Integrated Mathematics 3

Course overview

Find the topic you are stuck on

Integrated Mathematics 3 is the year the function families all arrive at once (polynomial, rational, radical, exponential, logarithmic and trigonometric) and the year statistics stops describing data and starts drawing conclusions from it. Three separate units here produce extraneous solutions, so checking becomes part of the method rather than an optional last step. The topics below follow the order most California Integrated III courses use, so you can go straight to whatever was covered in class today.

U1Unit 1: Polynomial Functions3 topics
U2Unit 2: Rational and Radical Functions3 topics
U3Unit 3: Exponential and Logarithmic Functions3 topics
U4Unit 4: Trigonometric Functions3 topics
U5Unit 5: Sequences, Series and Function Operations4 topics
U6Unit 6: Inference and Conclusions from Data2 topics

All six units are open, 18 topics in all. Every topic opens with the method, worked examples, and fifteen practice problems. Every problem has a full worked solution, so you can find the step where yours went wrong. Each unit closes with a ten-problem mixed review.

Free preview: open any 5 topics without an account. The counter on the left keeps track.

Topic 1.1 · Unit 1 · CA CCSS A-APR.2, A-APR.6

Polynomial division, remainder and factor theorems

Dividing polynomials is how you break a cubic or quartic down to something you can factor. Synthetic division is much faster than long division, but only when the divisor is linear, knowing which tool applies is half the topic.

The method

Long division works for any divisor. Divide the leading terms, multiply back, subtract, bring down. The subtraction step is where signs go wrong, write the subtracted line with its signs already flipped.
Synthetic division works only for divisors of the form $ x - c $. Write $ c $, then the coefficients. Bring down the first, multiply by $ c $, add, and repeat. The last number is the remainder; the rest are the quotient's coefficients, one degree lower.
Insert zeros for missing degrees. $ x^3 - 8 $ has coefficients $ 1, 0, 0, -8 $. Skipping a missing term shifts everything and silently ruins the answer.
Remainder theorem: the remainder on dividing $ P(x) $ by $ x - c $ is $ P(c) $. So evaluating is a one-line alternative to dividing when only the remainder is wanted.
Factor theorem: $ x - c $ is a factor exactly when $ P(c) = 0 $. This is what turns "is 3 a root?" into a substitution rather than a division.
Write the answer as $ \text{quotient} + \frac{\text{remainder}}{\text{divisor}} $.

Where marks are lost: the sign of $ c $. Dividing by $ x + 4 $ means $ c = -4 $, because $ x + 4 = x - (-4) $. Using $ +4 $ gives a wrong answer that still looks tidy.

Worked examples

Example 1: synthetic division. Divide $ 2x^3 - 5x^2 + 3x - 7 $ by $ x - 2 $.

Here $ c = 2 $ and the coefficients are $ 2, -5, 3, -7 $.

Bring down 2. Then $ 2 \times 2 = 4 $, $ -5 + 4 = -1 $. Then $ -1 \times 2 = -2 $, $ 3 - 2 = 1 $. Then $ 1 \times 2 = 2 $, $ -7 + 2 = -5 $.

Quotient $ 2x^2 - x + 1 $, remainder $ -5 $. So the answer is $ 2x^2 - x + 1 - \frac{5}{x-2} $.

Example 2: remainder theorem. Find the remainder when $ x^4 - 3x^2 + 5 $ is divided by $ x + 1 $.

$ x + 1 $ means $ c = -1 $. Evaluate: $ P(-1) = 1 - 3 + 5 = 3 $.

The remainder is 3, no division needed.

Example 3: factor theorem into a full factorization. Factor $ x^3 - 4x^2 + x + 6 $ given that $ x = 3 $ is a root.

Since $ P(3) = 27 - 36 + 3 + 6 = 0 $, $ x - 3 $ is a factor. Synthetic division with $ c = 3 $ on $ 1, -4, 1, 6 $ gives $ 1, -1, -2 $ with remainder 0.

So $ P(x) = (x-3)(x^2 - x - 2) = (x-3)(x-2)(x+1) $. Always try to factor the quotient further, stopping at the quadratic loses marks.

Practice · 15 problems

1–5 division, 6–10 the remainder and factor theorems, 11–15 full factorization and diagnosis.

Divide $ x^2 + 5x + 6 $ by $ x + 2 $.
Show the full solution

$ c = -2 $ on $ 1, 5, 6 $: bring down 1; $ 5 - 2 = 3 $; $ 6 - 6 = 0 $. $ x + 3 $
Divide $ x^3 - 2x^2 - 5x + 6 $ by $ x - 1 $.
Show the full solution

$ c = 1 $ on $ 1, -2, -5, 6 $: $ 1, -1, -6, 0 $. $ x^2 - x - 6 $
Divide $ 3x^3 + 2x - 5 $ by $ x - 1 $.
Show the full solution

Insert the missing $ x^2 $: coefficients $ 3, 0, 2, -5 $. Synthetic with $ c = 1 $: $ 3, 3, 5, 0 $. $ 3x^2 + 3x + 5 $
Divide $ 2x^3 + 3x^2 - 1 $ by $ x + 2 $.
Show the full solution

$ c = -2 $ on $ 2, 3, 0, -1 $: $ 2, -1, 2, -5 $. $ 2x^2 - x + 2 - \frac{5}{x+2} $
Divide $ x^3 + 2x^2 - x - 2 $ by $ x^2 - 1 $ using long division.
Show the full solution

$ x^3 \div x^2 = x $; subtracting $ x^3 - x $ leaves $ 2x^2 - 2 $. Then $ 2x^2 \div x^2 = 2 $; subtracting $ 2x^2 - 2 $ leaves 0. $ x + 2 $
Find the remainder when $ x^3 - 4x + 1 $ is divided by $ x - 2 $.
Show the full solution

$ P(2) = 8 - 8 + 1 $. 1
Find the remainder when $ 2x^3 + x^2 - 3 $ is divided by $ x + 1 $.
Show the full solution

$ c = -1 $: $ P(-1) = -2 + 1 - 3 = -4 $. $ -4 $
Is $ x - 3 $ a factor of $ x^3 - 2x^2 - 5x + 6 $?
Show the full solution

$ P(3) = 27 - 18 - 15 + 6 = 0 $. Yes
Is $ x + 2 $ a factor of $ x^3 + x^2 - 4x - 4 $?
Show the full solution

$ P(-2) = -8 + 4 + 8 - 4 = 0 $. Yes
Find $ k $ so that $ x - 2 $ is a factor of $ x^3 + kx - 6 $.
Show the full solution

Require $ P(2) = 0 $: $ 8 + 2k - 6 = 0 $, so $ 2k = -2 $. $ k = -1 $
Factor $ x^3 - 2x^2 - 5x + 6 $ completely.
Show the full solution

$ P(1) = 0 $, so divide by $ x - 1 $ to get $ x^2 - x - 6 $, which factors. $ (x-1)(x-3)(x+2) $
Factor $ x^3 + 3x^2 - 4x - 12 $ completely.
Show the full solution

Group: $ x^2(x+3) - 4(x+3) = (x+3)(x^2-4) $. $ (x+3)(x-2)(x+2) $
Solve $ x^3 - 7x + 6 = 0 $.
Show the full solution

$ P(1) = 0 $; dividing by $ x - 1 $ with coefficients $ 1, 0, -7, 6 $ gives $ x^2 + x - 6 = (x+3)(x-2) $. $ x = 1, 2, -3 $
A student divides $ x^3 - 8 $ by $ x - 2 $ using the coefficients $ 1, -8 $. Find the error.
Show the full solution

The $ x^2 $ and $ x $ terms are missing and must be entered as zeros: $ 1, 0, 0, -8 $. Synthetic division with $ c = 2 $ then gives $ 1, 2, 4 $ remainder 0. $ x^2 + 2x + 4 $
A student divides by $ x + 5 $ using $ c = 5 $. Explain why that is wrong.
Show the full solution

Synthetic division uses the value that makes the divisor zero. $ x + 5 = 0 $ gives $ x = -5 $, so $ c = -5 $. Using $ +5 $ divides by $ x - 5 $ instead. $ c = -5 $

Topic 1.2 · Unit 1 · CA CCSS A-APR.3, N-CN.9

Factoring and solving higher-degree polynomials

A cubic or quartic will not yield to the quadratic formula, so you find one root, divide it out, and repeat until what is left is quadratic. The rational root theorem tells you which candidates are worth testing.

The method

Common factor first, always. $ 2x^4 - 18x^2 $ becomes $ 2x^2(x^2 - 9) $ and the rest is easy.
Recognize the special forms. Sum and difference of cubes: $ a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2) $. Quadratic in disguise: $ x^4 - 5x^2 + 4 $ factors like $ u^2 - 5u + 4 $ with $ u = x^2 $.
Grouping handles four terms: pair them, factor each pair, and look for the shared bracket.
Rational root theorem: any rational root is $ \pm\frac{\text{factor of the constant}}{\text{factor of the leading coefficient}} $. Test candidates with the factor theorem, cheapest first.
Divide out each root found and continue on the quotient. Stop when the quotient is quadratic and finish with factoring or the formula.
Count the roots. A degree-$ n $ polynomial has exactly $ n $ roots counting multiplicity and complex ones. If you have fewer, you stopped early.

Where marks are lost: solving $ x^4 - 5x^2 + 4 = 0 $ and stopping at $ x^2 = 1, 4 $. Those are values of $ x^2 $, not $ x $; the four solutions are $ \pm 1 $ and $ \pm 2 $.

Worked examples

Example 1: quadratic in disguise. Solve $ x^4 - 13x^2 + 36 = 0 $.

Let $ u = x^2 $: $ u^2 - 13u + 36 = (u - 4)(u - 9) = 0 $, so $ u = 4 $ or $ u = 9 $.

Back-substitute: $ x^2 = 4 $ gives $ x = \pm 2 $; $ x^2 = 9 $ gives $ x = \pm 3 $. Four roots for a quartic, as expected.

Example 2: rational root theorem. Solve $ 2x^3 - 3x^2 - 8x + 12 = 0 $.

Candidates are $ \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12 $ over $ 1, 2 $. Testing $ x = 2 $: $ 16 - 12 - 16 + 12 = 0 $. ✓

Synthetic division by $ x - 2 $ on $ 2, -3, -8, 12 $ gives $ 2, 1, -6 $, so the quotient is $ 2x^2 + x - 6 = (2x - 3)(x + 2) $.

Roots: $ x = 2, \frac{3}{2}, -2 $.

Example 3: a cubic with complex roots. Solve $ x^3 + 8 = 0 $.

Sum of cubes: $ (x + 2)(x^2 - 2x + 4) = 0 $.

The linear factor gives $ x = -2 $. The quadratic gives $ x = \frac{2 \pm \sqrt{4 - 16}}{2} = \frac{2 \pm 2i\sqrt{3}}{2} = 1 \pm i\sqrt{3} $. Three roots for a cubic, answering only $ x = -2 $ is incomplete.

Practice · 15 problems

1–5 factoring, 6–11 solving, 12–15 harder cases and diagnosis.

Factor $ x^3 - 27 $.
Show the full solution

Difference of cubes with $ a = x $, $ b = 3 $. $ (x-3)(x^2+3x+9) $
Factor $ 8x^3 + 125 $.
Show the full solution

$ a = 2x $, $ b = 5 $. $ (2x+5)(4x^2-10x+25) $
Factor $ x^4 - 16 $ completely.
Show the full solution

$ (x^2-4)(x^2+4) $, and the first factors again. $ (x-2)(x+2)(x^2+4) $
Factor $ x^3 + 2x^2 - 9x - 18 $ by grouping.
Show the full solution

$ x^2(x+2) - 9(x+2) = (x+2)(x^2-9) $. $ (x+2)(x-3)(x+3) $
Factor $ 3x^4 - 12x^2 $.
Show the full solution

$ 3x^2(x^2-4) $. $ 3x^2(x-2)(x+2) $
Solve $ x^3 - 9x = 0 $.
Show the full solution

$ x(x^2-9) = 0 $. Dividing by $ x $ would lose the root at 0. $ x = 0, 3, -3 $
Solve $ x^4 - 5x^2 + 4 = 0 $.
Show the full solution

$ (x^2-1)(x^2-4) = 0 $, so $ x^2 = 1 $ or $ 4 $. $ x = \pm 1, \pm 2 $
List the possible rational roots of $ x^3 - 4x^2 + x + 6 $.
Show the full solution

Factors of 6 over factors of 1. $ \pm 1, \pm 2, \pm 3, \pm 6 $
Solve $ x^3 - 4x^2 + x + 6 = 0 $.
Show the full solution

$ P(-1) = -1 - 4 - 1 + 6 = 0 $. Dividing by $ x + 1 $ gives $ x^2 - 5x + 6 = (x-2)(x-3) $. $ x = -1, 2, 3 $
Solve $ 2x^3 + x^2 - 13x + 6 = 0 $.
Show the full solution

$ P(2) = 16 + 4 - 26 + 6 = 0 $. Dividing gives $ 2x^2 + 5x - 3 = (2x-1)(x+3) $. $ x = 2, \frac{1}{2}, -3 $
Solve $ x^3 - 1 = 0 $ completely.
Show the full solution

$ (x-1)(x^2+x+1) = 0 $; the quadratic gives $ x = \frac{-1 \pm i\sqrt{3}}{2} $. $ x = 1, \frac{-1 \pm i\sqrt{3}}{2} $
Write a polynomial of least degree with roots $ 2 $, $ -1 $ and $ 3 $.
Show the full solution

$ (x-2)(x+1)(x-3) $; expanding gives $ x^3 - 4x^2 + x + 6 $. $ x^3 - 4x^2 + x + 6 $
Write a real-coefficient polynomial of least degree with roots $ 3 $ and $ 2i $.
Show the full solution

Real coefficients force $ -2i $ as well, so $ (x-3)(x^2+4) = x^3 - 3x^2 + 4x - 12 $. $ x^3 - 3x^2 + 4x - 12 $
A student solves $ x^4 - 13x^2 + 36 = 0 $ and answers $ x = 4, 9 $. Find the error.
Show the full solution

Those are values of $ u = x^2 $, not of $ x $. Taking square roots of each gives four roots. $ x = \pm 2, \pm 3 $
A student solves a cubic and finds only one root. What should that prompt them to check?
Show the full solution

A cubic has exactly three roots counting multiplicity and complex ones, so the quadratic quotient was never solved; it either factors further or needs the formula, possibly giving a conjugate pair. Two roots are missing

Topic 1.3 · Unit 1 · CA CCSS F-IF.7c, A-APR.3

Graphing polynomial functions

You are not plotting points. A polynomial graph is determined by three things (where it crosses, what it does at each crossing, and where the ends go) and a sketch that gets those right earns the marks.

The method

Factor and read the zeros. Each factor $ (x - c) $ puts an $ x $-intercept at $ c $.
Multiplicity decides the behavior there. Odd multiplicity crosses the axis; even multiplicity touches and turns back. Multiplicity 3 or more flattens as it crosses.
End behavior comes from the leading term only. Even degree sends both ends the same way, odd degree sends them opposite ways; a positive leading coefficient sends the right end up, a negative one sends it down.
The $ y $-intercept is $ f(0) $, the constant term.
Turning points: a degree-$ n $ polynomial has at most $ n - 1 $ of them. Use that as a check on your sketch.
Sign chart: the sign can only change at a zero, so test one value in each interval to know whether the curve is above or below the axis there.

Where marks are lost: drawing every intercept as a crossing. In $ y = (x-1)^2(x+3) $, the graph touches at $ x = 1 $ and turns around; it does not pass through.

Worked examples

Example 1: a full sketch. Describe the graph of $ f(x) = (x+2)(x-1)^2 $.

Zeros at $ x = -2 $ (multiplicity 1, crosses) and $ x = 1 $ (multiplicity 2, touches).

Expanding mentally, the leading term is $ x^3 $: odd degree, positive coefficient, so the left end goes down and the right end goes up.

$ y $-intercept: $ f(0) = (2)(1) = 2 $. The curve rises from the lower left, crosses at $ -2 $, passes through $ (0, 2) $, comes down to touch the axis at $ x = 1 $, and rises again.

Example 2: end behavior. Describe the end behavior of $ f(x) = -2x^4 + 7x^3 - x + 5 $.

Only $ -2x^4 $ matters. Even degree means both ends behave alike; the negative coefficient sends them down.

As $ x \to \pm\infty $, $ f(x) \to -\infty $. The other terms change the middle of the graph, never the ends.

Example 3: sign chart. Where is $ f(x) = x(x-3)(x+1) $ positive?

Zeros at $ -1, 0, 3 $ split the line into four intervals. Test one point in each: $ f(-2) = (-2)(-5)(-1) = -10 $ negative; $ f(-0.5) = (-0.5)(-3.5)(0.5) $ positive; $ f(1) = (1)(-2)(2) = -4 $ negative; $ f(4) = (4)(1)(5) = 20 $ positive.

So $ f(x) > 0 $ on $ (-1, 0) $ and $ (3, \infty) $.

Practice · 15 problems

1–5 zeros and end behavior, 6–11 full descriptions, 12–15 reading graphs and diagnosis.

State the zeros of $ f(x) = (x-4)(x+2)(x-1) $.
Show the full solution

$ x = 4, -2, 1 $
State the degree of $ f(x) = (x-1)^2(x+3)^3 $.
Show the full solution

Add the multiplicities: $ 2 + 3 $. 5
Describe the end behavior of $ f(x) = 3x^5 - x^2 + 1 $.
Show the full solution

Odd degree, positive leading coefficient. Down on the left, up on the right
Describe the end behavior of $ f(x) = -x^6 + 4x $.
Show the full solution

Even degree, negative coefficient. Both ends down
Find the $ y $-intercept of $ f(x) = (x-2)(x+5) $.
Show the full solution

$ f(0) = (-2)(5) $. $ -10 $
At $ x = -3 $, does $ f(x) = (x+3)^2(x-1) $ cross or touch?
Show the full solution

Multiplicity 2, which is even. Touches and turns
At $ x = 2 $, does $ f(x) = (x-2)^3(x+1) $ cross or touch?
Show the full solution

Multiplicity 3 is odd, so it crosses, flattening as it does. Crosses
What is the maximum number of turning points of a degree-6 polynomial?
Show the full solution

$ n - 1 $. 5
Fully describe the graph of $ f(x) = (x-1)(x+2)^2 $.
Show the full solution

Degree 3, positive leading coefficient, so down-left and up-right. Crosses at $ x = 1 $, touches at $ x = -2 $, $ y $-intercept $ (-1)(4) = -4 $. See description
Fully describe the graph of $ f(x) = -x^2(x-3) $.
Show the full solution

Leading term $ -x^3 $: up-left, down-right. Touches at $ x = 0 $, crosses at $ x = 3 $, $ y $-intercept 0. See description
Where is $ f(x) = (x-2)(x+1) $ negative?
Show the full solution

An upward parabola is below the axis between its roots. $ -1 < x < 2 $
Solve $ x(x-4)(x+2) > 0 $.
Show the full solution

Zeros at $ -2, 0, 4 $. Testing: $ x=-3 $ negative, $ x=-1 $ positive, $ x=1 $ negative, $ x=5 $ positive. $ -2 < x < 0 $ or $ x > 4 $
A graph touches the axis at $ x = 2 $, crosses at $ x = -1 $ and at $ x = 5 $, and both ends go up. Write a possible equation.
Show the full solution

Touching needs even multiplicity, so $ (x-2)^2 $. Each crossing needs odd multiplicity, so $ (x+1) $ and $ (x-5) $. That is degree 4, even, and both ends go up with a positive leading coefficient, which matches. $ f(x) = (x-2)^2(x+1)(x-5) $
A student says $ f(x) = -x^4 + 100x^3 $ goes up on the right because of the $ 100x^3 $. Find the error.
Show the full solution

End behavior depends only on the highest-degree term. For large $ x $, $ x^4 $ dominates $ 100x^3 $ no matter how big the coefficient, the graph turns and both ends go down. Both ends down
A student draws $ y = (x-1)^2(x+3) $ crossing the axis at $ x = 1 $. Find the error.
Show the full solution

The factor is squared, so the multiplicity is even and the curve touches at $ x = 1 $ and turns back. Only $ x = -3 $ is a crossing. It touches, not crosses

Unit 1 mixed review · 10 problems · all topics

Unit 1 mixed review: Polynomial Functions

Check the root count on every solving problem, a cubic has three, a quartic has four, and a short answer means you stopped early.

Divide $ x^3 + 3x^2 - 4 $ by $ x - 1 $.
Show the full solution

Coefficients $ 1, 3, 0, -4 $ with $ c = 1 $: $ 1, 4, 4, 0 $. $ x^2 + 4x + 4 $
Find the remainder when $ 2x^3 - x + 4 $ is divided by $ x + 2 $.
Show the full solution

$ P(-2) = -16 + 2 + 4 $. $ -10 $
Is $ x - 2 $ a factor of $ x^3 - 3x^2 + 4 $?
Show the full solution

$ P(2) = 8 - 12 + 4 = 0 $. Yes
Factor $ x^3 + 64 $.
Show the full solution

Sum of cubes with $ a = x $, $ b = 4 $. $ (x+4)(x^2-4x+16) $
Solve $ x^4 - 10x^2 + 9 = 0 $.
Show the full solution

$ (x^2-1)(x^2-9) = 0 $, so $ x^2 = 1 $ or $ 9 $. $ x = \pm 1, \pm 3 $
Solve $ x^3 - 6x^2 + 11x - 6 = 0 $.
Show the full solution

$ P(1) = 0 $; dividing gives $ x^2 - 5x + 6 = (x-2)(x-3) $. $ x = 1, 2, 3 $
Describe the end behavior of $ f(x) = -3x^5 + 2x^2 $.
Show the full solution

Odd degree, negative leading coefficient. Up on the left, down on the right
At $ x = -2 $, does $ f(x) = (x+2)^4(x-1) $ cross or touch?
Show the full solution

Multiplicity 4 is even. Touches
Solve $ x(x-4)(x+1) \lt 0 $.
Show the full solution

Zeros at $ -1, 0, 4 $. Testing: $ x=-2 $ negative, $ x=-0.5 $ positive, $ x=1 $ negative, $ x=5 $ positive. $ x \lt -1 $ or $ 0 \lt x \lt 4 $
A student divides $ x^4 - 1 $ by $ x - 1 $ using coefficients $ 1, -1 $. Find the error.
Show the full solution

The missing degrees must be entered as zeros: $ 1, 0, 0, 0, -1 $. Synthetic division with $ c = 1 $ gives $ 1, 1, 1, 1 $ remainder 0. $ x^3 + x^2 + x + 1 $

Topic 2.1 · Unit 2 · CA CCSS A-APR.7, A-SSE.2

Simplifying and operating on rational expressions

A rational expression is a fraction with polynomials on top and bottom, and every operation you already know for numerical fractions carries over. The one new requirement is that you factor before you do anything else.

The method

Factor everything first: numerator and denominator of every fraction in the problem. Nothing else can be done reliably until this is complete.
Simplify by canceling matching factors. You may cancel $ (x-2) $ against $ (x-2) $; you may never cancel a term out of a sum.
Multiply by factoring, canceling across both fractions, then writing what is left. Divide by flipping the second fraction and multiplying.
Add and subtract require a common denominator, the least common multiple of the factored denominators, each distinct factor taken to its highest power. Rewrite each fraction, combine numerators, then factor again in case it simplifies.
Subtracting distributes. $ \frac{A}{D} - \frac{B}{D} = \frac{A - B}{D} $, and that minus applies to every term of $ B $.
State restrictions: any value making an original denominator zero is excluded, including ones that cancel away. They are invisible in the simplified form but still not allowed.

Where marks are lost: canceling terms instead of factors. In $ \frac{x + 3}{3} $ the 3s do not cancel. In $ \frac{x^2 - 9}{x - 3} $ you must factor first, giving $ \frac{(x-3)(x+3)}{x-3} = x + 3 $.

Worked examples

Example 1: simplifying with restrictions. Simplify $ \frac{x^2 - 4}{x^2 + x - 6} $.

Factor both: $ \frac{(x-2)(x+2)}{(x+3)(x-2)} $.

Cancel $ (x-2) $: the result is $ \frac{x+2}{x+3} $.

Restrictions come from the original denominator $ (x+3)(x-2) $, so $ x \neq -3 $ and $ x \neq 2 $. The 2 no longer appears anywhere in the simplified form but is still excluded, the graph has a hole there.

Example 2: dividing. Simplify $ \frac{x^2 - 1}{x + 4} \div \frac{x - 1}{x^2 - 16} $.

Flip and multiply: $ \frac{(x-1)(x+1)}{x+4} \times \frac{(x-4)(x+4)}{x-1} $.

$ (x-1) $ and $ (x+4) $ cancel, leaving $ (x+1)(x-4) $.

Example 3: subtracting. Simplify $ \frac{3}{x-2} - \frac{5}{x+1} $.

Common denominator $ (x-2)(x+1) $: $ \frac{3(x+1) - 5(x-2)}{(x-2)(x+1)} $.

The numerator is $ 3x + 3 - 5x + 10 $. Note the sign on the 10, the minus hits both terms of $ 5(x-2) $. That gives $ \frac{-2x + 13}{(x-2)(x+1)} $.

Practice · 15 problems

1–5 simplifying, 6–10 multiplying and dividing, 11–15 adding, subtracting and diagnosis.

Simplify $ \frac{6x^3}{9x} $.
Show the full solution

$ \frac{6}{9} = \frac{2}{3} $ and $ x^{3-1} = x^2 $. $ \frac{2x^2}{3} $
Simplify $ \frac{x^2 - 25}{x + 5} $.
Show the full solution

$ \frac{(x-5)(x+5)}{x+5} $. $ x - 5 $, $ x \neq -5 $
Simplify $ \frac{x^2 + 5x + 6}{x^2 - 9} $.
Show the full solution

$ \frac{(x+2)(x+3)}{(x-3)(x+3)} $. $ \frac{x+2}{x-3} $
Simplify $ \frac{2x^2 - 8}{x^2 - 4x + 4} $.
Show the full solution

$ \frac{2(x-2)(x+2)}{(x-2)^2} $. $ \frac{2(x+2)}{x-2} $
Simplify $ \frac{3 - x}{x - 3} $.
Show the full solution

Factor $ -1 $ from the top: $ \frac{-(x-3)}{x-3} $. $ -1 $
Multiply $ \frac{x}{x+1} \cdot \frac{x^2 - 1}{x^2} $.
Show the full solution

$ \frac{x(x-1)(x+1)}{(x+1)x^2} $; cancel $ x $ and $ (x+1) $. $ \frac{x-1}{x} $
Multiply $ \frac{x^2-4}{x+3} \cdot \frac{x^2+6x+9}{x-2} $.
Show the full solution

$ \frac{(x-2)(x+2)(x+3)^2}{(x+3)(x-2)} $. $ (x+2)(x+3) $
Divide $ \frac{x^2-9}{x} \div \frac{x+3}{2x} $.
Show the full solution

$ \frac{(x-3)(x+3)}{x} \cdot \frac{2x}{x+3} $. $ 2(x-3) $
Divide $ \frac{4x}{x-1} \div \frac{8x^2}{x^2-1} $.
Show the full solution

$ \frac{4x}{x-1} \cdot \frac{(x-1)(x+1)}{8x^2} = \frac{4x(x+1)}{8x^2} $. $ \frac{x+1}{2x} $
State the restrictions on $ \frac{x+1}{x^2 - 5x + 6} $.
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$ (x-2)(x-3) = 0 $. $ x \neq 2, 3 $
Add $ \frac{2}{x} + \frac{3}{x^2} $.
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Common denominator $ x^2 $: $ \frac{2x + 3}{x^2} $. $ \frac{2x+3}{x^2} $
Add $ \frac{1}{x-1} + \frac{2}{x+1} $.
Show the full solution

$ \frac{(x+1) + 2(x-1)}{(x-1)(x+1)} = \frac{3x - 1}{x^2-1} $. $ \frac{3x-1}{x^2-1} $
Subtract $ \frac{4}{x+2} - \frac{3}{x-2} $.
Show the full solution

$ \frac{4(x-2) - 3(x+2)}{(x+2)(x-2)} = \frac{4x - 8 - 3x - 6}{x^2-4} $. $ \frac{x-14}{x^2-4} $
A student simplifies $ \frac{x+4}{4} $ to $ x $. Find the error.
Show the full solution

The 4 on top is a term in a sum, not a factor, so it cannot be canceled. The expression is already simplified, or may be written $ \frac{x}{4} + 1 $. No cancellation is possible
A student simplifies $ \frac{x^2-4}{x-2} $ to $ x + 2 $ and states no restrictions. What is missing?
Show the full solution

The simplification is right, but $ x = 2 $ made the original denominator zero, so it is excluded even though nothing in $ x + 2 $ shows it. The graph has a hole there. $ x \neq 2 $

Topic 2.2 · Unit 2 · CA CCSS A-REI.2, F-IF.7d

Rational equations and their graphs

Solving a rational equation means clearing the denominators, which can manufacture solutions that do not exist. Checking is not optional here; it is part of the method.

The method

Note the restrictions before you start. Any value making a denominator zero cannot be a solution, however the algebra turns out.
Multiply every term by the least common denominator. Every fraction should vanish; if one survives, the LCD was wrong.
Solve the resulting polynomial equation as usual.
Discard any solution that hits a restriction. These are extraneous, introduced by the multiplication, not by the original equation. If every solution is extraneous, the answer is "no solution", and that is a legitimate answer.
Vertical asymptotes sit where the denominator is zero after simplifying. A factor that cancels gives a hole instead.
Horizontal asymptotes compare degrees: bottom bigger gives $ y = 0 $; equal degrees give the ratio of leading coefficients; top bigger gives no horizontal asymptote (a slant one instead).

Where marks are lost: not checking. A tidy answer of $ x = 3 $ is worth nothing if $ x - 3 $ was a denominator; that equation has no solution.

Worked examples

Example 1: a clean solve. Solve $ \frac{2}{x} + \frac{1}{3} = \frac{5}{x} $.

Restriction: $ x \neq 0 $. Multiply through by $ 3x $: $ 6 + x = 15 $.

So $ x = 9 $, which breaks no restriction. ✓

Example 2: extraneous solution. Solve $ \frac{x}{x-3} = \frac{3}{x-3} + 2 $.

Restriction: $ x \neq 3 $. Multiply by $ x - 3 $: $ x = 3 + 2(x-3) = 2x - 3 $, so $ x = 3 $.

That is exactly the restricted value. No solution. A student who skips the check writes $ x = 3 $ and loses the whole question.

Example 3: asymptotes. Describe the graph of $ f(x) = \frac{2x^2 + 3}{x^2 - 1} $.

Denominator zero at $ x = \pm 1 $; neither factor cancels, so both are vertical asymptotes.

Degrees are equal, so the horizontal asymptote is the ratio of leading coefficients, $ y = 2 $. The $ y $-intercept is $ \frac{3}{-1} = -3 $.

Practice · 15 problems

1–6 solving, 7–11 extraneous solutions, 12–15 graphs and diagnosis.

Solve $ \frac{x}{4} = \frac{3}{2} $.
Show the full solution

Cross multiply: $ 2x = 12 $. $ x = 6 $
Solve $ \frac{6}{x} = 3 $.
Show the full solution

$ 6 = 3x $. $ x = 2 $
Solve $ \frac{1}{x} + \frac{1}{2} = \frac{3}{4} $.
Show the full solution

Multiply by $ 4x $: $ 4 + 2x = 3x $. $ x = 4 $
Solve $ \frac{5}{x-1} = \frac{2}{x+2} $.
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$ 5(x+2) = 2(x-1) $, so $ 5x + 10 = 2x - 2 $ and $ 3x = -12 $. $ x = -4 $
Solve $ \frac{x}{x+1} = \frac{3}{4} $.
Show the full solution

$ 4x = 3x + 3 $. $ x = 3 $
Solve $ \frac{2}{x} + \frac{3}{x} = \frac{1}{4} $.
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$ \frac{5}{x} = \frac{1}{4} $, so $ x = 20 $. $ x = 20 $
Solve $ \frac{x}{x-2} = \frac{2}{x-2} + 1 $.
Show the full solution

Multiply by $ x-2 $: $ x = 2 + x - 2 = x $, true for all $ x $ except the restriction. So every value except 2 works. All $ x \neq 2 $
Solve $ \frac{1}{x-4} = \frac{x}{x-4} $.
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$ 1 = x $, and $ x = 1 $ breaks no restriction. $ x = 1 $
Solve $ \frac{x}{x-5} - \frac{5}{x-5} = 2 $.
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$ x - 5 = 2(x-5) $, so $ x - 5 = 2x - 10 $ and $ x = 5 $, which is restricted. No solution
Solve $ \frac{1}{x} + \frac{1}{x-1} = \frac{3}{2} $.
Show the full solution

Multiply by $ 2x(x-1) $: $ 2(x-1) + 2x = 3x(x-1) $, so $ 4x - 2 = 3x^2 - 3x $ and $ 3x^2 - 7x + 2 = 0 $. Factoring: $ (3x-1)(x-2) = 0 $. $ x = \frac{1}{3}, 2 $
Solve $ \frac{x}{x+3} = \frac{-3}{x+3} $.
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$ x = -3 $, which is the restriction. No solution
Find the vertical asymptotes of $ f(x) = \frac{x+1}{x^2-9} $.
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$ (x-3)(x+3) = 0 $, and neither cancels. $ x = 3 $ and $ x = -3 $
Find the horizontal asymptote of $ f(x) = \frac{3x+1}{x-5} $.
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Equal degrees, so take the ratio of leading coefficients. $ y = 3 $
$ f(x) = \frac{x^2-4}{x-2} $. Is there an asymptote at $ x = 2 $?
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The factor cancels, leaving $ x + 2 $, so the graph is a line with a single point missing rather than an asymptote. A hole at $ (2, 4) $
A student solves $ \frac{x}{x-3} = \frac{3}{x-3} $ and answers $ x = 3 $. Find the error.
Show the full solution

The algebra is right but the check was skipped: $ x = 3 $ makes both denominators zero, so it is extraneous, introduced by multiplying through. No solution

Topic 2.3 · Unit 2 · CA CCSS N-RN.1-2, A-REI.2

Radicals, rational exponents and radical equations

Rational exponents and radicals are the same thing written two ways, and converting to exponent form turns most "hard" radical problems into exponent-rule problems you already know.

The method

The conversion: $ a^{m/n} = \sqrt[n]{a^m} = (\sqrt[n]{a})^m $. The denominator is the root, the numerator is the power. Read it as "root on the bottom".
Simplifying a radical: pull out perfect $ n $th powers. $ \sqrt{72} = \sqrt{36 \cdot 2} = 6\sqrt{2} $.
Rationalize a denominator by multiplying by the radical (single term) or by the conjugate (two terms).
Solving a radical equation: isolate the radical, raise both sides to the matching power, solve, then check every answer.
Squaring creates extraneous solutions because it destroys sign information, $ x = -2 $ is false but $ x^2 = 4 $ is true. The check is where you catch it.
Two radicals: isolate one, square, then isolate the remaining one and square again. Squaring a sum needs the middle term.

Where marks are lost: squaring term by term. $ (\sqrt{x} + 3)^2 $ is $ x + 6\sqrt{x} + 9 $, not $ x + 9 $.

Worked examples

Example 1: rational exponents. Evaluate $ 27^{2/3} $.

The 3 on the bottom is the root: $ \sqrt[3]{27} = 3 $. The 2 on top is the power: $ 3^2 = 9 $.

Taking the root first keeps the numbers small, $ 27^2 = 729 $ then cube-rooting works but is needless arithmetic.

Example 2: solving with a check. Solve $ \sqrt{2x + 3} = x $.

Square both sides: $ 2x + 3 = x^2 $, so $ x^2 - 2x - 3 = 0 $ and $ (x-3)(x+1) = 0 $, giving $ x = 3 $ or $ x = -1 $.

Check $ x = 3 $: $ \sqrt{9} = 3 $ ✓. Check $ x = -1 $: $ \sqrt{1} = 1 \neq -1 $ ✗. A square root is never negative, so only $ x = 3 $ survives.

Example 3: isolating first. Solve $ \sqrt{x + 5} + 1 = x $.

Isolate: $ \sqrt{x+5} = x - 1 $. Square: $ x + 5 = x^2 - 2x + 1 $, so $ x^2 - 3x - 4 = 0 $ and $ (x-4)(x+1) = 0 $.

Check $ x = 4 $: $ 3 + 1 = 4 $ ✓. Check $ x = -1 $: $ 2 + 1 = 3 \neq -1 $ ✗. Squaring before isolating would have produced a mess with a radical still in it.

Practice · 15 problems

1–6 exponents and simplifying, 7–11 solving, 12–15 harder cases and diagnosis.

Write $ \sqrt[3]{x^2} $ with a rational exponent.
Show the full solution

$ x^{2/3} $
Evaluate $ 16^{3/4} $.
Show the full solution

Fourth root of 16 is 2, cubed is 8. 8
Evaluate $ 8^{-2/3} $.
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$ 8^{2/3} = 4 $, and the negative exponent reciprocates. $ \frac{1}{4} $
Simplify $ \sqrt{50} $.
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$ \sqrt{25 \cdot 2} $. $ 5\sqrt{2} $
Simplify $ \sqrt{12x^3} $.
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$ \sqrt{4x^2 \cdot 3x} $. $ 2x\sqrt{3x} $
Rationalize $ \frac{6}{\sqrt{3}} $.
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Multiply top and bottom by $ \sqrt{3} $: $ \frac{6\sqrt{3}}{3} $. $ 2\sqrt{3} $
Solve $ \sqrt{x} = 7 $.
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Square both sides; $ \sqrt{49} = 7 $ ✓. $ x = 49 $
Solve $ \sqrt{x - 3} = 4 $.
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$ x - 3 = 16 $; check $ \sqrt{16} = 4 $ ✓. $ x = 19 $
Solve $ \sqrt{3x + 1} = 5 $.
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$ 3x + 1 = 25 $, so $ x = 8 $; check $ \sqrt{25} = 5 $ ✓. $ x = 8 $
Solve $ \sqrt[3]{x + 1} = 2 $.
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Cube both sides: $ x + 1 = 8 $. Cube roots create no extraneous solutions, since cubing preserves sign. $ x = 7 $
Solve $ \sqrt{x + 7} = x - 5 $.
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$ x + 7 = x^2 - 10x + 25 $, so $ x^2 - 11x + 18 = 0 $ and $ (x-9)(x-2) = 0 $. Check $ x = 9 $: $ 4 = 4 $ ✓. Check $ x = 2 $: $ 3 \neq -3 $ ✗. $ x = 9 $
Solve $ \sqrt{2x + 5} - 1 = x $.
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Isolate: $ \sqrt{2x+5} = x + 1 $. Square: $ 2x + 5 = x^2 + 2x + 1 $, so $ x^2 = 4 $ and $ x = \pm 2 $. Check $ x = 2 $: $ 3 - 1 = 2 $ ✓. Check $ x = -2 $: $ 1 - 1 = 0 \neq -2 $ ✗. $ x = 2 $
Solve $ x^{2/3} = 9 $.
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Raise both sides to the $ \frac{3}{2} $: $ x = 9^{3/2} = 27 $. Since the exponent has an even numerator once cleared, $ x = -27 $ also satisfies it: $ (-27)^{2/3} = 9 $. $ x = \pm 27 $
A student expands $ (\sqrt{x} + 3)^2 $ as $ x + 9 $. Find the error.
Show the full solution

Squaring a binomial needs the middle term $ 2(\sqrt{x})(3) $. $ x + 6\sqrt{x} + 9 $
A student solves $ \sqrt{x+6} = x $ and answers $ x = 3 $ and $ x = -2 $. Find the error.
Show the full solution

The factoring is right but the check was skipped. $ x = 3 $: $ \sqrt{9} = 3 $ ✓. $ x = -2 $: $ \sqrt{4} = 2 \neq -2 $ ✗, a principal square root is never negative. $ x = 3 $ only

Unit 2 mixed review · 10 problems · all topics

Unit 2 mixed review: Rational and Radical Functions

Both halves of this unit manufacture extraneous solutions. Every solving problem here needs a check, and the check is part of the answer.

Simplify $ \frac{x^2 - 16}{x^2 + 8x + 16} $.
Show the full solution

$ \frac{(x-4)(x+4)}{(x+4)^2} $. $ \frac{x-4}{x+4} $
Multiply $ \frac{x^2-1}{x^2} \cdot \frac{x}{x-1} $.
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$ \frac{(x-1)(x+1)x}{x^2(x-1)} $. $ \frac{x+1}{x} $
Subtract $ \frac{5}{x-1} - \frac{2}{x+3} $.
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$ \frac{5(x+3) - 2(x-1)}{(x-1)(x+3)} = \frac{5x + 15 - 2x + 2}{(x-1)(x+3)} $. $ \frac{3x+17}{(x-1)(x+3)} $
Solve $ \frac{3}{x} + \frac{1}{2} = \frac{5}{x} $.
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Multiply by $ 2x $: $ 6 + x = 10 $. $ x = 4 $
Solve $ \frac{x}{x-4} = \frac{4}{x-4} + 3 $.
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$ x = 4 + 3(x-4) = 3x - 8 $, so $ x = 4 $, the restricted value. No solution
Find the vertical and horizontal asymptotes of $ f(x) = \frac{2x}{x-3} $.
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Denominator zero at 3; equal degrees give $ y = 2 $. $ x = 3 $, $ y = 2 $
Evaluate $ 81^{3/4} $.
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Fourth root 3, cubed. 27
Simplify $ \sqrt{18x^5} $.
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$ \sqrt{9x^4 \cdot 2x} $. $ 3x^2\sqrt{2x} $
Solve $ \sqrt{x + 9} = x - 3 $.
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$ x + 9 = x^2 - 6x + 9 $, so $ x^2 - 7x = 0 $ and $ x = 0 $ or $ 7 $. Check $ x = 7 $: $ 4 = 4 $ ✓. Check $ x = 0 $: $ 3 \neq -3 $ ✗. $ x = 7 $
A student cancels the $ x $ in $ \frac{x + 5}{x} $ to get 5. Find the error.
Show the full solution

Cancellation removes matching factors, and the $ x $ on top is a term in a sum. Split it instead: $ \frac{x}{x} + \frac{5}{x} $. $ 1 + \frac{5}{x} $

Topic 3.1 · Unit 3 · CA CCSS F-LE.1–2, F-IF.7e

Exponential growth and decay

Every exponential model in this course is $ y = a \cdot b^{\,x} $: $ a $ is where you start, $ b $ is what you multiply by each period. Most lost marks come from building $ b $ wrongly out of a percentage.

The method

$ a $ is the initial amount: the value when $ x = 0 $, since $ b^0 = 1 $.
$ b $ is the multiplier per period. Build it from a percentage: growth of $ r\% $ gives $ b = 1 + \frac{r}{100} $; decay of $ r\% $ gives $ b = 1 - \frac{r}{100} $. A 7% rise means $ b = 1.07 $, not 0.07; you keep the original amount and add 7%.
Read $ b $ backwards too. $ b = 1.25 $ is 25% growth; $ b = 0.85 $ is 15% decay, because you keep 85%. $ b > 1 $ grows, $ 0 < b < 1 $ decays.
Watch the period. If $ x $ is in years but the rate is monthly, either convert the rate or redefine $ x $. Mismatching these is a silent error the numbers will not flag.
The graph has a horizontal asymptote at $ y = 0 $; it approaches the axis without touching. Growth rises steeply to the right; decay falls toward the axis.

Where marks are lost: writing $ y = 500(0.03)^x $ for "3% growth". That model multiplies by 0.03 each year, destroying 97% of the value annually. The correct multiplier is 1.03. Sanity-check by computing one period by hand.

Worked examples

Example 1: building the model from a percentage. A $2000 investment grows 6% per year. Write the model and find its value after 10 years.

$ a = 2000 $ and $ b = 1 + 0.06 = 1.06 $, so $ V = 2000(1.06)^t $.

$ V(10) = 2000(1.06)^{10} = 2000 \times 1.7908 \approx \$3581.70 $. Check the direction: the money grew, as 6% growth requires.

Example 2: decay. A car worth $24 000 loses 15% of its value each year. Write the model and find its value after 4 years.

Losing 15% means keeping 85%, so $ b = 0.85 $ and $ V = 24000(0.85)^t $.

$ V(4) = 24000(0.85)^4 = 24000 \times 0.52200625 \approx \$12528 $. Note this is not $ 24000 - 4(15\%) $, each year's loss is taken from a smaller amount, which is exactly what makes it exponential rather than linear.

Example 3: reading a model backwards. A population is modeled by $ P = 4500(0.92)^t $. Describe what is happening.

$ a = 4500 $ is the starting population. $ b = 0.92 $ is less than 1, so this is decay, and $ 1 - 0.92 = 0.08 $ means it falls by 8% per year. Each year 92% of the population remains.

Practice · 15 problems

1–6 build or read models, 7–11 evaluate them, 12–15 apply and diagnose.

Write the multiplier $ b $ for 5% growth.
Show the full solution

$ 1 + 0.05 $. $ b = 1.05 $
Write the multiplier $ b $ for 20% decay.
Show the full solution

Losing 20% means keeping 80%. $ b = 0.80 $
Does $ y = 300(1.12)^x $ grow or decay, and at what rate?
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$ b > 1 $, so growth, at $ 1.12 - 1 = 0.12 $. Growth, 12% per period
Does $ y = 80(0.75)^x $ grow or decay, and at what rate?
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$ b < 1 $, so decay. Keeping 75% means losing 25%. Decay, 25% per period
Write a model for a $500 deposit growing 4% a year.
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$ V = 500(1.04)^t $
Write a model for a 900 mg dose decaying 30% each hour.
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Keeping 70%. $ A = 900(0.70)^h $
For $ y = 200(1.5)^x $, find $ y $ when $ x = 3 $.
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$ 1.5^3 = 3.375 $, so $ 200 \times 3.375 = 675 $. 675
For $ y = 640(0.5)^x $, find $ y $ when $ x = 4 $.
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$ 0.5^4 = 0.0625 $, so $ 640 \times 0.0625 = 40 $. Halving four times: 320, 160, 80, 40 ✓. 40
For $ y = 50(1.2)^x $, find the initial value and the value after 2 periods.
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Initial is 50 at $ x = 0 $. $ 1.2^2 = 1.44 $, so $ 50 \times 1.44 = 72 $. 50 and 72
A town of 8000 grows 3% a year. Find the population after 5 years.
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$ 8000(1.03)^5 = 8000 \times 1.159274 \approx 9274 $. about 9274
A $1200 laptop loses 25% of its value yearly. What is it worth after 3 years?
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$ 1200(0.75)^3 = 1200 \times 0.421875 = 506.25 $. $506.25
A culture triples every hour from 40 cells. Write a model and find the count after 4 hours.
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Tripling means $ b = 3 $: $ N = 40(3)^h $. $ 40 \times 81 = 3240 $. $ N = 40(3)^h $; 3240 cells
An investment doubles every 7 years. Write a model in terms of years $ t $.
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The doubling happens once per 7 years, so the exponent must count seven-year blocks: $ V = a \cdot 2^{\,t/7} $. Writing $ 2^t $ would double every year instead. $ V = a \cdot 2^{\,t/7} $
A student models 3% annual growth on $500 as $ y = 500(0.03)^t $. Find the error.
Show the full solution

Their multiplier destroys 97% of the value each year, after one year it gives $15. Growth of 3% keeps the original and adds 3%, so $ b = 1.03 $: $ y = 500(1.03)^t $, giving $515 after one year. $ y = 500(1.03)^t $
A student says a car losing 15% a year for 4 years has lost 60% of its value. Is that right?
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No; that treats it as linear. Each year's loss comes off a smaller amount: $ 0.85^4 \approx 0.522 $, so about 52% of the value remains and roughly 48% has been lost, not 60%. About 48% lost

Topic 3.2 · Unit 3 · CA CCSS F-BF.5, F-LE.4

Logarithms and their properties

A logarithm answers one question: what exponent do I need? Everything else in this topic is a consequence of that, including the three properties that let you take an unknown out of an exponent.

The method

The definition: $ \log_b a = c $ means exactly $ b^c = a $. Read it aloud as "the power that turns $ b $ into $ a $". Converting between the two forms is the single most useful move in this unit.
Notation: $ \log x $ with no base means base 10. $ \ln x $ means base $ e $, where $ e \approx 2.718 $.
Product: $ \log_b(MN) = \log_b M + \log_b N $. Multiplication inside becomes addition outside, because exponents add.
Quotient: $ \log_b\!\left(\frac{M}{N}\right) = \log_b M - \log_b N $.
Power: $ \log_b(M^p) = p\log_b M $. This is the one that solves equations; it brings a variable exponent down to the front.
Change of base: $ \log_b a = \frac{\log a}{\log b} $, which is how you evaluate an unusual base on a calculator.
Domain: you can only take the log of a positive number. $ \log_b 1 = 0 $ and $ \log_b b = 1 $ for every valid base.

Where marks are lost: inventing properties. $ \log(M + N) $ does not equal $ \log M + \log N $, the rules convert multiplication to addition, not addition to addition. And $ \frac{\log M}{\log N} $ is not $ \log M - \log N $.

Worked examples

Example 1: evaluating from the definition. Find $ \log_2 32 $.

Ask: 2 to what power is 32? Since $ 2^5 = 32 $, the answer is 5.

Same question for $ \log_9 3 $: $ 9^{1/2} = 3 $, so the answer is $ \frac{1}{2} $. Logs of numbers smaller than the base are fractions.

Example 2: expanding. Expand $ \log\!\left(\frac{x^3 y}{z^2}\right) $.

Quotient first: $ \log(x^3 y) - \log(z^2) $.

Then product and power: $ 3\log x + \log y - 2\log z $.

Example 3: condensing. Write $ 2\log x - 3\log y $ as a single logarithm.

Move the coefficients inside as powers: $ \log x^2 - \log y^3 $.

Subtraction becomes a quotient: $ \log\!\left(\frac{x^2}{y^3}\right) $. Condensing is expanding run backwards, and exam questions ask for both directions.

Practice · 15 problems

1–6 evaluating and converting, 7–11 expanding and condensing, 12–15 applications and diagnosis.

Evaluate $ \log_3 81 $.
Show the full solution

$ 3^4 = 81 $. 4
Evaluate $ \log_5 1 $.
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Any base to the power 0 is 1. 0
Evaluate $ \log_2 \frac{1}{8} $.
Show the full solution

$ 2^{-3} = \frac{1}{8} $. $ -3 $
Evaluate $ \log_{16} 4 $.
Show the full solution

$ 16^{1/2} = 4 $. $ \frac{1}{2} $
Write $ 4^3 = 64 $ in logarithmic form.
Show the full solution

The base stays the base, the exponent becomes the answer. $ \log_4 64 = 3 $
Write $ \log_7 49 = 2 $ in exponential form.
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$ 7^2 = 49 $
Expand $ \log(xy) $.
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$ \log x + \log y $
Expand $ \log\!\left(\frac{x^4}{y}\right) $.
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$ 4\log x - \log y $
Expand $ \ln\!\left(x^2\sqrt{y}\right) $.
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The root is a $ \frac{1}{2} $ power. $ 2\ln x + \frac{1}{2}\ln y $
Condense $ \log 5 + \log 4 $.
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$ \log 20 $, which is about 1.301. $ \log 20 $
Condense $ 3\log x + \frac{1}{2}\log y $.
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$ \log\!\left(x^3\sqrt{y}\right) $
Use change of base to evaluate $ \log_3 20 $ to three decimals.
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$ \frac{\log 20}{\log 3} = \frac{1.30103}{0.47712} \approx 2.727 $. about 2.727
State the domain of $ f(x) = \log(x - 4) $.
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The argument must be positive: $ x - 4 \gt 0 $. $ x \gt 4 $
A student writes $ \log(x + y) = \log x + \log y $. Find the error.
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The product rule converts multiplication inside to addition outside. Addition inside does not simplify at all, $ \log(x+y) $ must be left as it stands, while $ \log(xy) = \log x + \log y $. No such property
A student evaluates $ \log_2 0 $ as 0. Explain why there is no answer.
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$ \log_2 0 $ asks what power of 2 gives 0. No exponent does, $ 2^x $ approaches 0 but never reaches it, which is why $ x = 0 $ is a vertical asymptote of the log graph. Undefined

Topic 3.3 · Unit 3 · CA CCSS F-LE.4, A-REI.11

Solving exponential and logarithmic equations

Every problem in this topic comes down to one decision: can I make the bases match, or do I need to take a log? Choosing correctly turns a hard-looking equation into two lines of work.

The method

Same-base method. If both sides can be written with the same base, the exponents must be equal. $ 2^{x+1} = 8 $ becomes $ 2^{x+1} = 2^3 $, so $ x + 1 = 3 $.
Take a log when the bases will not match. Apply $ \log $ or $ \ln $ to both sides, use the power rule to bring the exponent down, then divide.
Isolate the exponential first. In $ 3 \cdot 2^x + 5 = 29 $, subtract and divide before touching the exponent.
Log equations: condense to a single log, convert to exponential form, solve.
Equal logs, equal arguments: if $ \log_b M = \log_b N $ then $ M = N $.
Always check log solutions. A value that makes any argument zero or negative is extraneous, no matter how the algebra went.

Where marks are lost: canceling the log. $ \frac{\log 20}{\log 3} $ is not $ \log\frac{20}{3} $; it is a single number, about 2.727. Evaluate it, do not simplify it.

Worked examples

Example 1: same base. Solve $ 9^{x} = 27^{x-1} $.

Write both as powers of 3: $ 3^{2x} = 3^{3(x-1)} $.

Equate exponents: $ 2x = 3x - 3 $, so $ x = 3 $.

Example 2: taking a log. Solve $ 5^x = 40 $.

The bases cannot match, so take logs of both sides: $ x\log 5 = \log 40 $.

$ x = \frac{\log 40}{\log 5} = \frac{1.60206}{0.69897} \approx 2.292 $. Check for plausibility: $ 5^2 = 25 $ and $ 5^3 = 125 $, so an answer between 2 and 3 is right.

Example 3: a log equation with an extraneous solution. Solve $ \log(x) + \log(x - 3) = 1 $.

Condense: $ \log(x(x-3)) = 1 $, which in exponential form is $ x^2 - 3x = 10^1 = 10 $.

So $ x^2 - 3x - 10 = 0 $ and $ (x-5)(x+2) = 0 $. Check $ x = 5 $: $ \log 5 + \log 2 = \log 10 = 1 $ ✓. Check $ x = -2 $: $ \log(-2) $ is undefined ✗.

The answer is $ x = 5 $ only.

Practice · 15 problems

1–5 same base, 6–10 taking logs, 11–15 log equations and applications.

Solve $ 2^x = 16 $.
Show the full solution

$ 2^4 = 16 $. $ x = 4 $
Solve $ 3^{x-1} = 81 $.
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$ 81 = 3^4 $, so $ x - 1 = 4 $. $ x = 5 $
Solve $ 4^x = 8 $.
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Base 2: $ 2^{2x} = 2^3 $, so $ 2x = 3 $. $ x = \frac{3}{2} $
Solve $ 5^{2x} = 125 $.
Show the full solution

$ 2x = 3 $. $ x = \frac{3}{2} $
Solve $ 2^{x} = \frac{1}{32} $.
Show the full solution

$ \frac{1}{32} = 2^{-5} $. $ x = -5 $
Solve $ 3^x = 20 $ to three decimals.
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$ x = \frac{\log 20}{\log 3} \approx 2.727 $. about 2.727
Solve $ 2^{x+1} = 7 $ to three decimals.
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$ x + 1 = \frac{\log 7}{\log 2} \approx 2.807 $, so $ x \approx 1.807 $. about 1.807
Solve $ 4 \cdot 3^x = 36 $.
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Isolate: $ 3^x = 9 $. $ x = 2 $
Solve $ e^x = 12 $ to three decimals.
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Take $ \ln $ of both sides: $ x = \ln 12 \approx 2.485 $. about 2.485
Solve $ 2 \cdot 5^x - 3 = 47 $.
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$ 5^x = 25 $. $ x = 2 $
Solve $ \log_2 x = 5 $.
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$ x = 2^5 $. $ x = 32 $
Solve $ \log(x + 3) = 2 $.
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$ x + 3 = 10^2 $. $ x = 97 $
Solve $ \log_3 x + \log_3 4 = 2 $.
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Condense: $ \log_3 4x = 2 $, so $ 4x = 9 $. $ x = \frac{9}{4} $
An investment of \$2000 grows at 6% compounded annually. How long until it reaches \$5000?
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$ 2000(1.06)^t = 5000 $, so $ 1.06^t = 2.5 $ and $ t = \frac{\log 2.5}{\log 1.06} = \frac{0.39794}{0.02531} \approx 15.7 $. about 15.7 years
A student solves $ \log x + \log(x-3) = 1 $ and answers $ x = 5 $ and $ x = -2 $. Find the error.
Show the full solution

The quadratic is right but the check was skipped. $ x = -2 $ makes $ \log x $ undefined, since the argument of a logarithm must be positive. $ x = 5 $ only

Unit 3 mixed review · 10 problems · all topics

Unit 3 mixed review: Exponential and Logarithmic Functions

Decide first whether the bases can be matched. If they can, no logarithm is needed; if they cannot, take a log immediately rather than guessing.

Evaluate $ \log_4 64 $.
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$ 4^3 = 64 $. 3
Evaluate $ \log_5 \frac{1}{25} $.
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$ 5^{-2} $. $ -2 $
Expand $ \log\!\left(\frac{x^2 y^3}{z}\right) $.
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$ 2\log x + 3\log y - \log z $
Condense $ \frac{1}{2}\log x + 2\log y $.
Show the full solution

$ \log\!\left(y^2\sqrt{x}\right) $
Solve $ 2^{3x} = 32 $.
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$ 32 = 2^5 $, so $ 3x = 5 $. $ x = \frac{5}{3} $
Solve $ 7^x = 30 $ to three decimals.
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$ x = \frac{\log 30}{\log 7} = \frac{1.47712}{0.84510} \approx 1.748 $. about 1.748
Solve $ \log_2(x - 1) = 4 $.
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$ x - 1 = 16 $. $ x = 17 $
Solve $ \log x + \log 4 = 2 $.
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$ \log 4x = 2 $, so $ 4x = 100 $. $ x = 25 $
A population of 800 grows 4% per year. How long until it doubles?
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$ 1.04^t = 2 $, so $ t = \frac{\log 2}{\log 1.04} = \frac{0.30103}{0.01703} \approx 17.7 $. about 17.7 years
A student writes $ \frac{\log 50}{\log 2} = \log 25 $. Find the error.
Show the full solution

The quotient rule turns a log of a quotient into a difference; it says nothing about a quotient of two logs. This expression is just a number: $ \frac{1.69897}{0.30103} \approx 5.644 $, while $ \log 25 \approx 1.398 $. No such property

Topic 4.1 · Unit 4 · CA CCSS F-TF.1-4

Angles, radians and the unit circle

Trigonometry stops being about triangles here and becomes about a point traveling around a circle. That shift is what lets the functions accept any angle, including negative ones and ones larger than a full turn.

The method

Radians: a full turn is $ 2\pi $ radians, so $ 180^\circ = \pi $. Convert by multiplying by $ \frac{\pi}{180} $ or $ \frac{180}{\pi} $, pick whichever cancels the unit you have.
The unit circle has radius 1 centered at the origin. For an angle $ \theta $ measured counterclockwise from the positive $ x $-axis, the point on the circle is $ (\cos\theta, \sin\theta) $. So cosine is the $ x $-coordinate and sine is the $ y $-coordinate; that one sentence replaces most memorization.
Reference angle: the acute angle to the nearest part of the $ x $-axis. Evaluate using the reference angle, then attach the sign from the quadrant.
Signs by quadrant: all positive in I; only sine in II; only tangent in III; only cosine in IV.
The special values are worth knowing cold: $ \sin\frac{\pi}{6} = \frac{1}{2} $, $ \sin\frac{\pi}{4} = \frac{\sqrt2}{2} $, $ \sin\frac{\pi}{3} = \frac{\sqrt3}{2} $, with cosine running the same list backwards.
Coterminal angles differ by whole turns. Add or subtract $ 2\pi $ (or $ 360^\circ $) until the angle lands in one turn, then work as usual.
Tangent is $ \frac{\sin\theta}{\cos\theta} $, undefined wherever cosine is zero.

Where marks are lost: leaving the calculator in the wrong mode. $ \sin(\frac{\pi}{6}) $ in degree mode returns about 0.0091 instead of 0.5. If an answer is wildly small, check the mode before checking the algebra.

Worked examples

Example 1: converting. Convert $ 135^\circ $ to radians and $ \frac{5\pi}{6} $ to degrees.

$ 135 \times \frac{\pi}{180} = \frac{135\pi}{180} = \frac{3\pi}{4} $.

$ \frac{5\pi}{6} \times \frac{180}{\pi} = \frac{5 \times 180}{6} = 150^\circ $.

Example 2: reference angles and signs. Evaluate $ \cos\frac{4\pi}{3} $.

$ \frac{4\pi}{3} $ is in quadrant III, past $ \pi $ by $ \frac{\pi}{3} $, so the reference angle is $ \frac{\pi}{3} $.

$ \cos\frac{\pi}{3} = \frac{1}{2} $, and cosine is negative in quadrant III. So $ \cos\frac{4\pi}{3} = -\frac{1}{2} $.

Example 3: coterminal. Evaluate $ \sin\frac{13\pi}{6} $.

Subtract a full turn: $ \frac{13\pi}{6} - \frac{12\pi}{6} = \frac{\pi}{6} $.

So $ \sin\frac{13\pi}{6} = \sin\frac{\pi}{6} = \frac{1}{2} $. Any whole number of turns lands on the same point of the circle, which is exactly why these functions repeat.

Practice · 15 problems

1–5 converting, 6–11 evaluating, 12–15 harder cases and diagnosis.

Convert $ 60^\circ $ to radians.
Show the full solution

$ 60 \cdot \frac{\pi}{180} $. $ \frac{\pi}{3} $
Convert $ 270^\circ $ to radians.
Show the full solution

$ \frac{270\pi}{180} $. $ \frac{3\pi}{2} $
Convert $ \frac{\pi}{4} $ to degrees.
Show the full solution

$ \frac{180}{4} $. $ 45^\circ $
Convert $ \frac{7\pi}{6} $ to degrees.
Show the full solution

$ \frac{7 \times 180}{6} $. $ 210^\circ $
Find an angle coterminal with $ \frac{9\pi}{4} $ between 0 and $ 2\pi $.
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$ \frac{9\pi}{4} - \frac{8\pi}{4} $. $ \frac{\pi}{4} $
Evaluate $ \sin\frac{\pi}{2} $.
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The point is $ (0, 1) $, and sine is the $ y $-coordinate. 1
Evaluate $ \cos\pi $.
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The point is $ (-1, 0) $. $ -1 $
Evaluate $ \sin\frac{5\pi}{6} $.
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Quadrant II, reference $ \frac{\pi}{6} $, sine positive there. $ \frac{1}{2} $
Evaluate $ \cos\frac{3\pi}{4} $.
Show the full solution

Quadrant II, reference $ \frac{\pi}{4} $, cosine negative there. $ -\frac{\sqrt2}{2} $
Evaluate $ \tan\frac{\pi}{4} $.
Show the full solution

$ \frac{\sqrt2/2}{\sqrt2/2} $. 1
Evaluate $ \sin\frac{7\pi}{6} $.
Show the full solution

Quadrant III, reference $ \frac{\pi}{6} $, sine negative there. $ -\frac{1}{2} $
Evaluate $ \tan\frac{\pi}{2} $.
Show the full solution

Cosine is 0 there, so the quotient has a zero denominator. Undefined
In which quadrant is $ \sin\theta < 0 $ and $ \cos\theta > 0 $?
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Negative $ y $, positive $ x $. Quadrant IV
$ \cos\theta = \frac{3}{5} $ with $ \theta $ in quadrant IV. Find $ \sin\theta $.
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$ \sin^2\theta = 1 - \frac{9}{25} = \frac{16}{25} $, so $ \sin\theta = \pm\frac{4}{5} $; quadrant IV makes it negative. $ -\frac{4}{5} $
A student evaluates $ \sin\frac{4\pi}{3} $ as $ +\frac{\sqrt3}{2} $. Find the error.
Show the full solution

The reference angle $ \frac{\pi}{3} $ is right, but $ \frac{4\pi}{3} $ is in quadrant III, where sine is negative. $ -\frac{\sqrt3}{2} $

Topic 4.2 · Unit 4 · CA CCSS F-IF.7e, F-TF.5

Graphing sine and cosine

Every sine and cosine graph is the same wave with four numbers applied to it. Read those four off the equation and the graph follows; you never need to plot points.

The method

The general form is $ y = a\sin\big(b(x - h)\big) + k $.
Amplitude is $ |a| $, half the distance from the maximum to the minimum. A negative $ a $ flips the wave vertically but does not change the amplitude.
Period is $ \frac{2\pi}{|b|} $. A larger $ b $ squeezes the wave into a shorter cycle.
Phase shift is $ h $, horizontally. Vertical shift is $ k $, which is also the midline $ y = k $.
Factor before reading $ h $. In $ y = \sin(2x - \pi) $ the shift is not $ \pi $; rewrite as $ \sin\big(2(x - \frac{\pi}{2})\big) $ to see it is $ \frac{\pi}{2} $.
Max and min: $ k + |a| $ and $ k - |a| $.
To sketch: draw the midline, mark one period on the axis, divide it into quarters, and place the five key points, sine starts at the midline going up, cosine starts at the maximum.

Where marks are lost: reading the phase shift straight off an unfactored equation. $ b $ multiplies the whole bracket, so the shift is always $ \frac{c}{b} $, not $ c $.

Worked examples

Example 1: reading all four. Describe $ y = 3\sin(2x) - 1 $.

Amplitude 3. Period $ \frac{2\pi}{2} = \pi $. No phase shift. Midline $ y = -1 $.

Maximum $ -1 + 3 = 2 $, minimum $ -1 - 3 = -4 $. The wave completes a full cycle in $ \pi $ rather than $ 2\pi $.

Example 2: a phase shift hidden by a factor. Find the phase shift of $ y = \cos(3x + \pi) $.

Factor the 3 out of the bracket: $ \cos\big(3(x + \frac{\pi}{3})\big) $.

So the shift is $ \frac{\pi}{3} $ to the left, not $ \pi $. The period is $ \frac{2\pi}{3} $.

Example 3: writing an equation from a description. A cosine wave has maximum 9, minimum 1, and period 4.

Midline is the average: $ k = \frac{9+1}{2} = 5 $. Amplitude is half the range: $ a = \frac{9-1}{2} = 4 $.

Period 4 gives $ \frac{2\pi}{b} = 4 $, so $ b = \frac{\pi}{2} $.

$ y = 4\cos\!\left(\frac{\pi}{2}x\right) + 5 $.

Practice · 15 problems

1–6 reading the four numbers, 7–11 harder forms, 12–15 writing equations and diagnosis.

State the amplitude of $ y = 5\sin x $.
Show the full solution

5
State the amplitude of $ y = -2\cos x $.
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Amplitude is $ |a| $; the sign only reflects the graph. 2
State the period of $ y = \sin(4x) $.
Show the full solution

$ \frac{2\pi}{4} $. $ \frac{\pi}{2} $
State the period of $ y = \cos\!\left(\frac{x}{3}\right) $.
Show the full solution

$ b = \frac{1}{3} $, so the period is $ 2\pi \div \frac{1}{3} $. $ 6\pi $
State the midline of $ y = \sin x + 4 $.
Show the full solution

$ y = 4 $
Find the maximum of $ y = 3\cos x - 2 $.
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$ k + |a| = -2 + 3 $. 1
Find the minimum of $ y = 6\sin x + 1 $.
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$ 1 - 6 $. $ -5 $
State the amplitude, period and midline of $ y = 2\sin(3x) + 5 $.
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$ |a| = 2 $; period $ \frac{2\pi}{3} $; midline $ y = 5 $. 2, $ \frac{2\pi}{3} $, $ y = 5 $
Find the phase shift of $ y = \sin\!\left(x - \frac{\pi}{4}\right) $.
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$ \frac{\pi}{4} $ right
Find the phase shift of $ y = \cos(2x - \pi) $.
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Factor: $ \cos\big(2(x - \frac{\pi}{2})\big) $. $ \frac{\pi}{2} $ right
Fully describe $ y = -4\cos\!\left(\frac{1}{2}x\right) + 3 $.
Show the full solution

Amplitude 4, reflected vertically; period $ 4\pi $; midline $ y = 3 $; maximum 7, minimum $ -1 $. Because of the reflection it starts at its minimum rather than its maximum. See description
Write a sine equation with amplitude 3, period $ \pi $ and midline $ y = -2 $.
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$ b = \frac{2\pi}{\pi} = 2 $. $ y = 3\sin(2x) - 2 $
A wave has maximum 12, minimum 4 and period 6. Write a cosine equation.
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Midline 8, amplitude 4, $ b = \frac{2\pi}{6} = \frac{\pi}{3} $. $ y = 4\cos\!\left(\frac{\pi}{3}x\right) + 8 $
A Ferris wheel has radius 20 m, center 25 m above ground, and one turn takes 40 s. Write a height equation starting at the bottom.
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Amplitude 20, midline 25, $ b = \frac{2\pi}{40} = \frac{\pi}{20} $. Starting at the bottom means a negative cosine. $ h = -20\cos\!\left(\frac{\pi}{20}t\right) + 25 $
A student says $ y = \sin(3x - \pi) $ is shifted $ \pi $ units right. Find the error.
Show the full solution

The coefficient must be factored out of the bracket first: $ \sin\big(3(x - \frac{\pi}{3})\big) $. The shift is $ \frac{c}{b} $. $ \frac{\pi}{3} $ right

Topic 4.3 · Unit 4 · CA CCSS F-TF.8

Identities and solving trigonometric equations

A trigonometric equation has infinitely many solutions, so the question always tells you which interval to report. Finding the first solution is usually easy; finding all of them in the interval is where the marks are.

The method

Pythagorean identity: $ \sin^2\theta + \cos^2\theta = 1 $. Its rearrangements, $ \sin^2\theta = 1 - \cos^2\theta $ and the reverse, are how you convert an equation into a single function.
Quotient identity: $ \tan\theta = \frac{\sin\theta}{\cos\theta} $.
To solve: isolate the trig function, find the reference angle from the absolute value, then place solutions in every quadrant where the sign matches.
Sine is positive in quadrants I and II, cosine in I and IV, tangent in I and III. Each equation therefore usually has two solutions per turn.
Mixed functions: use an identity to get everything in terms of one function, then factor or use the quadratic formula.
Factor, do not divide. Dividing $ \sin\theta\cos\theta = \sin\theta $ by $ \sin\theta $ deletes every solution where $ \sin\theta = 0 $.
A multiple angle needs a wider interval. For $ \sin(2\theta) $ on $ [0, 2\pi) $, solve $ 2\theta $ on $ [0, 4\pi) $ and then halve, which is why these have four solutions, not two.

Where marks are lost: reporting only the calculator's answer. $ \sin\theta = 0.5 $ gives $ \frac{\pi}{6} $ from the inverse, but $ \frac{5\pi}{6} $ is equally valid and the inverse function will never show it to you.

Worked examples

Example 1: all solutions in an interval. Solve $ 2\cos\theta = 1 $ on $ [0, 2\pi) $.

$ \cos\theta = \frac{1}{2} $, so the reference angle is $ \frac{\pi}{3} $.

Cosine is positive in quadrants I and IV, giving $ \theta = \frac{\pi}{3} $ and $ \theta = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3} $.

Example 2: using an identity. Solve $ 2\sin^2\theta + \cos\theta = 2 $ on $ [0, 2\pi) $.

Replace $ \sin^2\theta $ with $ 1 - \cos^2\theta $: $ 2 - 2\cos^2\theta + \cos\theta = 2 $, so $ -2\cos^2\theta + \cos\theta = 0 $.

Factor: $ \cos\theta(1 - 2\cos\theta) = 0 $. So $ \cos\theta = 0 $, giving $ \frac{\pi}{2} $ and $ \frac{3\pi}{2} $; or $ \cos\theta = \frac{1}{2} $, giving $ \frac{\pi}{3} $ and $ \frac{5\pi}{3} $.

Four solutions. Dividing by $ \cos\theta $ at the factoring step would have lost the first two.

Example 3: verifying an identity. Show that $ \frac{\sin^2\theta}{1 - \cos\theta} = 1 + \cos\theta $.

Work on the left only. Replace $ \sin^2\theta $ with $ 1 - \cos^2\theta $, which is a difference of squares: $ (1-\cos\theta)(1+\cos\theta) $.

$ \frac{(1-\cos\theta)(1+\cos\theta)}{1-\cos\theta} = 1 + \cos\theta $. ✓ Verifying means transforming one side into the other, never moving terms across the equals sign.

Practice · 15 problems

All solving is on $ [0, 2\pi) $ unless stated. 1–5 identities, 6–11 basic equations, 12–15 harder equations and diagnosis.

Simplify $ 1 - \cos^2\theta $.
Show the full solution

$ \sin^2\theta $
Simplify $ \frac{\sin\theta}{\cos\theta} $.
Show the full solution

$ \tan\theta $
Given $ \sin\theta = \frac{3}{5} $ in quadrant I, find $ \cos\theta $.
Show the full solution

$ \cos^2\theta = 1 - \frac{9}{25} $, positive in quadrant I. $ \frac{4}{5} $
Simplify $ \sin^2\theta + \cos^2\theta + \tan^2\theta $ when $ \tan\theta = 1 $.
Show the full solution

The first two sum to 1, plus $ 1^2 $. 2
Verify $ \cos\theta\tan\theta = \sin\theta $.
Show the full solution

$ \cos\theta \cdot \frac{\sin\theta}{\cos\theta} $; the cosines cancel. Verified
Solve $ \sin\theta = \frac{1}{2} $.
Show the full solution

Reference $ \frac{\pi}{6} $; sine positive in I and II. $ \frac{\pi}{6}, \frac{5\pi}{6} $
Solve $ \cos\theta = -\frac{\sqrt2}{2} $.
Show the full solution

Reference $ \frac{\pi}{4} $; cosine negative in II and III. $ \frac{3\pi}{4}, \frac{5\pi}{4} $
Solve $ \tan\theta = 1 $.
Show the full solution

Reference $ \frac{\pi}{4} $; tangent positive in I and III. $ \frac{\pi}{4}, \frac{5\pi}{4} $
Solve $ 2\sin\theta + 1 = 0 $.
Show the full solution

$ \sin\theta = -\frac{1}{2} $; reference $ \frac{\pi}{6} $, negative in III and IV. $ \frac{7\pi}{6}, \frac{11\pi}{6} $
Solve $ \cos\theta = 0 $.
Show the full solution

The $ x $-coordinate is 0 at the top and bottom of the circle. $ \frac{\pi}{2}, \frac{3\pi}{2} $
Solve $ \sin^2\theta = \frac{1}{4} $.
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$ \sin\theta = \pm\frac{1}{2} $, so all four quadrants contribute. $ \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6} $
Solve $ 2\cos^2\theta - \cos\theta - 1 = 0 $.
Show the full solution

Factor as $ (2\cos\theta + 1)(\cos\theta - 1) = 0 $. So $ \cos\theta = -\frac{1}{2} $, giving $ \frac{2\pi}{3} $ and $ \frac{4\pi}{3} $; or $ \cos\theta = 1 $, giving $ 0 $. $ 0, \frac{2\pi}{3}, \frac{4\pi}{3} $
Solve $ \sin\theta\cos\theta = \sin\theta $.
Show the full solution

Move everything over and factor: $ \sin\theta(\cos\theta - 1) = 0 $. So $ \sin\theta = 0 $ giving $ 0, \pi $; or $ \cos\theta = 1 $ giving $ 0 $. $ 0, \pi $
Solve $ \sin(2\theta) = 1 $.
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Let $ u = 2\theta $ and solve on $ [0, 4\pi) $: $ u = \frac{\pi}{2} $ and $ \frac{5\pi}{2} $. Halving gives the answers. $ \frac{\pi}{4}, \frac{5\pi}{4} $
A student solves $ \sin\theta\cos\theta = \sin\theta $ by dividing by $ \sin\theta $ and answers $ \theta = 0 $. Find the error.
Show the full solution

Dividing by $ \sin\theta $ assumes it is non-zero and silently deletes the solutions where it is. Factoring instead keeps them. $ 0 $ and $ \pi $

Unit 4 mixed review · 10 problems · all topics

Unit 4 mixed review: Trigonometric Functions

Solve on $ [0, 2\pi) $ unless told otherwise, and remember that each equation usually has two solutions per turn, not one.

Convert $ 150^\circ $ to radians.
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$ \frac{150\pi}{180} $. $ \frac{5\pi}{6} $
Convert $ \frac{4\pi}{3} $ to degrees.
Show the full solution

$ \frac{4 \times 180}{3} $. $ 240^\circ $
Evaluate $ \cos\frac{5\pi}{3} $.
Show the full solution

Quadrant IV, reference $ \frac{\pi}{3} $, cosine positive there. $ \frac{1}{2} $
Evaluate $ \sin\frac{4\pi}{3} $.
Show the full solution

Quadrant III, reference $ \frac{\pi}{3} $, sine negative there. $ -\frac{\sqrt3}{2} $
State the amplitude and period of $ y = 6\sin(4x) $.
Show the full solution

$ |a| = 6 $; period $ \frac{2\pi}{4} $. 6 and $ \frac{\pi}{2} $
State the midline and maximum of $ y = 2\cos x + 7 $.
Show the full solution

Midline $ y = 7 $; maximum $ 7 + 2 $. $ y = 7 $, max 9
Write a cosine equation with maximum 10, minimum 2 and period $ \pi $.
Show the full solution

Midline 6, amplitude 4, $ b = 2 $. $ y = 4\cos(2x) + 6 $
Solve $ 2\sin\theta = \sqrt3 $.
Show the full solution

$ \sin\theta = \frac{\sqrt3}{2} $; reference $ \frac{\pi}{3} $, positive in I and II. $ \frac{\pi}{3}, \frac{2\pi}{3} $
Solve $ 2\cos^2\theta + \cos\theta = 0 $.
Show the full solution

$ \cos\theta(2\cos\theta + 1) = 0 $. So $ \cos\theta = 0 $ giving $ \frac{\pi}{2}, \frac{3\pi}{2} $; or $ \cos\theta = -\frac{1}{2} $ giving $ \frac{2\pi}{3}, \frac{4\pi}{3} $. Four solutions
A student solves $ \cos\theta = \frac{1}{2} $ and answers $ \frac{\pi}{3} $ only. Find the error.
Show the full solution

The inverse cosine returns one angle, but cosine is positive in quadrants I and IV, so $ \frac{5\pi}{3} $ is equally valid on $ [0, 2\pi) $. $ \frac{\pi}{3} $ and $ \frac{5\pi}{3} $

Topic 5.1 · Unit 5 · CA CCSS F-BF.2, F-LE.2

Arithmetic and geometric sequences

Arithmetic sequences add the same amount each step; geometric sequences multiply by the same amount. Identifying which you have takes one subtraction and one division, and everything after that follows a formula.

The method

Identify the type. Subtract consecutive terms: if the difference is constant it is arithmetic, and that constant is $ d $. Divide consecutive terms: if the ratio is constant it is geometric, and that constant is $ r $.
Arithmetic explicit rule: $ a_n = a_1 + (n-1)d $. The $ (n-1) $ is there because the first term has had no steps added to it yet.
Geometric explicit rule: $ a_n = a_1 \cdot r^{\,n-1} $, for the same reason, the first term has been multiplied zero times.
Recursive rules describe the step rather than the position: $ a_n = a_{n-1} + d $ or $ a_n = a_{n-1} \cdot r $. Either form must state $ a_1 $ as well, or it describes nothing.
Connect to functions. An arithmetic sequence is a linear function on whole-number inputs, with $ d $ as its slope. A geometric sequence is an exponential function, with $ r $ as its base.

Where marks are lost: the $ n-1 $. To find the 10th term of an arithmetic sequence starting at 5 with $ d = 3 $, it is $ 5 + 9(3) = 32 $, not $ 5 + 10(3) = 35 $. Count the gaps, not the terms, ten terms have nine gaps between them.

Worked examples

Example 1: arithmetic, explicit rule and a distant term. For $ 7, 11, 15, 19, \dots $ write the rule and find the 25th term.

Differences: $ 11-7 = 4 $, $ 15-11 = 4 $. Constant, so arithmetic with $ d = 4 $ and $ a_1 = 7 $. Rule: $ a_n = 7 + (n-1)4 $, which simplifies to $ a_n = 4n + 3 $.

$ a_{25} = 4(25) + 3 = 103 $. Check with the unsimplified form: $ 7 + 24(4) = 103 $ ✓.

Example 2: geometric. For $ 3, 12, 48, 192, \dots $ write the rule and find the 7th term.

Ratios: $ 12 \div 3 = 4 $, $ 48 \div 12 = 4 $. Geometric with $ r = 4 $, $ a_1 = 3 $. Rule: $ a_n = 3 \cdot 4^{\,n-1} $.

$ a_7 = 3 \cdot 4^6 = 3 \times 4096 = 12288 $. The exponent is 6, not 7, six multiplications get you from the first term to the seventh.

Example 3: working backwards. An arithmetic sequence has $ a_4 = 14 $ and $ a_9 = 34 $. Find $ a_1 $ and the rule.

From the 4th term to the 9th is five gaps, and the value rose by $ 34 - 14 = 20 $, so $ d = 20 \div 5 = 4 $. To get back from $ a_4 $ to $ a_1 $ is three gaps down: $ 14 - 3(4) = 2 $. Rule: $ a_n = 2 + (n-1)4 = 4n - 2 $.

Check: $ a_4 = 4(4) - 2 = 14 $ ✓ and $ a_9 = 4(9) - 2 = 34 $ ✓.

Practice · 15 problems

1–6 identify and continue, 7–11 build rules, 12–15 work backwards or into context.

Is $ 2, 5, 8, 11, \dots $ arithmetic or geometric? Give $ d $ or $ r $.
Show the full solution

Differences are all 3; ratios are not constant. Arithmetic, $ d = 3 $
Is $ 2, 6, 18, 54, \dots $ arithmetic or geometric? Give $ d $ or $ r $.
Show the full solution

Ratios are all 3. Geometric, $ r = 3 $
Write the next two terms of $ 20, 17, 14, \dots $
Show the full solution

$ d = -3 $. 11 and 8
Write the next two terms of $ 80, 40, 20, \dots $
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$ r = \frac{1}{2} $. 10 and 5
Find the 10th term of an arithmetic sequence with $ a_1 = 5 $ and $ d = 3 $.
Show the full solution

$ 5 + (10-1)3 = 5 + 27 = 32 $. Nine gaps, not ten. 32
Find the 6th term of a geometric sequence with $ a_1 = 2 $ and $ r = 3 $.
Show the full solution

$ 2 \cdot 3^5 = 2 \times 243 = 486 $. 486
Write the explicit rule for $ 4, 9, 14, 19, \dots $
Show the full solution

$ d = 5 $, $ a_1 = 4 $: $ a_n = 4 + (n-1)5 = 5n - 1 $. Check $ n=3 $: $ 14 $ ✓. $ a_n = 5n - 1 $
Write the explicit rule for $ 5, 15, 45, 135, \dots $
Show the full solution

$ r = 3 $, $ a_1 = 5 $. $ a_n = 5 \cdot 3^{\,n-1} $
Write a recursive rule for $ 100, 93, 86, \dots $
Show the full solution

$ d = -7 $. The rule needs the starting value too. $ a_1 = 100 $, $ a_n = a_{n-1} - 7 $
Write a recursive rule for $ 6, 18, 54, \dots $
Show the full solution

$ r = 3 $. $ a_1 = 6 $, $ a_n = 3a_{n-1} $
An arithmetic sequence has $ a_1 = 12 $ and $ a_5 = 32 $. Find $ d $.
Show the full solution

Four gaps between the 1st and 5th terms, and the value rose by 20, so $ d = 20 \div 4 = 5 $. $ d = 5 $
An arithmetic sequence has $ a_3 = 11 $ and $ a_8 = 31 $. Find the explicit rule.
Show the full solution

Five gaps, rise of 20, so $ d = 4 $. Back three gaps from $ a_3 $: $ a_1 = 11 - 3(4) = -1 $. Rule: $ a_n = -1 + (n-1)4 = 4n - 5 $. Check $ n=8 $: $ 27 $… recompute: $ 4(8) - 5 = 27 $, but we need 31. The error is the back-step: from $ a_3 $ to $ a_1 $ is two gaps, not three. $ a_1 = 11 - 2(4) = 3 $, giving $ a_n = 4n - 1 $. Check: $ a_3 = 11 $ ✓, $ a_8 = 31 $ ✓. $ a_n = 4n - 1 $
A theater has 18 seats in row 1 and 2 more in each row after. How many in row 15?
Show the full solution

Arithmetic with $ a_1 = 18 $, $ d = 2 $: $ 18 + 14(2) = 46 $. 46 seats
A bacterial culture doubles hourly, starting at 300. How many after 6 hours?
Show the full solution

Geometric with $ r = 2 $. Careful with indexing: 300 is the count at hour 0, so after 6 hours it has doubled six times: $ 300 \times 2^6 = 19200 $. 19 200
A student finds the 12th term of $ 4, 7, 10, \dots $ as $ 4 + 12(3) = 40 $. Find the error.
Show the full solution

They used 12 gaps instead of 11. Twelve terms have eleven gaps between them: $ 4 + 11(3) = 37 $. Listing a few terms confirms the pattern, the $ n $th term is $ 3n + 1 $, so the 12th is 37. 37

Topic 5.2 · Unit 5 · CA CCSS A-SSE.4, F-BF.2

Series, summation notation and the sum formulas

A sequence lists terms; a series adds them. The formulas below replace adding fifty numbers by hand, but they only work if you have correctly identified which kind of series you have.

The method

Identify the type. A constant difference between terms means arithmetic; a constant ratio means geometric. Divide consecutive terms to test for a ratio, subtract to test for a difference.
Arithmetic sum: $ S_n = \frac{n}{2}(a_1 + a_n) $, the number of terms times the average of the first and last. If $ a_n $ is unknown, find it first with $ a_n = a_1 + (n-1)d $.
Geometric sum: $ S_n = a_1\frac{1 - r^n}{1 - r} $ for $ r \neq 1 $.
Infinite geometric series: if $ |r| < 1 $ the sum converges to $ S = \frac{a_1}{1 - r} $. If $ |r| \geq 1 $ the terms do not shrink and the series diverges; there is no sum.
Summation notation: $ \sum_{k=1}^{n} f(k) $ means substitute $ k = 1, 2, \dots, n $ and add. Count the terms as $ \text{upper} - \text{lower} + 1 $, not just the upper limit.
Find $ n $ when it is not given by solving $ a_n = a_1 + (n-1)d $ for $ n $. A non-integer answer means the stated last term is not in the sequence.

Where marks are lost: miscounting terms. $ \sum_{k=3}^{10} $ has $ 10 - 3 + 1 = 8 $ terms, not 7 and not 10. Off-by-one here corrupts every subsequent step.

Worked examples

Example 1: arithmetic series. Find the sum of the first 30 terms of $ 4, 7, 10, \dots $

$ a_1 = 4 $, $ d = 3 $. The 30th term is $ 4 + 29(3) = 91 $.

$ S_{30} = \frac{30}{2}(4 + 91) = 15 \times 95 = 1425 $.

Example 2: finite geometric series. Find $ \sum_{k=1}^{6} 3(2)^{k-1} $.

This is geometric with $ a_1 = 3 $, $ r = 2 $, $ n = 6 $.

$ S_6 = 3 \cdot \frac{1 - 2^6}{1 - 2} = 3 \cdot \frac{-63}{-1} = 189 $. The double negative is where sign errors creep in, both the numerator and denominator are negative when $ r > 1 $.

Example 3: infinite series. Find the sum of $ 8 + 4 + 2 + 1 + \dots $

$ r = \frac{4}{8} = \frac{1}{2} $, and $ |r| < 1 $, so it converges.

$ S = \frac{8}{1 - \frac{1}{2}} = \frac{8}{\frac{1}{2}} = 16 $. The sum is finite even though there are infinitely many terms, because the terms shrink fast enough.

Practice · 15 problems

1–5 notation and identification, 6–11 finite sums, 12–15 infinite series and diagnosis.

Evaluate $ \sum_{k=1}^{4} (2k + 1) $.
Show the full solution

$ 3 + 5 + 7 + 9 $. 24
How many terms are in $ \sum_{k=3}^{10} k^2 $?
Show the full solution

$ 10 - 3 + 1 $. 8
Is $ 5, 9, 13, 17, \dots $ arithmetic or geometric?
Show the full solution

Constant difference of 4. Arithmetic
Is $ 2, 6, 18, 54, \dots $ arithmetic or geometric?
Show the full solution

Constant ratio of 3. Geometric
Find the 20th term of $ 7, 11, 15, \dots $
Show the full solution

$ 7 + 19(4) = 83 $. Using $ 20d $ instead of $ 19d $ is the standard slip. 83
Find the sum of the first 25 terms of $ 3, 8, 13, \dots $
Show the full solution

$ a_{25} = 3 + 24(5) = 123 $, so $ S = \frac{25}{2}(3 + 123) = 25 \times 63 = 1575 $. 1575
Find the sum of the first 100 positive integers.
Show the full solution

$ \frac{100}{2}(1 + 100) = 50 \times 101 $. 5050
Find the sum of the first 8 terms of $ 2, 6, 18, \dots $
Show the full solution

$ r = 3 $: $ S_8 = 2 \cdot \frac{1 - 6561}{1 - 3} = 2 \cdot \frac{-6560}{-2} $. 6560
Find the sum of the first 6 terms of $ 64, 32, 16, \dots $
Show the full solution

$ r = \frac{1}{2} $: $ S_6 = 64 \cdot \frac{1 - \frac{1}{64}}{\frac{1}{2}} = 128 \cdot \frac{63}{64} $. 126
Evaluate $ \sum_{k=1}^{5} 2(3)^{k-1} $.
Show the full solution

$ a_1 = 2 $, $ r = 3 $, $ n = 5 $: $ 2 \cdot \frac{1 - 243}{-2} = 242 $. 242
How many terms of $ 5, 9, 13, \dots $ are needed to reach 101?
Show the full solution

$ 5 + (n-1)4 = 101 $, so $ 4(n-1) = 96 $ and $ n - 1 = 24 $. 25 terms
Find the sum of $ 12 + 6 + 3 + \dots $ to infinity.
Show the full solution

$ r = \frac{1}{2} $, so $ S = \frac{12}{\frac{1}{2}} $. 24
Does $ 3 + 6 + 12 + \dots $ have an infinite sum?
Show the full solution

$ r = 2 $, and $ |r| \geq 1 $, so the terms grow rather than shrink. No; it diverges
Write $ 0.\overline{7} $ as a fraction using an infinite geometric series.
Show the full solution

It is $ 0.7 + 0.07 + 0.007 + \dots $ with $ a_1 = 0.7 $ and $ r = 0.1 $, so $ S = \frac{0.7}{0.9} = \frac{7}{9} $. $ \frac{7}{9} $
A student computes $ \sum_{k=4}^{12} $ using 12 terms. Find the error.
Show the full solution

The count is $ \text{upper} - \text{lower} + 1 = 12 - 4 + 1 = 9 $. The upper limit alone is only the term count when the sum starts at $ k = 1 $. 9 terms

Topic 5.3 · Unit 5 · CA CCSS F-BF.1b-c

Combining functions and composition

Composition feeds one function's output into another as its input. The notation looks like multiplication and is not, which is where most of the trouble in this topic comes from.

The method

Arithmetic combinations are what they look like: $ (f+g)(x) = f(x) + g(x) $, and likewise for subtraction, multiplication and division. For $ \frac{f}{g} $, exclude any $ x $ where $ g(x) = 0 $.
Composition: $ (f \circ g)(x) = f(g(x)) $. Work from the inside out, evaluate $ g $ first, then feed that result into $ f $.
Order matters. $ f(g(x)) $ and $ g(f(x)) $ are generally different functions. Read the notation carefully; the function on the left is applied last.
To compose algebraically, substitute the whole expression for $ g(x) $ wherever $ x $ appears in $ f $. Bracket it, $ f(x) = x^2 $ composed with $ g(x) = x + 3 $ gives $ (x+3)^2 $, not $ x^2 + 3 $.
To evaluate at a number, compute the inner value first, then apply the outer function to it. This is faster than building the composite formula.
Domain of a composite: $ x $ must be in the domain of $ g $, and $ g(x) $ must be in the domain of $ f $. The second condition is the one that gets forgotten.

Where marks are lost: reading $ (f \circ g)(x) $ as $ f(x) \cdot g(x) $. The open circle means composition; a dot or nothing at all means multiplication. They give different answers for almost every pair of functions.

Worked examples

Example 1: evaluating. With $ f(x) = 2x + 1 $ and $ g(x) = x^2 $, find $ f(g(3)) $.

Inside first: $ g(3) = 9 $.

Then $ f(9) = 2(9) + 1 = 19 $.

The other order: $ g(f(3)) = g(7) = 49 $. Different answer, same functions.

Example 2: building the formula. With the same functions, find $ (f \circ g)(x) $ and $ (g \circ f)(x) $.

$ f(g(x)) = 2(x^2) + 1 = 2x^2 + 1 $.

$ g(f(x)) = (2x+1)^2 = 4x^2 + 4x + 1 $. The bracket is essential: the entire inner expression gets squared.

Example 3: domain of a composite. With $ f(x) = \sqrt{x} $ and $ g(x) = x - 5 $, find $ (f \circ g)(x) $ and its domain.

$ f(g(x)) = \sqrt{x - 5} $.

$ g $ accepts every real number, but $ f $ requires a non-negative input, so we need $ x - 5 \geq 0 $. The domain is $ x \geq 5 $.

Practice · 15 problems

Throughout, $ f(x) = 3x - 2 $, $ g(x) = x^2 + 1 $ and $ h(x) = \sqrt{x} $ unless a problem says otherwise.

Find $ (f + g)(x) $.
Show the full solution

$ 3x - 2 + x^2 + 1 $. $ x^2 + 3x - 1 $
Find $ (g - f)(x) $.
Show the full solution

$ x^2 + 1 - (3x - 2) $; the minus hits both terms. $ x^2 - 3x + 3 $
Find $ (fg)(2) $.
Show the full solution

$ f(2) = 4 $, $ g(2) = 5 $. 20
Find $ f(g(0)) $.
Show the full solution

$ g(0) = 1 $, then $ f(1) = 1 $. 1
Find $ g(f(0)) $.
Show the full solution

$ f(0) = -2 $, then $ g(-2) = 5 $. 5
Find $ f(g(2)) $.
Show the full solution

$ g(2) = 5 $, then $ f(5) = 13 $. 13
Find $ (f \circ f)(4) $.
Show the full solution

$ f(4) = 10 $, then $ f(10) = 28 $. 28
Find $ (f \circ g)(x) $.
Show the full solution

$ 3(x^2+1) - 2 = 3x^2 + 1 $. $ 3x^2 + 1 $
Find $ (g \circ f)(x) $.
Show the full solution

$ (3x-2)^2 + 1 = 9x^2 - 12x + 4 + 1 $. $ 9x^2 - 12x + 5 $
Find $ (h \circ g)(x) $.
Show the full solution

$ \sqrt{x^2 + 1} $
State the domain of $ (h \circ f)(x) = \sqrt{3x - 2} $.
Show the full solution

Need $ 3x - 2 \geq 0 $, so $ x \geq \frac{2}{3} $. $ x \geq \frac{2}{3} $
State the domain of $ \left(\frac{f}{g}\right)(x) $.
Show the full solution

$ x^2 + 1 $ is never zero, so nothing is excluded. All real numbers
A shop takes 20% off, then applies a \$5 coupon. Write the final price as a composition of $ d(p) = 0.8p $ and $ c(p) = p - 5 $, in the right order.
Show the full solution

The discount happens first, so it is the inner function: $ c(d(p)) = 0.8p - 5 $. The other order, $ d(c(p)) = 0.8p - 4 $, charges more, order genuinely changes the price. $ c(d(p)) = 0.8p - 5 $
A student computes $ (f \circ g)(x) $ as $ f(x) \cdot g(x) $. Find the error.
Show the full solution

The open circle is composition, not multiplication. $ f(x)g(x) = (3x-2)(x^2+1) $ is a cubic, while $ f(g(x)) = 3x^2 + 1 $ is a quadratic. $ 3x^2 + 1 $
A student computes $ g(f(x)) $ as $ 9x^2 + 1 $. Find the error.
Show the full solution

They squared only the first term of $ 3x - 2 $. Squaring a binomial needs the middle term: $ (3x-2)^2 = 9x^2 - 12x + 4 $. $ 9x^2 - 12x + 5 $

Topic 5.4 · Unit 5 · CA CCSS F-BF.4

Inverse functions

An inverse undoes what the original function did. That is the whole idea, and it explains every rule in this topic, including why the graphs mirror each other and why some functions need their domain restricted first.

The method

To find an inverse: write $ y = f(x) $, swap $ x $ and $ y $, then solve for $ y $. Rename the result $ f^{-1}(x) $.
The check: $ f(f^{-1}(x)) = x $ and $ f^{-1}(f(x)) = x $. If composing in either order does not return $ x $, the inverse is wrong.
Notation: $ f^{-1} $ means the inverse function, not $ \frac{1}{f} $. The $ -1 $ is a label here, not an exponent.
Domain and range swap. The domain of $ f^{-1} $ is the range of $ f $, and vice versa. A point $ (a, b) $ on $ f $ becomes $ (b, a) $ on the inverse.
Graphs are reflections in the line $ y = x $.
Only one-to-one functions have inverses. Use the horizontal line test: if a horizontal line meets the graph twice, the inverse would fail the vertical line test. Restrict the domain to fix it, $ f(x) = x^2 $ has an inverse of $ \sqrt{x} $ only once you restrict to $ x \geq 0 $.
Exponentials and logarithms are inverses of each other, which is why $ \log_b(b^x) = x $.

Where marks are lost: reading $ f^{-1}(x) $ as $ \frac{1}{f(x)} $. For $ f(x) = x + 3 $, the inverse is $ x - 3 $, not $ \frac{1}{x+3} $.

Worked examples

Example 1: a linear inverse. Find the inverse of $ f(x) = 3x - 7 $.

Write $ y = 3x - 7 $, then swap: $ x = 3y - 7 $.

Solve: $ 3y = x + 7 $, so $ f^{-1}(x) = \frac{x+7}{3} $.

Check: $ f\!\left(\frac{x+7}{3}\right) = 3 \cdot \frac{x+7}{3} - 7 = x $. ✓

Example 2: restricting the domain. Find the inverse of $ f(x) = x^2 $.

Swapping gives $ x = y^2 $, so $ y = \pm\sqrt{x} $, which is not a function, because each input would have two outputs.

The cause is that $ f(x) = x^2 $ is not one-to-one: both 3 and $ -3 $ map to 9. Restrict the domain to $ x \geq 0 $ and the inverse is $ f^{-1}(x) = \sqrt{x} $.

Example 3: a rational inverse. Find the inverse of $ f(x) = \frac{2x}{x - 1} $.

Swap: $ x = \frac{2y}{y-1} $. Multiply out: $ x(y - 1) = 2y $, so $ xy - x = 2y $.

Collect the $ y $ terms on one side: $ xy - 2y = x $, so $ y(x - 2) = x $ and $ f^{-1}(x) = \frac{x}{x-2} $. Gathering the $ y $ terms and factoring is the step this kind of problem is really testing.

Practice · 15 problems

1–6 finding inverses, 7–11 properties and graphs, 12–15 harder cases and diagnosis.

Find the inverse of $ f(x) = x + 9 $.
Show the full solution

Undo the addition. $ f^{-1}(x) = x - 9 $
Find the inverse of $ f(x) = 4x $.
Show the full solution

$ f^{-1}(x) = \frac{x}{4} $
Find the inverse of $ f(x) = 2x + 5 $.
Show the full solution

$ x = 2y + 5 $, so $ y = \frac{x-5}{2} $. $ f^{-1}(x) = \frac{x-5}{2} $
Find the inverse of $ f(x) = \frac{x}{3} - 1 $.
Show the full solution

$ x = \frac{y}{3} - 1 $, so $ y = 3(x + 1) $. $ f^{-1}(x) = 3x + 3 $
Find the inverse of $ f(x) = x^3 $.
Show the full solution

Cubing is one-to-one, so no restriction is needed. $ f^{-1}(x) = \sqrt[3]{x} $
Find the inverse of $ f(x) = \sqrt{x - 2} $.
Show the full solution

$ x = \sqrt{y-2} $, so $ y = x^2 + 2 $, with $ x \geq 0 $ since the original output was never negative. $ f^{-1}(x) = x^2 + 2, \; x \geq 0 $
$ f(2) = 7 $. Find $ f^{-1}(7) $.
Show the full solution

The inverse reverses the pair. 2
The point $ (3, 8) $ is on $ f $. Name a point on $ f^{-1} $.
Show the full solution

$ (8, 3) $
$ f $ has domain $ x \geq 1 $ and range $ y \geq 4 $. State the domain of $ f^{-1} $.
Show the full solution

Domain and range swap. $ x \geq 4 $
In which line are a function and its inverse reflections?
Show the full solution

$ y = x $
Does $ f(x) = x^2 - 4 $ have an inverse on all real numbers?
Show the full solution

It fails the horizontal line test, since $ f(2) = f(-2) = 0 $. Restricting to $ x \geq 0 $ fixes it. Not without a restriction
Find the inverse of $ f(x) = \frac{1}{x + 4} $.
Show the full solution

$ x = \frac{1}{y+4} $, so $ y + 4 = \frac{1}{x} $. $ f^{-1}(x) = \frac{1}{x} - 4 $
Find the inverse of $ f(x) = 2^x $.
Show the full solution

Swap and solve: $ x = 2^y $ means $ y = \log_2 x $ by the definition of a logarithm. $ f^{-1}(x) = \log_2 x $
A student says the inverse of $ f(x) = x + 3 $ is $ \frac{1}{x+3} $. Find the error.
Show the full solution

The $ -1 $ in $ f^{-1} $ is notation, not an exponent; it means the function that undoes $ f $, which is subtracting 3. Checking confirms it: $ f(x - 3) = x $. $ f^{-1}(x) = x - 3 $
A student finds the inverse of $ f(x) = 5x + 2 $ to be $ 5x - 2 $. Show the check that catches this.
Show the full solution

Compose: $ f(5x - 2) = 5(5x-2) + 2 = 25x - 8 $, which is not $ x $. Solving properly, $ x = 5y + 2 $ gives $ y = \frac{x-2}{5} $, and $ f\!\left(\frac{x-2}{5}\right) = x $ ✓. $ f^{-1}(x) = \frac{x-2}{5} $

Unit 5 mixed review · 10 problems · all topics

Unit 5 mixed review: Sequences, Series and Function Operations

Identify the type before reaching for a formula, count terms carefully, and read composition notation as composition rather than multiplication.

Find the 15th term of $ 6, 10, 14, \dots $
Show the full solution

$ 6 + 14(4) $. 62
Find the sum of the first 20 terms of $ 2, 7, 12, \dots $
Show the full solution

$ a_{20} = 2 + 19(5) = 97 $, so $ S = 10(2 + 97) $. 990
Find the sum of $ 27 + 9 + 3 + \dots $ to infinity.
Show the full solution

$ r = \frac{1}{3} $, so $ S = \frac{27}{\frac{2}{3}} $. 40.5
Evaluate $ \sum_{k=1}^{5} (3k - 2) $.
Show the full solution

$ 1 + 4 + 7 + 10 + 13 $. 35
With $ f(x) = x - 4 $ and $ g(x) = x^2 $, find $ f(g(3)) $.
Show the full solution

$ g(3) = 9 $, then $ f(9) = 5 $. 5
With the same functions, find $ g(f(3)) $.
Show the full solution

$ f(3) = -1 $, then $ g(-1) = 1 $. 1
With the same functions, find $ (g \circ f)(x) $.
Show the full solution

$ (x-4)^2 = x^2 - 8x + 16 $. $ x^2 - 8x + 16 $
Find the inverse of $ f(x) = 6x - 3 $.
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$ x = 6y - 3 $, so $ y = \frac{x+3}{6} $. $ f^{-1}(x) = \frac{x+3}{6} $
$ f(5) = -2 $. Find $ f^{-1}(-2) $.
Show the full solution

The inverse reverses the pair. 5
A student says the inverse of $ f(x) = x - 7 $ is $ \frac{1}{x-7} $. Find the error.
Show the full solution

The $ -1 $ is notation, not an exponent, the inverse undoes the function, so it adds 7. Check: $ f(x+7) = x $ ✓. $ f^{-1}(x) = x + 7 $

Topic 6.1 · Unit 6 · CA CCSS S-ID.4, S-IC.1-6

Normal distributions, z-scores and sampling

This is the statistics strand of Algebra 2, and it rewards precision about vocabulary as much as about arithmetic. Most lost marks here come from confusing a sample with a population, or an experiment with an observational study.

The method

The normal curve is symmetric about its mean, and mean, median and mode coincide there.
Empirical rule: about 68% of data lies within one standard deviation of the mean, 95% within two, and 99.7% within three. Sketch the curve and label the boundaries before computing anything.
z-score: $ z = \frac{x - \mu}{\sigma} $, how many standard deviations a value sits from the mean. Negative means below the mean. z-scores let you compare values from different distributions.
Percentages between marks: halve the symmetric percentages. Within one SD is 68%, so each side is 34%; between one and two SDs on one side is $ \frac{95 - 68}{2} = 13.5\% $.
Sample against population: a statistic describes a sample, a parameter describes a population. A random sample is what lets you generalize; a convenience or voluntary sample does not.
Study type decides the conclusion. Only a randomized experiment, where the researcher assigns the treatment, can establish cause. An observational study can only establish association.
Margin of error shrinks as the sample grows, roughly by a factor of $ \sqrt{n} $, so quadrupling the sample halves the margin.

Where marks are lost: claiming causation from an observational study. "Students who eat breakfast score higher" from a survey supports association only, the students chose whether to eat breakfast, so a lurking variable could explain both.

Worked examples

Example 1: the empirical rule. Test scores are normal with mean 70 and standard deviation 8. What percentage scored between 62 and 86?

62 is one SD below the mean; 86 is two SDs above.

One SD below to the mean is 34%. The mean to two SDs above is $ 34 + 13.5 = 47.5\% $. Total: 81.5%.

Example 2: comparing with z-scores. Ana scored 82 on a test with mean 75 and SD 5. Ben scored 88 on a test with mean 80 and SD 10. Who did better relative to their class?

Ana: $ z = \frac{82 - 75}{5} = 1.4 $. Ben: $ z = \frac{88 - 80}{10} = 0.8 $.

Ana is further above her mean in standard deviations, so she did better relative to her class, despite the lower raw score.

Example 3: reading a study. A researcher surveys 500 randomly selected adults and finds that coffee drinkers report better mood. What can be concluded?

The sample is random, so the result generalizes to the population it was drawn from. But nobody was assigned to drink coffee, the participants chose.

So this supports an association between coffee and mood, not causation. To claim cause you would need to randomly assign coffee and a placebo.

Practice · 15 problems

1–5 the empirical rule, 6–10 z-scores, 11–15 study design and diagnosis.

In a normal distribution, what percentage lies within 2 standard deviations of the mean?
Show the full solution

About 95%
Heights are normal with mean 66 in and SD 3 in. What percentage is between 63 and 69?
Show the full solution

One SD either side. About 68%
With the same distribution, what percentage is above 66 in?
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The curve is symmetric about the mean. 50%
What percentage is above 72 in?
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72 is two SDs above. Outside two SDs is 5%, split between both tails. 2.5%
What percentage is between 66 and 72 in?
Show the full solution

Mean to two SDs above is half of 95%. 47.5%
A value of 85 comes from a distribution with mean 70, SD 5. Find its z-score.
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$ \frac{15}{5} $. 3
A value of 62 comes from a distribution with mean 70, SD 4. Find its z-score.
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$ \frac{-8}{4} $. $ -2 $
A distribution has mean 50 and SD 6. Find the value with a z-score of 1.5.
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$ x = 50 + 1.5(6) $. 59
Which is more unusual: a z-score of $ -2.4 $ or one of $ 1.9 $?
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Distance from the mean is what matters, not sign, and $ 2.4 \gt 1.9 $. $ -2.4 $
Maria scored 90 (mean 85, SD 2.5); Luis scored 78 (mean 70, SD 5). Who did better relative to their group?
Show the full solution

Maria: $ z = 2 $. Luis: $ z = 1.6 $. Maria
A radio show asks listeners to call in with opinions. Name the sampling flaw.
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Only listeners who feel strongly call, so the sample is self-selected and not representative. Voluntary response bias
Is the mean of a sample a statistic or a parameter?
Show the full solution

It describes the sample, not the population. A statistic
Researchers randomly assign patients to a drug or a placebo. What kind of study is this, and what can it show?
Show the full solution

Random assignment of a treatment makes it a randomized experiment, which is the only design that supports a causal conclusion. An experiment; it can show cause
A poll of 400 people has a margin of error of 5%. Roughly what sample would halve it?
Show the full solution

The margin shrinks with $ \sqrt{n} $, so halving it needs four times the sample. About 1600
A survey finds students who sleep more have higher grades; a news report says sleep raises grades. Find the error.
Show the full solution

A survey is observational (nobody was assigned a sleep schedule) so it shows association only. A lurking variable such as workload or household stability could drive both. Causation claimed from an observational study

Topic 6.2 · Unit 6 · CA CCSS S-IC.2-6

Simulation, randomization and margin of error

This topic asks a single question in several disguises: is the result I am looking at bigger than chance variation would produce anyway? Simulation is how you answer it without any formula.

The method

Simulation logic: assume nothing is going on, generate many random results under that assumption, and see how often chance alone produces something as extreme as what you observed. Rare under chance means the result is meaningful.
Setting one up: state the assumption, choose a random device that matches the probability, define one trial, run many trials, and count.
Margin of error gives an interval: $ \text{estimate} \pm \text{margin} $. A 52% result with a 3% margin means the plausible range is 49% to 55%.
Interpreting a lead: if the interval contains 50%, the poll cannot distinguish the result from a tie. "Too close to call" is a statement about the interval, not about the point estimate.
Sample size: the margin shrinks with $ \sqrt{n} $, so quadrupling the sample halves the margin. Doubling the sample does noticeably less than people expect.
Randomization test for two groups: if the treatment did nothing, the group labels are arbitrary. Reshuffle the labels many times, recompute the difference each time, and see where the real difference falls in that distribution.
Sample size does not fix bias. A biased method stays biased at any size, it just produces a confidently wrong answer.

Where marks are lost: treating a rare-under-chance result as proof. Simulation shows the outcome would be unusual if nothing were going on; it never proves the cause, and only random assignment licenses a causal claim.

Worked examples

Example 1: designing a simulation. A basketball player claims a 70% free throw rate but makes 12 of 20. Is that surprising?

Assume the 70% claim is true. One trial: generate 20 random digits 0–9, counting 0–6 as a make (7 of 10 digits, so 70%). Record how many makes.

Repeat 200 times and count how often 12 or fewer makes occur. If that happens in, say, 10% of trials, 12 of 20 is unremarkable. If it happens in 2%, the claim looks doubtful.

Example 2: reading a margin of error. A poll has candidate A at 52% with a margin of error of 3%. Is A ahead?

The interval is 49% to 55%. Candidate B's interval is 45% to 51%.

These overlap, and A's interval includes values below 50%, so the poll cannot establish a lead. The honest conclusion is that it is too close to call.

Example 3: sample size. A poll of 500 has a margin of 4%. What sample halves it?

Since the margin behaves like $ \frac{1}{\sqrt{n}} $, halving requires four times the sample.

$ 4 \times 500 = 2000 $. Note that going from 500 to 1000 would only reduce the margin to about 2.8%, not 2%.

Practice · 15 problems

1–5 simulation design, 6–10 margin of error, 11–15 interpretation and diagnosis.

What assumption does a simulation start from?
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That nothing unusual is happening, the claim is true, or the treatment has no effect. The no-effect assumption
Describe a device to simulate a fair coin toss.
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Any equally likely two-outcome device: random digits with 0–4 heads and 5–9 tails works. Two equally likely outcomes
How would you simulate a 25% chance using random digits?
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Use pairs 00–99 and count 00–24 as success, which is exactly 25 of 100. 00–24 of 00–99
In a simulation of 20 free throws at 60%, what counts as one trial?
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A complete set of 20 simulated shots, with the number of makes recorded. All 20 shots
Why run many trials rather than a few?
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A handful of trials cannot show how often an outcome occurs by chance; many trials build the distribution you compare against. To estimate the chance reliably
A poll reports 45% with a 4% margin. State the interval.
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41% to 49%
A poll reports 58% with a 3% margin. Can you conclude a majority supports it?
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The interval is 55% to 61%, entirely above 50%. Yes
A poll reports 51% with a 3% margin. Can you conclude a majority?
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The interval is 48% to 54%, which includes values below 50%. No
A poll of 900 has a margin of 3%. Roughly what sample halves it?
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Four times the sample. About 3600
A sample is doubled from 400 to 800. Does the margin halve?
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It shrinks by a factor of $ \sqrt{2} \approx 1.41 $, so to about 71% of its previous size. No, quadrupling is needed
In 200 simulated trials, a result as extreme as the observed one occurs 4 times. What does that suggest?
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2% of trials, so chance alone rarely produces it, evidence against the no-effect assumption. The result is unusual under chance
In 200 simulated trials, the result occurs 62 times. What does that suggest?
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31%, entirely ordinary chance variation, so no evidence of anything unusual. Consistent with chance
Two groups differ by 6 points; reshuffled labels produce a difference that large in 40% of reshuffles. What do you conclude?
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A gap that size arises easily from how people happened to be split, so the treatment has not been shown to do anything. No evidence of an effect
A website surveys its own visitors and gets 50 000 responses. Is the large sample reassuring?
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No. Visitors are self-selected, and size does not remove bias; it narrows the margin around a wrong center. Still biased
A student concludes from a simulation that a coach's new drill causes better shooting. Find the error.
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A simulation only shows the result would be unlikely under chance. Establishing cause needs random assignment to drill and no-drill groups; without it, the players who did the drill may simply have differed to begin with. Causation not established

Unit 6 mixed review · 10 problems · all topics

Unit 6 mixed review: Inference and Conclusions from Data

Most of the marks here are for the conclusion, not the arithmetic. Say exactly what the evidence supports, and no more.

Scores are normal with mean 500 and SD 100. What percentage is between 400 and 700?
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One SD below to the mean is 34%; the mean to two SDs above is 47.5%. 81.5%
Find the z-score of 640 in that distribution.
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$ \frac{140}{100} $. 1.4
What percentage lies more than two standard deviations above the mean?
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5% lies outside two SDs, split between two tails. 2.5%
Is the mean of a population a statistic or a parameter?
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It describes the population. A parameter
A poll reports 47% with a 4% margin. State the interval.
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43% to 51%
Can that poll establish a majority against the measure?
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The interval crosses 50%, so no conclusion either way. No
A poll of 600 has a 4% margin. Roughly what sample halves it?
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Four times the sample. About 2400
In 500 simulated trials, a result as extreme as the observed one occurs 7 times. What does that suggest?
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1.4%, rare under the no-effect assumption, so evidence against it. Unusual under chance
Researchers randomly assign students to two study methods. What kind of conclusion is available?
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Random assignment makes it an experiment, which supports a causal conclusion. Causal
A survey of 80 000 self-selected app users finds strong support for a feature. A student says the huge sample makes it reliable. Find the error.
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Self-selection is a bias in the method, and no sample size removes it, a larger sample only narrows the margin around an estimate that is centered in the wrong place. Size does not cure bias

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