6/2 = 3, and the sequence runs 3, 6, 12, 24, 48, confirming a₅ = 48.

Worked example 6: which term is it?

In the sequence 3, 7, 11, ..., which term equals 99? Set the explicit formula equal to 99 and solve for n:

1. 3 + (n - 1)(4) = 99.
2. (n - 1)(4) = 96, so n - 1 = 24.
3. n = 25. The 25th term is 99.

If solving gives a non-whole n, the target value is simply not a term of the sequence, which is itself a meaningful answer.

Negative and fractional ratios

The common ratio need not be a whole number or positive. The sequence 16, -8, 4, -2, ... has r = -1/2: each term is half the size and opposite in sign, so the terms alternate while shrinking toward zero. Alternating signs are the fingerprint of a negative ratio.

Real-world applications

Arithmetic sequences model steady addition: a salary starting at 40,000 dollars with 2,000 dollar annual raises reaches 40000 + (6 - 1)(2000) = 50,000 dollars in year 6, and stadium rows, fence posts, and simple interest follow the same pattern. Geometric sequences model proportional change: compound interest, population growth, radioactive decay, and the halving thickness of folded paper. Deciding which model fits (constant difference or constant ratio) is the first step of every application problem.

Common misconceptions

• "The nth term formula uses n, so multiply d by n." The factor is (n - 1): the first term has taken zero steps.
• "Ratios must be greater than 1." Ratios can be fractions (decay) or negative (alternating signs).
• "If it grows, it is geometric." Arithmetic sequences grow too; test whether the difference or the ratio is constant.
• "Recursive and explicit formulas are rivals." They describe the same sequence; recursive is easy to state, explicit is fast to evaluate at large n.
• "Sequences must start at a small number." Any first term works; the pattern, not the start, determines the type.

Recap

A sequence is an ordered list. Arithmetic sequences add a common difference d with explicit formula aₙ = a₁ + (n - 1)d; geometric sequences multiply by a common ratio r with aₙ = a₁ r to the power n - 1. Recursive rules define each term from its predecessor. Two known terms recover d (divide the gap by the number of steps) or r (take the appropriate root of the quotient), and setting the explicit formula equal to a target value finds which term it is.

Sources

1. OpenStax, Precalculus 2e, Section 11.2: Arithmetic Sequences and Section 11.3: Geometric Sequences. Free at openstax.org/details/books/precalculus-2e.
2. OpenStax, Algebra and Trigonometry 2e, Chapter 13: Sequences, Probability, and Counting Theory. Free at openstax.org/details/books/algebra-and-trigonometry-2e.
3. Khan Academy, Algebra and Precalculus: arithmetic and geometric sequences. khanacademy.org/math/precalculus.

From sequence to series

A series is the sum of the terms of a sequence. We write sums compactly with sigma notation, the Greek capital sigma, giving a starting index, an ending index, and a formula. For example, the sum from k = 1 to 4 of k means 1 + 2 + 3 + 4 = 10.

Sum of a finite arithmetic series

Pairing the first and last terms leads to a clean formula. The sum of the first n terms of an arithmetic sequence is:

Sₙ = (n/2)(a₁ + aₙ)

In words: the number of terms times the average of the first and last term.

Worked example 1: arithmetic sum

Add the integers from 1 to 100. This is arithmetic with a₁ = 1, a₁₀₀ = 100, and n = 100: S = (100/2)(1 + 100) = 50 · 101 = 5050.

Sum of a 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: geometric sum

Add 2 + 6 + 18 + 54 + 162. Here a₁ = 2, r = 3, n = 5: S = 2(1 - 3⁵)/(1 - 3) = 2(1 - 243)/(-2) = 2(-242)/(-2) = 242. A quick add of the five numbers confirms 242.

Infinite geometric series

If the common ratio satisfies |r| < 1, the terms shrink toward zero and the infinite sum converges to a finite value:

S = a₁ / (1 - r)

If |r| ≥ 1, the terms do not shrink and the series diverges (has no finite sum).

Worked example 3: infinite sum

Sum 1 + 1/2 + 1/4 + 1/8 + .... Here a₁ = 1 and r = 1/2, and since |1/2| < 1 it converges: S = 1/(1 - 1/2) = 1/(1/2) = 2. Adding endlessly many positive pieces yet reaching a finite total is a first taste of the limit ideas at the end of the course.

Reading sigma notation carefully

Evaluate the sum from k = 1 to 5 of (2k + 1). Write out the terms by substituting each k: 3 + 5 + 7 + 9 + 11 = 35. A faster route splits the sum: twice the sum of k (which is 2 · 15 = 30) plus five 1s (which is 5), again 35. The index k is a counter, not an unknown: the letter can change to i or j without changing the sum, and the limits tell you exactly which terms appear.

Another form of the arithmetic sum

When the last term is not handy, substitute the explicit formula into Sₙ = (n/2)(a₁ + aₙ) to get Sₙ = (n/2)(2a₁ + (n - 1)d). Example: the sum of the first 20 even numbers 2 + 4 + ... + 40 is (20/2)(2 + 40) = 10 · 42 = 420, or with the second form, (20/2)(2 · 2 + 19 · 2) = 10(4 + 38) = 420. Same answer, two routes.

Worked example 4: watch a series converge

The partial sums of 1 + 1/2 + 1/4 + 1/8 + ... march toward the limit 2:

Terms added	1	2	3	4	5
Partial sum	1	1.5	1.75	1.875	1.9375

Each partial sum halves the remaining gap to 2 and never overshoots. "The series sums to 2" means precisely that these partial sums approach 2 as their limit.

Worked example 5: a repeating decimal is a geometric series

Write 0.7777... as a fraction. The decimal unpacks into 7/10 + 7/100 + 7/1000 + ..., a geometric series with a₁ = 7/10 and r = 1/10. Since |r| < 1: S = (7/10)/(1 - 1/10) = (7/10)/(9/10) = 7/9. Check by dividing: 7 divided by 9 is 0.777... exactly.

Worked example 6: the bouncing ball

A ball drops 10 m and rebounds to 60 percent of each previous height. What total distance does it travel?

1. The rebound heights are geometric: 6, 3.6, 2.16, ... with r = 0.6, summing to 6/(1 - 0.6) = 15 m.
2. Each rebound height is traveled twice, up and down: 2 · 15 = 30 m.
3. Add the initial 10 m drop: 10 + 30 = 40 m total.

An infinite number of bounces, a finite total distance: convergence made physical.

Real-world applications

Finite geometric sums price annuities and loan payments (each payment's value forms a geometric sequence), and arithmetic sums count stacked logs, theater seats, and total payments under linear raises. Infinite geometric series compute repeating decimals, total medicine accumulation under repeated doses, and the total effect of economic stimulus (the multiplier effect is a geometric series in disguise).

Common misconceptions

• "Infinitely many terms means an infinite sum." If |r| < 1 the pieces shrink fast enough for a finite total.
• "The infinite sum formula always applies." It requires |r| < 1; for r = 3 or r = -2 the series diverges and the formula is meaningless.
• "(n/2)(a₁ + aₙ) works with any term at the end." The second entry must be the nth term, the last one actually added.
• "A sequence and a series are the same." A sequence is the list; a series is the sum of the list.
• "Partial sums equal the limit." They approach it; each partial sum of a positive series still falls short of the total.

Recap

Sigma notation compresses sums: read the limits and substitute each index value. Arithmetic series sum by pairing: S = (n/2)(first + last) = (n/2)(2a₁ + (n - 1)d). Finite geometric series use S = a₁(1 - rⁿ)/(1 - r), and infinite ones converge exactly when |r| < 1, to S = a₁/(1 - r). Repeating decimals, bouncing balls, and annuities are geometric series in costume, and convergence of partial sums previews the limit concept.

Sources

1. OpenStax, Precalculus 2e, Section 11.4: Series and Their Notations. Free at openstax.org/details/books/precalculus-2e.
2. OpenStax, Algebra and Trigonometry 2e, Chapter 13: Sequences, Probability, and Counting Theory. Free at openstax.org/details/books/algebra-and-trigonometry-2e.
3. Khan Academy, Precalculus: series and sigma notation. khanacademy.org/math/precalculus.
4. Paul Dawkins, Paul's Online Math Notes, Calculus II: Series basics (for the convergence idea). tutorial.math.lamar.edu.

What a vector is

A vector has both magnitude (length) and direction, unlike a plain number, which has only size. In the plane we write a vector by its components, the horizontal and vertical amounts, as v = ⟨a, b⟩. You can picture it as an arrow from the origin to the point (a, b).

Magnitude

The magnitude (length) of v = ⟨a, b⟩ comes from the Pythagorean theorem:

|v| = √(a² + b²)

Worked example 1: magnitude

For v = ⟨3, 4⟩, the magnitude is |v| = √(9 + 16) = √25 = 5.

Adding, subtracting, and scaling

Vector operations work component by component:

• Addition: ⟨a, b⟩ + ⟨c, d⟩ = ⟨a + c, b + d⟩.
• Subtraction: ⟨a, b⟩ - ⟨c, d⟩ = ⟨a - c, b - d⟩.
• Scalar multiplication: k⟨a, b⟩ = ⟨ka, kb⟩, which stretches (or flips, if k is negative) the vector.

Worked example 2: combine vectors

Let u = ⟨2, -1⟩ and v = ⟨3, 5⟩. Then u + v = ⟨5, 4⟩ and 2u - v = ⟨4, -2⟩ - ⟨3, 5⟩ = ⟨1, -7⟩.

The dot product

The dot product of two vectors is a single number: u · v = a₁a₂ + b₁b₂. It measures how much the vectors point the same way and connects to the angle θ between them by u · v = |u| |v| cosθ. A key consequence: two nonzero vectors are perpendicular exactly when their dot product is 0 (because cos 90 = 0).

Worked example 3: dot product and angle

Let u = ⟨1, 2⟩ and v = ⟨3, -1⟩. The dot product is u · v = (1)(3) + (2)(-1) = 3 - 2 = 1. Magnitudes: |u| = √5, |v| = √10. So cosθ = 1 / (√5 · √10) = 1/√50 ≈ 0.141, giving θ ≈ 81.9 degrees. Since the dot product is positive but small, the vectors point in broadly similar but nearly perpendicular directions, which the angle confirms.

Unit vectors and i, j notation

A unit vector has magnitude 1 and records pure direction. To find the unit vector pointing the same way as v, divide v by its own length: for v = ⟨3, 4⟩ with |v| = 5, the unit vector is ⟨3/5, 4/5⟩. Check: √(9/25 + 16/25) = √(25/25) = 1. Two special unit vectors get names: i = ⟨1, 0⟩ and j = ⟨0, 1⟩, so any vector can be written either way: ⟨3, 4⟩ = 3i + 4j.

Vectors between points

The vector from point P(1, 2) to point Q(4, 6) is found by subtracting coordinates, head minus tail: ⟨4 - 1, 6 - 2⟩ = ⟨3, 4⟩, with magnitude 5. Note the distinction: points name locations, vectors name displacements. Two different trips (from (0,0) to (3,4), or from (10, 10) to (13, 14)) are the same vector because the displacement is identical.

Direction angles

A vector's direction angle θ is measured counterclockwise from the positive x-axis, found from tanθ = b/a with quadrant care, exactly as in polar-style thinking. For ⟨3, 3⟩: tanθ = 1 in Quadrant I, so θ = 45 degrees, and the magnitude is √(9 + 9) = 3√2. Conversely, a vector with magnitude m and direction θ has components ⟨m cosθ, m sinθ⟩.

Worked example 4: an airplane in wind

A plane flies east at 200 mph while the wind blows north at 50 mph. The ground velocity is the vector sum ⟨200, 0⟩ + ⟨0, 50⟩ = ⟨200, 50⟩.

1. Ground speed: |v| = √(200² + 50²) = √(40000 + 2500) = √42500 ≈ 206.2 mph.
2. Direction: tanθ = 50/200 = 0.25, so θ ≈ 14.0 degrees north of east.

The plane moves faster than its airspeed and slightly off its heading; adding one more vector (the pilot's correction) is how real navigation is planned.

Worked example 5: work as a dot product

In physics, the work done by a force F over a displacement d is W = F · d. If F = ⟨10, 5⟩ newtons and d = ⟨4, 0⟩ meters, then W = (10)(4) + (5)(0) = 40 joules. Only the component of force along the motion does work, which is exactly what the dot product measures.

Real-world applications

Vectors carry every quantity with both size and direction: velocities of planes and boats in current, forces in statics and tug-of-war problems, displacement in navigation, and even video game character movement. Dot products test perpendicularity in graphics engines, compute lighting angles on 3D surfaces, and, as above, define mechanical work.

Common misconceptions

• "A vector is a point." A vector is a displacement; it has no fixed location and can be drawn anywhere.
• "|u + v| = |u| + |v|." Only when the vectors point the same way; in general the triangle inequality makes the sum's length shorter.
• "The dot product is a vector." It is a single number (a scalar); do not expect components.
• "tan gives the direction angle directly." The calculator's arctan needs a quadrant adjustment when the x-component is negative.
• "Scalar multiplication changes direction." A positive scalar rescales length only; a negative scalar also flips the vector 180 degrees.

Recap

A vector ⟨a, b⟩ encodes magnitude √(a² + b²) and direction (from tanθ = b/a with quadrant care). Add, subtract, and scale component by component, build vectors between points by head minus tail, and normalize to a unit vector by dividing by the magnitude. The dot product a₁a₂ + b₁b₂ equals |u||v|cosθ, detects perpendicularity at zero, and computes work in physics. Vector addition solves navigation and force problems by turning geometry into arithmetic.

Sources

1. OpenStax, Precalculus 2e, Section 8.8: Vectors. Free at openstax.org/details/books/precalculus-2e.
2. OpenStax, Algebra and Trigonometry 2e, Chapter 10: Further Applications of Trigonometry. Free at openstax.org/details/books/algebra-and-trigonometry-2e.
3. Khan Academy, Precalculus: Vectors unit. khanacademy.org/math/precalculus.
4. Paul Dawkins, Paul's Online Math Notes, Calculus II: Vectors, the basics and the dot product. tutorial.math.lamar.edu.

One family, four shapes

Slicing a double cone with a plane at different angles produces the conic sections: the circle, ellipse, parabola, and hyperbola. Each has a standard equation you can read features from.

Circles

A circle is all points a fixed distance (the radius r) from a center (h, k):

(x - h)² + (y - k)² = r²

Worked example: (x - 2)² + (y + 3)² = 25 has center (2, -3) and radius √25 = 5. Note the sign flip: (y + 3) means k = -3.

Parabolas

A parabola is the set of points equally far from a focus point and a directrix line. A vertical parabola with vertex (h, k) has the form y = a(x - h)² + k. The vertex is the turning point, and the sign of a decides whether it opens up or down. Worked example: y = 2(x - 1)² - 4 has vertex (1, -4) and opens upward because a = 2 > 0.

Ellipses

An ellipse is an oval, the set of points whose distances to two foci add to a constant. Centered at the origin its standard form is:

x²/a² + y²/b² = 1

The graph reaches a units left and right and b units up and down. Worked example: x²/25 + y²/9 = 1 stretches to x = ±5 and y = ±3, so it is wider than it is tall.

Hyperbolas

A hyperbola has two separate branches; centered at the origin one standard form is x²/a² - y²/b² = 1. The minus sign is the giveaway that separates a hyperbola from an ellipse.

Identifying a conic quickly

Clue in the equation	Conic
x squared and y squared, equal coefficients, added	Circle
only one variable is squared	Parabola
x squared and y squared, different positive coefficients, added	Ellipse
x squared and y squared subtracted	Hyperbola

Reading these clues lets you name a conic before doing any other work, then jump straight to the matching standard form to extract its features.

Completing the square to reach standard form

Conics often arrive expanded. Rebuild the standard form by completing the square in x and y separately. Analyze x² + y² - 6x + 4y - 12 = 0:

1. Group: (x² - 6x) + (y² + 4y) = 12.
2. Complete each square: half of -6 is -3, squared is 9; half of 4 is 2, squared is 4. Add both to each side: (x² - 6x + 9) + (y² + 4y + 4) = 12 + 9 + 4.
3. Rewrite: (x - 3)² + (y + 2)² = 25.

The mystery equation is a circle with center (3, -2) and radius √25 = 5. Every expanded conic yields to this same regrouping.

Foci of an ellipse

The two foci of an ellipse sit inside it on the longer axis, at distance c from the center where c² = a² - b² (a is the larger denominator's square root). For x²/25 + y²/9 = 1: c² = 25 - 9 = 16, so c = 4 and the foci are at (4, 0) and (-4, 0). The defining property, distances to the two foci adding to a constant (here 2a = 10), is what makes ellipses the shape of orbits.

Asymptotes of a hyperbola

A hyperbola x²/a² - y²/b² = 1 opens left and right and hugs two crossing guide lines, the asymptotes y = (b/a)x and y = -(b/a)x. For x²/9 - y²/4 = 1: a = 3, b = 2, so the asymptotes are y = ±(2/3)x and the vertices are at (±3, 0). Sketch the asymptotes first and the branches practically draw themselves.

The focus of a parabola

A vertical parabola with vertex at the origin can be written x² = 4py, where p is the distance from the vertex to the focus (and also from the vertex to the directrix, on the opposite side). For x² = 8y: 4p = 8 gives p = 2, so the focus is (0, 2) and the directrix is the line y = -2. Every point of the curve stays equidistant from that point and that line.

Real-world applications

Each conic earns its keep through its reflection property. A parabolic satellite dish reflects incoming parallel signals to its focus, where the receiver sits; car headlights run the trick in reverse. An elliptical whispering gallery carries a murmur from one focus to the other, and medical lithotripsy machines focus shock waves the same way; planets follow elliptical orbits with the Sun at one focus. Hyperbolas appear in LORAN-style navigation (constant difference of distances), sonic boom ground curves, and the silhouettes of cooling towers.

Common misconceptions

• "(x + 4) means the center is at +4." The standard form uses (x - h), so (x + 4) means h = -4; signs flip.
• "The radius is the number on the right side." The right side is r squared; take its square root.
• "In an ellipse, a goes with x automatically." a² is the larger denominator wherever it sits; if it is under y, the ellipse is tall rather than wide.
• "Ellipses and hyperbolas are close cousins of circles only." The sign between the squared terms is the whole story: plus is an ellipse, minus is a hyperbola.
• "Completing the square changes the graph." Adding the same amount to both sides only rewrites the equation; the curve never moves.

Recap

The four conic sections come from slicing a cone and are identified by their squared terms: both squared and equal coefficients (circle), both squared and added with different coefficients (ellipse), one squared (parabola), squared terms subtracted (hyperbola). Standard forms expose the features: center and radius for circles, vertices and foci with c² = a² - b² for ellipses, vertex and focus with x² = 4py for parabolas, and vertices with asymptotes y = plus or minus (b/a)x for hyperbolas. Completing the square converts expanded equations back to standard form.

Sources

1. OpenStax, Precalculus 2e, Chapter 10: Analytic Geometry (Sections 10.1 to 10.3: The Ellipse, The Hyperbola, The Parabola). Free at openstax.org/details/books/precalculus-2e.
2. OpenStax, Algebra and Trigonometry 2e, Chapter 12: Analytic Geometry. Free at openstax.org/details/books/algebra-and-trigonometry-2e.
3. Khan Academy, Precalculus: Conic sections unit. khanacademy.org/math/precalculus.
4. Paul Dawkins, Paul's Online Math Notes, Algebra: Circles, Ellipses, Parabolas, and Hyperbolas. tutorial.math.lamar.edu.

A different address system

Rectangular coordinates (x, y) locate a point by how far right and up it is. Polar coordinates (r, θ) instead give a distance r from the origin (the pole) and an angle θ measured from the positive x-axis. It is like giving directions as "go 5 units at 30 degrees" rather than "go 4 right and 3 up."

Plotting a polar point

To plot (4, 60 degrees), face along the positive x-axis, rotate 60 degrees counterclockwise, and walk 4 units. If r is negative, walk backward: (-2, 30 degrees) points opposite the 30-degree direction.

Converting polar to rectangular

Basic right-triangle trigonometry gives the bridge:

x = r cosθ and y = r sinθ

Worked example 1: polar to rectangular

Convert (4, 60 degrees). Then x = 4 cos 60 = 4(0.5) = 2 and y = 4 sin 60 = 4(√3/2) = 2√3 ≈ 3.46. So the point is (2, 2√3).

Converting rectangular to polar

Going the other way:

r = √(x² + y²) and tanθ = y/x

Worked example 2: rectangular to polar

Convert (3, 3). The distance is r = √(9 + 9) = √18 = 3√2. The angle has tanθ = 3/3 = 1, and since the point is in Quadrant I, θ = 45 degrees. So the polar form is (3√2, 45 degrees).

Simple polar equations

Some curves are far simpler in polar form:

• r = 5 is a circle of radius 5 centered at the origin, since every point is distance 5 from the pole regardless of angle.
• θ = 45 degrees is a line through the origin at that fixed angle.
• Equations like r = 2 + 2 cosθ trace elegant shapes (this one a heart-like curve called a cardioid) that would be clumsy to write in x and y.

Choosing coordinates that fit the symmetry of a problem, circular symmetry especially, can turn a hard equation into an easy one. That is the practical reason polar coordinates matter.

Many names for one point

Unlike rectangular coordinates, polar names are not unique. The same point answers to (r, θ), to (r, θ + 360°) (extra full turns), and to (-r, θ + 180°) (walk backward while facing the opposite way). For example, (3, 45°), (3, 405°), and (-3, 225°) all label the same location. Exams love this fact, and so do polar graphing utilities, which is why a curve can pass through a point whose "coordinates" do not obviously satisfy the equation.

Worked example 3: convert a negative-r point

Convert (-2, 30°) to rectangular coordinates. The formulas do not care about the sign of r:

1. x = r cosθ = -2 cos 30° = -2(√3/2) = -√3 ≈ -1.73.
2. y = r sinθ = -2 sin 30° = -2(1/2) = -1.

The point is (-√3, -1), in Quadrant III, exactly opposite the 30 degree direction, as a negative radius demands.

Worked example 4: convert equations, not just points

Convert the polar equation r = 4 cosθ to rectangular form. Multiply both sides by r to create convertible pieces:

1. r² = 4 r cosθ.
2. Substitute r² = x² + y² and r cosθ = x: so x² + y² = 4x.
3. Complete the square: x² - 4x + y² = 0 becomes (x - 2)² + y² = 4.

The graph is a circle of radius 2 centered at (2, 0). Going the other way is often easier: the circle x² + y² = 9 is simply r = 3, and the line y = x is θ = π/4.

Worked example 5: quadrant care in rectangular-to-polar

Convert (-3, 3) to polar form. The distance is r = √(9 + 9) = 3√2. Naively, tanθ = 3/(-3) = -1 suggests θ = -45°, but the point sits in Quadrant II (x negative, y positive), so add 180 degrees: θ = 135°. The polar form is (3√2, 135°). Always plot the point mentally before trusting the arctangent button.

A gallery of polar graphs

• r = a: circle centered at the pole; r = 2a cosθ: circle through the pole centered on the x-axis.
• r = a ± a cosθ (or with sine): cardioids, heart-shaped curves.
• r = a cos(nθ): rose curves with n petals when n is odd and 2n petals when n is even.
• r = a ± b cosθ with a ≠ b: limacons, with an inner loop when a < b.

Plotting a few strategic angles (0, 90, 180, 270 degrees) reveals which family an equation belongs to before you compute anything else.

Real-world applications

Anything measured by distance and bearing is already polar: radar and sonar displays, air traffic control scopes, and compass navigation. Directional microphones advertise a cardioid pickup pattern, literally the polar graph r = a + a cosθ, strong in front and dead behind. Antenna engineers publish radiation patterns as polar plots, and robotic arms and CNC machines often think in radius-angle terms around a rotation axis.

Common misconceptions

• "Each point has one polar name." Every point has infinitely many, via extra turns and negative radii.
• "r must be positive." Negative r is legal and means walk backward along the opposite ray.
• "theta = arctan(y/x), always." The arctangent needs a quadrant adjustment whenever x is negative.
• "Polar and rectangular graphs cannot describe the same curve." They are two languages for one geometry; convert with the standard bridges.
• "Degree and radian mode do not matter here." Evaluating r cos theta in the wrong mode silently ruins every conversion.

Recap

Polar coordinates locate points by distance r from the pole and angle theta from the positive x-axis. Convert with x = r cosθ, y = r sinθ, r = √(x² + y²), and tanθ = y/x plus a quadrant check. Points have many polar names, including negative-r versions. Equations convert too, often via multiplying by r, and the classic polar families (circles, cardioids, roses, limacons) model real radiation and pickup patterns.

Sources

1. OpenStax, Precalculus 2e, Section 8.3: Polar Coordinates and Section 8.4: Polar Coordinates: Graphs. Free at openstax.org/details/books/precalculus-2e.
2. OpenStax, Algebra and Trigonometry 2e, Chapter 10: Further Applications of Trigonometry. Free at openstax.org/details/books/algebra-and-trigonometry-2e.
3. Khan Academy, Precalculus: polar coordinates and polar graphs. khanacademy.org/math/precalculus.
4. Paul Dawkins, Paul's Online Math Notes, Calculus II: Polar Coordinates. tutorial.math.lamar.edu.

The idea behind a limit

A limit describes the value a function approaches as the input approaches some number, whether or not the function is actually defined there. We write lim (x approaches c) f(x) = L to mean "as x gets closer and closer to c, f(x) gets closer and closer to L." The limit is about the journey toward c, not the single point at c. This idea is the doorway to all of calculus.

Estimating from a table

Consider f(x) = (x² - 1)/(x - 1) near x = 1. At exactly x = 1 the formula gives 0/0, which is undefined. But we can sneak up on 1 from both sides:

x	0.9	0.99	1.01	1.1
f(x)	1.9	1.99	2.01	2.1

The outputs close in on 2 from both sides, so lim (x approaches 1) f(x) = 2, even though the function has a hole exactly at x = 1.

Confirming with algebra

Factor the top: x² - 1 = (x - 1)(x + 1). So for every x ≠ 1, f(x) = (x - 1)(x + 1)/(x - 1) = x + 1. As x approaches 1, x + 1 approaches 2. The algebra confirms the table exactly. This is a first example of resolving an indeterminate form (the 0/0 shape) by simplifying first.

Direct substitution when it works

For functions with no break at c (polynomials, for instance), you can simply substitute. lim (x approaches 3) (2x + 1) = 2(3) + 1 = 7, because the line is perfectly continuous there.

One-sided limits

Sometimes a function approaches different values from the left and the right. If the left-hand limit and the right-hand limit disagree, the two-sided limit does not exist. For a step that jumps from 2 up to 5 at x = 0, the limit from the left is 2 and from the right is 5, so lim (x approaches 0) does not exist.

Why this matters

Limits let mathematics handle "approaching without arriving," which is exactly what you need to define instantaneous speed (a slope over a vanishing interval) and the exact area under a curve (a sum of infinitely thin strips). Everything you learned about functions, graphs, and asymptotes in this course feeds directly into that next step, calculus.

Limit laws

Limits respect arithmetic: the limit of a sum is the sum of the limits, and the same holds for differences, products, constant multiples, and quotients (when the bottom limit is not zero). These limit laws justify direct substitution for polynomials: lim (x approaches 2) (x² + 3x) = 2² + 3(2) = 4 + 6 = 10, computed one piece at a time.

Worked example: another 0/0 rescue

Evaluate lim (x approaches 2) (x² - 5x + 6)/(x - 2).

1. Substituting x = 2 gives (4 - 10 + 6)/0 = 0/0, indeterminate, so simplify first.
2. Factor the numerator: x² - 5x + 6 = (x - 2)(x - 3).
3. Cancel: for x ≠ 2, the function equals x - 3.
4. Now substitute: 2 - 3 = -1. The limit is -1.

A table confirms it: at x = 1.99 the function is -1.01, and at x = 2.01 it is -0.99, closing in on -1 from both sides.

Limits at infinity

We can also ask where a function heads as x grows without bound, written lim (x approaches infinity). The basic fact is lim (x approaches infinity) 1/x = 0: the reciprocal of a huge number is nearly zero. Dividing top and bottom by the highest power handles rational functions: lim (x approaches infinity) (3x² + 1)/(x² + 2) = (3 + 1/x²)/(1 + 2/x²), and the little fractions vanish, leaving 3/1 = 3. Notice this is exactly the horizontal asymptote rule from the rational functions lesson; limits at infinity are what horizontal asymptotes were secretly describing all along.

Infinite limits and vertical asymptotes

Near a vertical asymptote, outputs grow without bound. As x approaches 0 from the right, 1/x exceeds every bound, written lim (x approaches 0 from the right) 1/x = infinity; from the left it plunges to negative infinity. Saying a limit "equals infinity" describes the unbounded behavior; as a finite number, the limit does not exist. Vertical asymptotes and one-sided infinite limits are two views of the same phenomenon.

Continuity

A function is continuous at a point c when three things line up: f(c) is defined, the limit as x approaches c exists, and the two are equal. Informally, you can draw the graph through that point without lifting your pencil. The hole in (x² - 1)/(x - 1) at x = 1 is a removable discontinuity (the limit exists but the value is missing), a step function has a jump discontinuity (one-sided limits disagree), and 1/x at 0 has an infinite discontinuity. Polynomials are continuous everywhere, which is precisely why substitution works for them.

A first taste of the derivative

Here is where all of this leads. The slope of the secant line through (1, 1) and a nearby point (1 + h, (1 + h)²) on the parabola f(x) = x² is:

1. ((1 + h)² - 1)/h = (1 + 2h + h² - 1)/h = (2h + h²)/h.
2. Factor and cancel h (legal since h is small but not zero): 2 + h.
3. Let h approach 0: the slope approaches 2.

The parabola has an exact steepness of 2 at the point (1, 1), computed by a limit of an indeterminate 0/0 form, resolved by algebra, exactly the skills of this lesson. That number is called the derivative, and it is the opening act of calculus.

Real-world applications

A speedometer reports a limit: average speeds over shrinking time intervals approaching instantaneous speed. Drug concentrations approach steady-state levels (limits at infinity), populations approach carrying capacities, and compound interest compounded ever more often approaches the continuous limit e. Whenever a process stabilizes, saturates, or needs an exact instantaneous rate, a limit is doing the work.

Common misconceptions

• "The limit is the value of the function at the point." The limit describes the approach; the function may be undefined there or defined with a different value.
• "0/0 means the limit does not exist." It means the form is indeterminate; algebra often reveals a perfectly finite limit.
• "If f(c) is undefined, the limit cannot exist." Holes are the classic counterexample: undefined value, existing limit.
• "A limit of infinity is an ordinary answer." It reports unbounded growth; as a finite number the limit does not exist.
• "Checking one side is enough." The two-sided limit requires the left and right approaches to agree.

Recap

A limit is the value a function approaches as its input approaches a point, independent of the value at the point. Estimate limits with tables and graphs, compute them with substitution when continuous, and rescue 0/0 forms by factoring and canceling. One-sided limits must agree for a two-sided limit to exist, limits at infinity explain horizontal asymptotes, infinite limits explain vertical ones, and continuity means the limit equals the value. The derivative, a limit of slopes, is where calculus begins.

Sources

1. OpenStax, Precalculus 2e, Chapter 12: Introduction to Calculus (Sections 12.1 to 12.4). Free at openstax.org/details/books/precalculus-2e.
2. Khan Academy, Precalculus and AP Calculus: Limits and continuity unit. khanacademy.org/math/precalculus.
3. Paul Dawkins, Paul's Online Math Notes, Calculus I: Limits (tangent lines, one-sided limits, continuity). tutorial.math.lamar.edu.