Final Cram — You Got This, Pema

A 4-hour study plan built around the stuff you missed and the stuff most likely to show up tomorrow. Each section has a worked example and a problem to try. Work through it top to bottom — earlier blocks matter most.

⏱ Only have 30 min? Jump here ↓ Memorize tonight ↓ Top mistakes ↓ Morning warm-up

📋 Tomorrow morning — read this first

Five moves that are easily worth 5-10 extra points on the exam. Skim tonight, re-read in the car / on the train before you walk in.

1. Brain dump first

Before reading a single problem, write your most-feared formulas at the top of your scratch paper: unit circle values, log rules, the binomial theorem, quadratic formula, sequence/series formulas. Two minutes, no panic later.

2. Three-pass strategy

Pass 1: every problem you immediately know. Pass 2: ones you can work out with effort. Pass 3: hard ones with whatever time is left. Easy points are worth the same as hard ones — collect them first.

3. Write the formula, then plug in

Start every problem by writing the relevant formula on its own line, THEN substitute numbers. This earns partial credit even if your arithmetic slips, and it stops you from doing the wrong problem.

4. Don't get stuck

If you've stared at a problem for 3 minutes with no progress, circle it and move on. You can come back. Three correct easy problems > grinding on one hard one.

5. Check your domain

Got an answer for a log equation? Plug it back in — if you'd be taking log of a negative or zero, that root is extraneous, cross it out. Same trap with square roots.

The 4-Hour Plan

Total: 4 hr. Take a short break after each block. Don't skip — the order matters.

Block 1· 1 hr 45 min⚠ start here — you missed this exam

Exam 4 — Conics, Sequences, Series, Binomial

You missed this exam entirely. Every minute here is brand-new points. For each section: read the formulas, work the example on paper, then redo the practice problem WITHOUT looking. That's the test.

▶circlesCirclesfull lesson →

Formulas

Standard form of a circle

(x - h)^2 + (y - k)^2 = r^2

center $(h, k)$ , radius $r$

General form to standard form

x^2 + y^2 + Dx + Ey + F = 0 \;\Longrightarrow\; \left(x + \tfrac{D}{2}\right)^2 + \left(y + \tfrac{E}{2}\right)^2 = \tfrac{D^2 + E^2 - 4F}{4}

center $\left(-\tfrac{D}{2},\, -\tfrac{E}{2}\right)$ , radius $\tfrac{\sqrt{D^2 + E^2 - 4F}}{2}$

Distance formula (for finding radius)

r = \sqrt{(x_1 - h)^2 + (y_1 - k)^2}

where $(x_1, y_1)$ is any point on the circle

Key Points

Standard form: $(x - h)^2 + (y - k)^2 = r^2$ — the center is $(h, k)$ and the right side is $r^2$ , not $r$
Signs flip: $(x + 3)$ means $h = -3$ — always take the opposite sign of what's inside the parentheses
To convert general form to standard form, complete the square for both the $x$ and $y$ groups
Given diameter endpoints, use the midpoint formula for the center and the distance formula for the radius

Don't do this

Using the diameter as the radius — if the problem gives diameter $= 10$ , the radius is $5$ and $r^2 = 25$ , not $100$
Reading the right side of $(x-h)^2 + (y-k)^2 = r^2$ as the radius — the right side is $r^2$ ; take the square root to get the actual radius
Getting the center signs backwards — $(x + 3)$ means $h = -3$ , not $h = 3$ ; the signs in the equation are always opposite the center coordinates

Strategy

If the equation is already in the form $(x-h)^2 + (y-k)^2 = r^2$ , read off the center $(h,k)$ and radius $r$ directly. If it's expanded (like $x^2 + y^2 + Dx + Ey + F = 0$ ), complete the square for both $x$ and $y$ to convert it. Remember: the signs in the equation are opposite the center coordinates.

Your turn — work on paper, then check

Try it· easy

Write the equation of the circle with center

(4, -1)

and radius

3

Try it· easy

Identify the center and radius of the circle

(x + 2)^2 + (y - 5)^2 = 36

▶10.1The Ellipsefull lesson →

Formulas

Standard Form (Horizontal Major Axis)

\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1, \quad a > b

Center (h, k), vertices at (h ± a, k), co-vertices at (h, k ± b)

Standard Form (Vertical Major Axis)

\frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1, \quad a > b

Center (h, k), vertices at (h, k ± a), co-vertices at (h ± b, k)

Foci Distance

c^2 = a^2 - b^2

c = distance from center to each focus; foci lie on the major axis

Eccentricity

e = \frac{c}{a}, \quad 0 < e < 1

e near 0 → nearly circular; e near 1 → elongated

Key Points

$a$ is always the larger denominator — whichever variable has $a^2$ underneath determines the major axis direction
Foci formula: $c^2 = a^2 - b^2$ (minus for ellipses) — foci lie on the major axis, inside the ellipse
Eccentricity $e = c/a$ ranges from $0$ (circle) to just below $1$ (very elongated)
When completing the square, factor out leading coefficients first, then balance both sides carefully

Don't do this

Assuming the larger denominator is always under the $x$ -term — $a^2$ is the LARGER denominator regardless of which variable it's under; check both
Using $c^2 = a^2 + b^2$ (that's for hyperbolas!) — for ellipses it's $c^2 = a^2 - b^2$ with a minus sign
When completing the square, forgetting to multiply the added value by the factored-out coefficient — if you add $4$ inside $9(x^2 + \ldots)$ , you're really adding $9 \times 4 = 36$ to the right side

Strategy

If the equation is already in standard form, find the larger denominator — that's $a^2$ , and whichever variable it's under tells you horizontal vs. vertical major axis. Then compute $c$ from $c^2 = a^2 - b^2$ . If you're given a general-form equation, complete the square for both $x$ and $y$ , divide to get $1$ on the right, and read off center, $a$ , $b$ .

Your turn — work on paper, then check

Try it· easy

Find the center, vertices, co-vertices, and foci of

\frac{x^2}{16} + \frac{y^2}{49} = 1

▶10.2The Hyperbolafull lesson →

Formulas

Standard Form (Horizontal Transverse Axis)

\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1

Center (h, k), vertices at (h ± a, k), branches open left and right

Standard Form (Vertical Transverse Axis)

\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1

Center (h, k), vertices at (h, k ± a), branches open up and down

Asymptotes

\text{Horizontal: } y - k = \pm\frac{b}{a}(x - h) \\[4pt] \text{Vertical: } y - k = \pm\frac{a}{b}(x - h)

Slopes depend on which axis is the transverse axis — don't mix them up!

Foci Distance

c^2 = a^2 + b^2

PLUS, not minus — opposite of ellipses. Foci lie on the transverse axis, beyond the vertices.

Eccentricity

e = \frac{c}{a}, \quad e > 1

e near 1 → narrow branches; larger e → wider, more open branches

Key Points

$a^2$ is always under the positive (first) term — it's NOT necessarily the larger denominator
Foci formula: $c^2 = a^2 + b^2$ (plus for hyperbolas) — the opposite sign from ellipses
Asymptote slopes: $\pm b/a$ for horizontal, $\pm a/b$ for vertical — draw the reference rectangle to sketch quickly
Eccentricity $e = c/a > 1$ always; the positive term tells you which way the branches open

Don't do this

Assuming the larger denominator is $a^2$ — that's an ellipse rule! For hyperbolas, $a^2$ is under the POSITIVE term, regardless of whether it's bigger or smaller
Using $c^2 = a^2 - b^2$ instead of $c^2 = a^2 + b^2$ — hyperbolas use plus, ellipses use minus
Swapping the asymptote slope formulas — horizontal hyperbolas use slopes $\pm b/a$ , vertical hyperbolas use slopes $\pm a/b$ ; mix them up and your sketch is wrong

Strategy

Look at which variable's term is positive (the one being subtracted *from*) — that's where $a^2$ is, and it tells you the transverse axis direction. Use $c^2 = a^2 + b^2$ (plus, not minus). For asymptotes, remember the slopes: $\pm b/a$ for horizontal, $\pm a/b$ for vertical. If given general form, complete the square and watch the negative coefficient carefully.

▶11.1Sequences & Notationfull lesson →

Formulas

Explicit Formula

a_n = f(n)

Gives the nth term directly as a function of n — no previous terms needed

Recursive Formula

a_n = f(a_{n-1}), \quad a_1 = \text{(given)}

Each term depends on the previous term; you must know the starting value

Factorial

n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1, \qquad 0! = 1

Grows extremely fast; 0! = 1 by definition

Key Points

Explicit formulas ( $a_n = f(n)$ ) let you jump to any term; recursive formulas require building from the start
Factorials grow faster than any exponential — $n!$ wins over $a^n$ eventually
$0! = 1$ by definition — this makes combination formulas work correctly
To find a formula from a list, check for a constant difference (arithmetic) or constant ratio (geometric)

Don't do this

Confusing explicit and recursive formulas — if the formula references $a_{n-1}$ , it's recursive and you need the previous term; if it only has $n$ , it's explicit
Starting with the wrong index — $a_1$ means plug in $n = 1$ , not $n = 0$ , unless the problem specifically says otherwise
Forgetting that $0! = 1$ by definition — it looks weird but it's a convention that makes all the combination and binomial formulas work

Strategy

When you see "find the first $N$ terms," just plug in $n = 1, 2, 3, \ldots$ — that's it. If given a list and asked for a formula, check for a constant difference (arithmetic) or constant ratio (geometric). For recursive formulas, you must build term by term — there's no shortcut to skip ahead. For factorial problems, cancel before you multiply out.

▶11.2Arithmetic Sequencesfull lesson →

Formulas

Explicit (nth term) formula

a_n = a_1 + (n - 1)d

Jumps directly to the $n$ th term without computing every term before it.

Recursive formula

a_n = a_{n-1} + d

Defines each term from the previous one; you also need the first term $a_1$ .

Sum of an arithmetic series (version 1)

S_n = \frac{n}{2}(a_1 + a_n)

Use when you know both the first and last term.

Sum of an arithmetic series (version 2)

S_n = \frac{n}{2}\bigl(2a_1 + (n - 1)d\bigr)

Use when you know $a_1$ , $d$ , and $n$ but not $a_n$ .

Key Points

Explicit formula: $a_n = a_1 + (n-1)d$ — note it's $(n-1)$ , not $n$ , because $a_1$ gets $d$ added zero times
Sum formula: $S_n = \frac{n}{2}(a_1 + a_n)$ — it's the average of first and last, times the count
Given two terms, subtract the equations to find $d$ — then back-substitute for $a_1$
If terms go down by a constant, $d$ is negative — the formulas still work the same way

Don't do this

Using $n$ instead of $(n-1)$ in the explicit formula — $a_n = a_1 + (n-1)d$ , not $a_n = a_1 + nd$ ; the first term gets $d$ added zero times
When given two terms like $a_4$ and $a_{10}$ , trying to guess $d$ instead of setting up two equations and subtracting to solve for $d$ systematically
Mixing up the two sum formulas — use $S_n = \frac{n}{2}(a_1 + a_n)$ when you know the last term; use $S_n = \frac{n}{2}(2a_1 + (n-1)d)$ when you don't

Strategy

First, identify $a_1$ and $d$ by subtracting consecutive terms. For a specific term, plug into $a_n = a_1 + (n-1)d$ . For a sum, use $S_n = \frac{n}{2}(a_1 + a_n)$ if you know the last term, or $S_n = \frac{n}{2}(2a_1 + (n-1)d)$ if you don't. If given two terms like $a_4$ and $a_{10}$ , set up two equations and subtract to find $d$ .

Your turn — work on paper, then check

Try it· easy

Determine whether the sequence

7, 3, -1, -5, -9, \ldots

is arithmetic. If so, find the common difference.

Try it· easy

Find the 35th term of the arithmetic sequence whose first term is

-8

and common difference is

3

▶11.3Geometric Sequencesfull lesson →

Formulas

Explicit (general) term

a_n = a_1 \cdot r^{n-1}

jumps directly to the $n$ th term; $a_1$ is the first term, $r$ is the common ratio

Recursive definition

a_n = r \cdot a_{n-1}, \quad a_1 \text{ given}

each term equals the previous term times $r$

Finite geometric series

S_n = a_1 \cdot \frac{1 - r^n}{1 - r}, \quad r \neq 1

sum of the first $n$ terms

Infinite geometric series

S = \frac{a_1}{1 - r}, \quad |r| < 1

only converges when $|r| < 1$ ; otherwise the series diverges

Key Points

Explicit formula: $a_n = a_1 \cdot r^{n-1}$ — the exponent is $n-1$ , not $n$
Infinite series converges only when $|r| < 1$ ; use $S = \frac{a_1}{1 - r}$
A negative $r$ makes terms alternate in sign — check $|r|$ (absolute value) for convergence
For finite sums, use $S_n = a_1 \cdot \frac{1 - r^n}{1 - r}$ — the two negatives often cancel

Don't do this

Using $r^n$ in the explicit formula instead of $r^{n-1}$ — the first term is $a_1 \cdot r^0 = a_1$ , not $a_1 \cdot r^1$
Forgetting to check $|r| < 1$ before using the infinite series formula — if $|r| \geq 1$ , the series diverges and the formula doesn't apply
Ignoring the sign of $r$ when checking convergence — $r = -\frac{1}{2}$ still converges because $|{-\frac{1}{2}}| = \frac{1}{2} < 1$

Strategy

Find $r$ by dividing any term by the previous one. For a specific term, use $a_n = a_1 \cdot r^{n-1}$ (the exponent is $n-1$ , not $n$ ). For infinite sums, check $|r| < 1$ first — if it fails, just write "diverges." If it converges, use $S = \frac{a_1}{1 - r}$ . For two given terms, divide them to eliminate $a_1$ and solve for $r$ .

Your turn — work on paper, then check

Try it· easy

Find the common ratio of the geometric sequence

4, -12, 36, -108, \ldots

Try it· easy

Write the explicit formula for the geometric sequence

2, 10, 50, 250, \ldots

and find

a_6

▶9.3Summation Notationfull lesson →

Formulas

Sigma Notation

\sum_{i=m}^{n} a_i = a_m + a_{m+1} + \cdots + a_n

i is the index, m is the start, n is the end, aᵢ is the expression

Sum of a Constant

\sum_{i=1}^{n} c = c \cdot n

Adding the same number n times

Sum of First n Positive Integers

\sum_{i=1}^{n} i = \frac{n(n+1)}{2}

The classic Gauss formula: 1 + 2 + 3 + ⋯ + n

Sum of First n Squares

\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}

Linearity of Summation

\sum_{i=1}^{n}(a_i + b_i) = \sum_{i=1}^{n} a_i + \sum_{i=1}^{n} b_i, \qquad \sum_{i=1}^{n} c \cdot a_i = c \cdot \sum_{i=1}^{n} a_i

Split sums apart and pull constants out — just like distributing

Key Points

Three formulas to memorize: $\sum c = cn$ , $\sum i = \frac{n(n+1)}{2}$ , $\sum i^2 = \frac{n(n+1)(2n+1)}{6}$
Linearity: split sums apart and pull constants out — $\sum(a_i + b_i) = \sum a_i + \sum b_i$
Always check the starting index — formulas assume $i = 1$ ; adjust if the sum starts at $0$ or $2$
For small upper limits (4 or 5), just expand and add — it's faster than using formulas

Don't do this

Forgetting to check the starting index — most closed-form formulas assume $i = 1$ , so if your sum starts at $i = 0$ or $i = 2$ , you need to adjust or compute the extra/missing terms separately
Trying to pull out a variable as if it were a constant — $\sum 3i^2$ lets you pull out the $3$ , but $\sum i \cdot i^2$ does NOT let you pull out the $i$ because $i$ changes with each term
Mixing up the closed-form formulas — $\sum i = \frac{n(n+1)}{2}$ vs. $\sum i^2 = \frac{n(n+1)(2n+1)}{6}$ ; double-check which one you need before plugging in

Strategy

For small upper limits ( $n \leq 5$ ), just expand and add — it's faster than formulas. For larger or variable-bound sums, split using linearity ( $\sum(a_i + b_i) = \sum a_i + \sum b_i$ ), pull out constants, and apply closed-form formulas: $\sum i = \frac{n(n+1)}{2}$ , $\sum i^2 = \frac{n(n+1)(2n+1)}{6}$ . Always check whether the sum starts at $i = 1$ .

▶11.4Series & Their Notationsfull lesson →

Formulas

Arithmetic series (partial sum)

S_n = \frac{n(a_1 + a_n)}{2}

Average of first and last term, times the number of terms.

Geometric series (partial sum)

S_n = a_1 \cdot \frac{1 - r^n}{1 - r}, \quad r \neq 1

Multiply by $r$ , subtract, and almost everything cancels.

Infinite geometric series

S_\infty = \frac{a_1}{1 - r}, \quad |r| < 1

Only converges when $|r| < 1$ ; diverges otherwise.

Sigma notation

\sum_{k=m}^{n} a_k = a_m + a_{m+1} + \cdots + a_n

Key Points

Arithmetic series → common difference → use $S_n = \frac{n}{2}(a_1 + a_n)$
Geometric series → common ratio → use $S_n = a_1 \cdot \frac{1 - r^n}{1 - r}$
Infinite geometric series: converges only if $|r| < 1$ , then $S_\infty = \frac{a_1}{1-r}$
First step on any series problem: identify whether it's arithmetic or geometric

Don't do this

Mixing up the arithmetic and geometric sum formulas — arithmetic uses $S_n = \frac{n}{2}(a_1 + a_n)$ ; geometric uses $S_n = a_1 \cdot \frac{1 - r^n}{1 - r}$
Jumping into a formula without first identifying whether the series is arithmetic (constant difference) or geometric (constant ratio)
Using the infinite series formula $S = \frac{a_1}{1 - r}$ without checking that $|r| < 1$ first — if $|r| \geq 1$ , just write "diverges"

Strategy

First, decide: is it arithmetic (constant difference) or geometric (constant ratio)? For arithmetic, use $S_n = \frac{n}{2}(a_1 + a_n)$ . For geometric, use $S_n = a_1 \cdot \frac{1 - r^n}{1 - r}$ . For infinite geometric series, check $|r| < 1$ before applying $S = \frac{a_1}{1 - r}$ . Bouncing ball problems: initial drop $+$ $2 \times$ (sum of bounce heights).

Your turn — work on paper, then check

Try it· easy

Find the sum:

\displaystyle\sum_{k=1}^{5} (2k + 1)

▶11.6Binomial Theorem★ top pickfull lesson →

Formulas

Binomial Coefficient

\binom{n}{k} = \frac{n!}{k!\,(n-k)!}

Also written $C(n,k)$ or ${}_nC_k$ . Counts the ways to choose $k$ items from $n$ .

Binomial Theorem

(a + b)^n = \sum_{k=0}^{n} \binom{n}{k}\, a^{n-k}\, b^{k}

Specific Term Formula

\text{The } (k+1)\text{th term} = \binom{n}{k}\, a^{n-k}\, b^{k}

Use $k = 0$ for the 1st term, $k = 1$ for the 2nd, etc.

Key Points

$(a+b)^n$ has $n+1$ terms; in each term, the exponents on $a$ and $b$ add up to $n$
The $(k+1)$ th term is $\binom{n}{k} a^{n-k} b^k$ — the 1st term uses $k = 0$
When $b$ is negative (e.g., $(x - 3)^n$ ), set $b = -3$ so signs are handled by the powers automatically
Raise ALL of $a$ to its power: $(2x)^3 = 8x^3$ , not $2x^3$

Don't do this

Forgetting the negative sign when $b$ is negative — in $(x - 3)^4$ , set $b = -3$ so signs alternate automatically via powers of $(-3)$
Not raising the ENTIRE term to the power — $(2x)^3 = 8x^3$ , NOT $2x^3$ ; the coefficient gets raised too
Off-by-one error in the specific term formula — the 4th term uses $k = 3$ (not $k = 4$ ), because the 1st term uses $k = 0$

Strategy

For a full expansion, use Pascal's Triangle for coefficients and write each term with $a$ 's exponent counting down and $b$ 's counting up. For **"find the $k$ th term"**, use the formula: the $(k+1)$ th term is $\binom{n}{k} a^{n-k} b^k$ . If $b$ is negative (like $x - 3$ ), set $b = -3$ so the signs are handled by the powers automatically.

Worked example — read this carefully

Expand

(x + 2)^4

using the Binomial Theorem.

01
Identify the pieces: $a = x$ , $b = 2$ , $n = 4$ .
We'll apply the theorem $(a + b)^n = \sum \binom{n}{k}\, a^{n-k}\, b^k$ with these values.
02
Write out each term for $k = 0, 1, 2, 3, 4$ .
$\binom{4}{0}x^4(2)^0 + \binom{4}{1}x^3(2)^1 + \binom{4}{2}x^2(2)^2 + \binom{4}{3}x^1(2)^3 + \binom{4}{4}x^0(2)^4$
There are $n + 1 = 5$ terms. In each one, the exponents on $x$ and $2$ add up to 4.
03
Evaluate the binomial coefficients and powers of 2.
$1 \cdot x^4 + 4 \cdot 2x^3 + 6 \cdot 4x^2 + 4 \cdot 8x + 1 \cdot 16$
The coefficients from row 4 of Pascal's Triangle are $1, 4, 6, 4, 1$ . Powers of 2: $1, 2, 4, 8, 16$ .
04
Simplify.
$x^4 + 8x^3 + 24x^2 + 32x + 16$
Multiply the binomial coefficient by the power of 2 in each term to get the final coefficients.

Your turn — work on paper, then check

Try it· easy

Expand

(a + b)^5

using the Binomial Theorem.

Try it· medium

Find the 3rd term of

(2x + y)^6

Block 2· 1 hr 15 min

Trig essentials (Unit Circle + Identities + Equations)

The unit circle and reference-angle moves show up on almost every trig problem. Lock these in and you'll catch most of the trig points on the final.

The unit circle — burn this image into your brain

First-quadrant sine values: 0, ½, √2/2, √3/2, 1 for 0°, 30°, 45°, 60°, 90°. Cosine uses the same values in reverse. Everywhere else, find the reference angle and apply ASTC for the sign.

▶5.1Angles & Radian Measurefull lesson →

Formulas

Degrees to Radians

\theta_{\text{rad}} = \theta_{\text{deg}} \times \frac{\pi}{180}

Multiply the degree measure by π/180

Radians to Degrees

\theta_{\text{deg}} = \theta_{\text{rad}} \times \frac{180}{\pi}

Multiply the radian measure by 180/π

Arc Length

s = r\theta

Where s is arc length, r is radius, and θ is the angle in radians

Area of a Sector

A = \frac{1}{2}r^2\theta

Only works when θ is in radians

Coterminal Angles

\theta \pm 360°n \quad \text{or} \quad \theta \pm 2\pi n \quad (n = 1, 2, 3, \ldots)

Add or subtract full rotations to find angles that land in the same position

Key Points

To convert: degrees → radians multiply by $\frac{\pi}{180}$ ; radians → degrees multiply by $\frac{180}{\pi}$
Arc length $s = r\theta$ and sector area $A = \frac{1}{2}r^2\theta$ require $\theta$ in radians
Coterminal angles differ by full rotations: $\theta \pm 360°$ or $\theta \pm 2\pi$
$180° = \pi$ radians — this single fact drives every conversion

Don't do this

Forgetting to convert degrees to radians before using $s = r\theta$ or $A = \frac{1}{2}r^2\theta$ — these formulas only work with radians
Converting the wrong direction: multiplying by $\frac{\pi}{180}$ goes degrees → radians, not the other way around
Not simplifying the fraction after converting — always reduce (e.g., $\frac{60\pi}{180} = \frac{\pi}{3}$ , not leaving it unsimplified)

Strategy

For conversions, multiply by $\frac{\pi}{180}$ (degrees → radians) or $\frac{180}{\pi}$ (radians → degrees). For arc length or sector area, convert to radians first — the formulas $s = r\theta$ and $A = \frac{1}{2}r^2\theta$ only work in radians. For coterminal angles, add or subtract $360°$ (or $2\pi$ ) until you land in the requested range.

▶5.2Unit Circle — Sine & Cosine★ top pickfull lesson →

Formulas

Unit circle definitions

\cos\theta = x, \quad \sin\theta = y

where $(x, y)$ is the point on the unit circle at angle $\theta$

Pythagorean identity

\cos^2\theta + \sin^2\theta = 1

follows directly from $x^2 + y^2 = 1$

Key first-quadrant values

\sin 0 = 0,\; \sin\frac{\pi}{6} = \frac{1}{2},\; \sin\frac{\pi}{4} = \frac{\sqrt{2}}{2},\; \sin\frac{\pi}{3} = \frac{\sqrt{3}}{2},\; \sin\frac{\pi}{2} = 1

Reference angle

\text{ref}(\theta) = \text{acute angle between terminal side and } x\text{-axis}

use the reference angle to find trig values in any quadrant

Even/odd identities

\cos(-\theta) = \cos\theta, \quad \sin(-\theta) = -\sin\theta

cosine is even, sine is odd

Key Points

On the unit circle: $\cos\theta = x$ -coordinate, $\sin\theta = y$ -coordinate
Sine values for key angles follow the pattern $\frac{\sqrt{0}}{2}, \frac{\sqrt{1}}{2}, \frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{2}, \frac{\sqrt{4}}{2}$ ; cosine is the reverse order
All Students Take Calculus → QI all positive, QII sine positive, QIII tangent positive, QIV cosine positive
Reference angle is always measured to the $x$ -axis: QII subtract from $\pi$ , QIII subtract $\pi$ , QIV subtract from $2\pi$

Don't do this

Measuring the reference angle to the $y$ -axis instead of the $x$ -axis — it's always the acute angle to the nearest part of the $x$ -axis
Getting the ASTC signs wrong: cosine is the $x$ -coordinate (negative on the left side), sine is the $y$ -coordinate (negative below)
Confusing $\sin\frac{\pi}{3}$ and $\sin\frac{\pi}{6}$ : $\sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}$ (bigger angle, bigger sine) and $\sin\frac{\pi}{6} = \frac{1}{2}$

Strategy

When asked for an exact trig value like $\sin\frac{5\pi}{4}$ , find the reference angle first, look up the QI value, then apply the correct sign using ASTC. When given a value like $\cos\theta = -\frac{\sqrt{3}}{2}$ , find the reference angle from the positive version, then place it in every quadrant where that function has the given sign.

Worked example — read this carefully

Find the exact value of

\sin\frac{5\pi}{6}

01
Identify the quadrant and reference angle.
$\frac{5\pi}{6} \text{ is in Quadrant II; reference angle } = \pi - \frac{5\pi}{6} = \frac{\pi}{6}$
$\frac{5\pi}{6}$ is between $\frac{\pi}{2}$ and $\pi$ , so it's in Quadrant II. Subtract from $\pi$ to get the reference angle.
02
Recall the sine of the reference angle.
$\sin\frac{\pi}{6} = \frac{1}{2}$
This is one of the key first-quadrant values to memorize.
03
Apply the correct sign for Quadrant II.
$\sin\frac{5\pi}{6} = +\frac{1}{2}$
Sine is positive in Quadrant II (the $y$ -coordinate is positive there), so the answer stays positive.

Your turn — work on paper, then check

Try it· easy

Find the exact value of

\cos\frac{5\pi}{4}

Try it· medium

Find all angles

\theta

[0, 2\pi)

such that

\sin\theta = \frac{\sqrt{2}}{2}

▶5.3The Other Trigonometric Functionsfull lesson →

Formulas

Tangent and cotangent

\tan\theta = \frac{\sin\theta}{\cos\theta}, \quad \cot\theta = \frac{\cos\theta}{\sin\theta}

Secant and cosecant

\sec\theta = \frac{1}{\cos\theta}, \quad \csc\theta = \frac{1}{\sin\theta}

Reciprocal pairs (product form)

\tan\theta \cdot \cot\theta = 1, \quad \sec\theta \cdot \cos\theta = 1, \quad \csc\theta \cdot \sin\theta = 1

Each pair multiplies to 1 — useful for simplifying expressions

Pythagorean identity (tangent form)

1 + \tan^2\theta = \sec^2\theta

divide $\cos^2\theta + \sin^2\theta = 1$ by $\cos^2\theta$

Pythagorean identity (cotangent form)

1 + \cot^2\theta = \csc^2\theta

divide $\cos^2\theta + \sin^2\theta = 1$ by $\sin^2\theta$

Key Points

All six trig functions come from sin and cos: $\tan = \frac{\sin}{\cos}$ , $\sec = \frac{1}{\cos}$ , $\csc = \frac{1}{\sin}$ , $\cot = \frac{\cos}{\sin}$
Reciprocal pairs: sec goes with cos, csc goes with sin — the names are counterintuitive
Three Pythagorean identities: $\sin^2 + \cos^2 = 1$ , $1 + \tan^2 = \sec^2$ , $1 + \cot^2 = \csc^2$
When stuck simplifying, rewrite everything in terms of sine and cosine first

Don't do this

Mixing up reciprocal pairs: $\sec$ goes with $\cos$ (not $\sin$ ) and $\csc$ goes with $\sin$ (not $\cos$ ) — the names are counterintuitive
Forgetting to rationalize denominators — e.g., writing $\frac{1}{\sqrt{3}}$ instead of $\frac{\sqrt{3}}{3}$
Getting the sign wrong when the angle is in QIII or QIV — always check the quadrant sign rules for each function before computing

Strategy

When asked to find all six trig values, start with $\sin$ and $\cos$ from the unit circle, then build the rest by dividing and flipping: $\tan = \sin/\cos$ , $\cot = \cos/\sin$ , $\sec = 1/\cos$ , $\csc = 1/\sin$ . When simplifying expressions with sec, csc, tan, or cot, rewrite everything in terms of $\sin$ and $\cos$ first.

▶5.4Right Triangle Trigonometryfull lesson →

Formulas

SOH-CAH-TOA

\sin\theta = \frac{\text{opp}}{\text{hyp}}, \quad \cos\theta = \frac{\text{adj}}{\text{hyp}}, \quad \tan\theta = \frac{\text{opp}}{\text{adj}}

Cofunction identity (sine–cosine)

\sin\theta = \cos\!\left(\frac{\pi}{2} - \theta\right), \quad \cos\theta = \sin\!\left(\frac{\pi}{2} - \theta\right)

Cofunction identity (tangent–cotangent)

\tan\theta = \cot\!\left(\frac{\pi}{2} - \theta\right), \quad \cot\theta = \tan\!\left(\frac{\pi}{2} - \theta\right)

Cofunction identity (secant–cosecant)

\sec\theta = \csc\!\left(\frac{\pi}{2} - \theta\right), \quad \csc\theta = \sec\!\left(\frac{\pi}{2} - \theta\right)

Pythagorean theorem

a^2 + b^2 = c^2

where $c$ is the hypotenuse

Key Points

SOH-CAH-TOA: Sine = Opposite/Hypotenuse, Cosine = Adjacent/Hypotenuse, Tangent = Opposite/Adjacent
"Opposite" and "adjacent" are relative to the angle you're working with — always label the triangle first
Cofunction identities: $\sin\theta = \cos(90° - \theta)$ — the 'co' in cosine literally means 'complement'
Angle of elevation (looking up) equals angle of depression (looking down) by alternate interior angles

Don't do this

Labeling 'opposite' and 'adjacent' from the wrong angle — these labels change depending on which angle you're working with; always ask 'opposite to *which* angle?'
Using the wrong trig ratio — double-check that the two sides you're connecting match SOH, CAH, or TOA before solving
Forgetting to use the inverse trig function when solving for an angle — e.g., writing $\theta = \frac{3}{8}$ instead of $\theta = \arctan\!\left(\frac{3}{8}\right)$

Strategy

Draw and label the triangle first — identify which side is opposite, adjacent, and hypotenuse relative to the angle in question. Then pick the SOH-CAH-TOA ratio that connects the known side to the unknown. For word problems with angles of elevation or depression, the angle is always measured from the horizontal.

▶6.3Inverse Trigonometric Functionsfull lesson →

Formulas

Arcsine

y = \arcsin(x): \quad \text{Domain } [-1, 1], \quad \text{Range } \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]

Arccosine

y = \arccos(x): \quad \text{Domain } [-1, 1], \quad \text{Range } [0, \pi]

Arctangent

y = \arctan(x): \quad \text{Domain } (-\infty, \infty), \quad \text{Range } \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)

Cancellation (inner inverse)

\sin(\arcsin(x)) = x \;\text{ for } x \in [-1,1]; \quad \arcsin(\sin(x)) = x \;\text{ for } x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]

Same pattern holds for cos/arccos and tan/arctan with their respective domains

Composition via Triangle

\sin(\arccos(x)) = \sqrt{1 - x^2}, \quad \cos(\arcsin(x)) = \sqrt{1 - x^2}

Derived from a right triangle with hypotenuse 1

Key Points

Restricted ranges: $\arcsin$ returns $[-\frac{\pi}{2}, \frac{\pi}{2}]$ , $\arccos$ returns $[0, \pi]$ , $\arctan$ returns $(-\frac{\pi}{2}, \frac{\pi}{2})$
$\arcsin(\sin(x)) = x$ only when $x$ is already in the restricted range — otherwise find the equivalent angle
For compositions like $\sin(\arccos(x))$ : draw a right triangle, label sides, use Pythagorean theorem
Input to inverse trig is a ratio (number); output is an angle in the restricted range

Don't do this

Assuming $\arcsin(\sin(x)) = x$ always — it only equals $x$ when $x$ is already in the restricted range $[-\frac{\pi}{2}, \frac{\pi}{2}]$
Returning an angle outside the restricted range — e.g., giving $\frac{5\pi}{6}$ for $\arcsin(\frac{1}{2})$ instead of $\frac{\pi}{6}$
Mixing up the three restricted ranges: $\arcsin$ returns $[-\frac{\pi}{2}, \frac{\pi}{2}]$ , $\arccos$ returns $[0, \pi]$ , $\arctan$ returns $(-\frac{\pi}{2}, \frac{\pi}{2})$

Strategy

When evaluating $\arcsin$ , $\arccos$ , or $\arctan$ , ask: is my answer in the restricted range? ( $\arcsin$ : $[-\frac{\pi}{2}, \frac{\pi}{2}]$ , $\arccos$ : $[0, \pi]$ , $\arctan$ : $(-\frac{\pi}{2}, \frac{\pi}{2})$ ). For compositions like $\sin(\arccos(x))$ , draw a right triangle, label the sides from the inner function, then read off the outer function.

▶7.1Trig Identitiesfull lesson →

Formulas

Pythagorean Identity

\sin^2(\theta) + \cos^2(\theta) = 1

The most important identity. Rearranges to give you sin² or cos² alone.

Pythagorean Identity (tan/sec)

1 + \tan^2(\theta) = \sec^2(\theta)

Divide the main Pythagorean identity by cos²(θ) to get this one.

Pythagorean Identity (cot/csc)

1 + \cot^2(\theta) = \csc^2(\theta)

Divide the main Pythagorean identity by sin²(θ) to get this one.

Quotient Identities

\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}, \quad \cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}

Reciprocal Identities

\csc(\theta) = \frac{1}{\sin(\theta)}, \quad \sec(\theta) = \frac{1}{\cos(\theta)}, \quad \cot(\theta) = \frac{1}{\tan(\theta)}

Key Points

The Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$ is the most-used identity — know its rearranged forms ( $\sin^2 = 1 - \cos^2$ and $\cos^2 = 1 - \sin^2$ )
To simplify: rewrite everything in terms of $\sin$ and $\cos$ , then look for cancellations or Pythagorean patterns
To verify an identity: work on one side only — never move terms across the equals sign
Common strategies: factor, multiply by a conjugate ( $1 + \sin\theta$ ), or convert to sin/cos

Don't do this

Working on both sides of an identity at once — when verifying, only transform one side; treat the equals sign as a wall you can't cross
Moving terms across the equals sign — this assumes the identity is true, which is exactly what you're trying to prove
Forgetting the Pythagorean identity rearrangements: $\sin^2\theta = 1 - \cos^2\theta$ and $\cos^2\theta = 1 - \sin^2\theta$ are used just as often as the original

Strategy

When you see "verify the identity," pick the more complicated side and simplify toward the simpler one — never move terms across the equals sign. If you see $\tan$ or $\sec$ , rewrite as $\frac{\sin}{\cos}$ or $\frac{1}{\cos}$ . Look for Pythagorean identity patterns ( $\sin^2 + \cos^2 = 1$ ) or try multiplying by a conjugate.

Your turn — work on paper, then check

Try it· easy

Simplify

\cot(\theta)\sin(\theta)

▶7.2Sum & Difference Identitiesfull lesson →

Formulas

Sine of a Sum

\sin(A + B) = \sin A \cos B + \cos A \sin B

Sine of a Difference

\sin(A - B) = \sin A \cos B - \cos A \sin B

Cosine of a Sum

\cos(A + B) = \cos A \cos B - \sin A \sin B

Watch the sign — it's minus for the sum, which is the opposite of what you might guess.

Cosine of a Difference

\cos(A - B) = \cos A \cos B + \sin A \sin B

Tangent of a Sum/Difference

\tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}

The sign in the denominator is opposite to the sign in the numerator.

Key Points

$\sin(A + B) \neq \sin A + \sin B$ — you must use the formula: $\sin A \cos B + \cos A \sin B$
Cosine formulas have the opposite sign: $\cos(A - B)$ has a plus, $\cos(A + B)$ has a minus
Key decompositions to memorize: $75° = 45° + 30°$ , $15° = 45° - 30°$ , $\frac{\pi}{12} = \frac{\pi}{3} - \frac{\pi}{4}$
If given $\sin A$ and $\cos B$ with quadrant info, find the missing values via $\sin^2 + \cos^2 = 1$ first

Don't do this

Thinking $\sin(A + B) = \sin A + \sin B$ — this is wrong; you must use the full formula $\sin A\cos B + \cos A\sin B$
Getting the sign wrong in the cosine formula — $\cos(A + B)$ uses minus and $\cos(A - B)$ uses plus, which is the opposite of what you'd guess
Not knowing how to decompose non-standard angles — memorize these: $75° = 45° + 30°$ , $15° = 45° - 30°$ , $\frac{\pi}{12} = \frac{\pi}{3} - \frac{\pi}{4}$

Strategy

When asked for an exact value of a non-standard angle, decompose it into two familiar angles (e.g., $75° = 45° + 30°$ , $\frac{\pi}{12} = \frac{\pi}{3} - \frac{\pi}{4}$ ), then apply the sum/difference formula. If given $\sin A$ and $\cos B$ with quadrant info, use $\sin^2 + \cos^2 = 1$ to find the missing values before plugging into the formula.

▶7.3Double-Angle & Half-Angle Formulasfull lesson →

Formulas

Sine Double-Angle

\sin(2\theta) = 2\sin\theta\cos\theta

Cosine Double-Angle (3 forms)

\cos(2\theta) = \cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta

Choose the form that best fits what you already know (sin only, cos only, or both).

Tangent Double-Angle

\tan(2\theta) = \frac{2\tan\theta}{1 - \tan^2\theta}

Half-Angle: Sine

\sin\frac{\alpha}{2} = \pm\sqrt{\frac{1 - \cos\alpha}{2}}

The ± depends on the quadrant of α/2.

Half-Angle: Cosine

\cos\frac{\alpha}{2} = \pm\sqrt{\frac{1 + \cos\alpha}{2}}

The ± depends on the quadrant of α/2.

Power-Reducing: Sine

\sin^2\theta = \frac{1 - \cos(2\theta)}{2}

Derived from the cosine double-angle formula; replaces a square with a first-power expression.

Power-Reducing: Cosine

\cos^2\theta = \frac{1 + \cos(2\theta)}{2}

Derived from the cosine double-angle formula; replaces a square with a first-power expression.

Half-Angle: Tangent

\tan\frac{\alpha}{2} = \frac{1 - \cos\alpha}{\sin\alpha} = \frac{\sin\alpha}{1 + \cos\alpha}

Two equivalent forms — pick whichever avoids a zero denominator.

Key Points

$\sin(2\theta) = 2\sin\theta\cos\theta$ — you need both sin and cos, so find the missing one first
$\cos(2\theta)$ has three forms — pick the one matching what you know: sin only → $1 - 2\sin^2\theta$ , cos only → $2\cos^2\theta - 1$
Half-angle formulas have a $\pm$ that depends on the quadrant of the half-angle, not the original angle
Power-reducing formulas turn squares into first-power expressions: $\sin^2\theta = \frac{1 - \cos(2\theta)}{2}$

Don't do this

Writing $\sin(2\theta) = 2\sin\theta$ instead of $2\sin\theta\cos\theta$ — you need both sine AND cosine
Picking the wrong form of $\cos(2\theta)$ — use $1 - 2\sin^2\theta$ when you only know sine, $2\cos^2\theta - 1$ when you only know cosine
Forgetting to choose $+$ or $-$ in half-angle formulas — the sign depends on the quadrant of the half-angle $\frac{\alpha}{2}$ , not the full angle $\alpha$

Strategy

For $\sin(2\theta)$ , you need both $\sin\theta$ and $\cos\theta$ — find the missing one first via $\sin^2 + \cos^2 = 1$ . For $\cos(2\theta)$ , pick the form matching what you know: use $1 - 2\sin^2\theta$ if you only have sine, $2\cos^2\theta - 1$ if you only have cosine. For half-angle problems, the $\pm$ depends on the quadrant of the half-angle, not the original.

▶7.5Solving Trigonometric Equationsfull lesson →

Formulas

General Solution (Sine/Cosine)

x = x_0 + 2n\pi, \quad n \in \mathbb{Z}

Sine and cosine repeat every 2π.

General Solution (Tangent)

x = x_0 + n\pi, \quad n \in \mathbb{Z}

Tangent repeats every π.

Zero Product Property

AB = 0 \implies A = 0 \text{ or } B = 0

Factor and set each factor to zero — never divide by a trig expression.

Quadratic Substitution

a\sin^2(x) + b\sin(x) + c = 0 \implies \text{let } u = \sin(x)

Treat it like au² + bu + c = 0, solve for u, then find x.

Key Points

Strategy: get everything down to one trig function, then solve — use identities, factoring, or substitution
Never divide by a trig expression — factor instead, or you'll lose solutions where that expression equals zero
Don't forget $\pm$ when taking a square root — $\sin^2 x = \frac{1}{4}$ gives $\sin x = \pm\frac{1}{2}$
General solutions: add $+ 2n\pi$ for sin/cos equations, $+ n\pi$ for tangent equations ( $n \in \mathbb{Z}$ )

Don't do this

Dividing both sides by $\sin(x)$ or $\cos(x)$ instead of factoring — this loses solutions where that function equals zero
Forgetting the $\pm$ when taking a square root: $\sin^2(x) = \frac{1}{4}$ means $\sin(x) = \frac{1}{2}$ OR $\sin(x) = -\frac{1}{2}$ (four solutions, not two)
Only checking one or two quadrants — after finding the reference angle, check ALL quadrants where the trig function has the correct sign

Strategy

Reduce to one trig function using identities, then isolate it and solve with the unit circle. If you get a quadratic in $\sin x$ or $\cos x$ , factor — never divide by a trig expression or you'll lose solutions. Don't forget $\pm$ when square-rooting, and add $+2n\pi$ (sin/cos) or $+n\pi$ (tan) for general solutions.

Your turn — work on paper, then check

Try it· easy

Solve

\cos(x) = -1

[0, 2\pi)

Block 3· 40 min

Logs & Exponentials (with rational/inequality refresh)

Final exams love log/exponential equations and log rules. Get the three log rules cold and you can grind through anything they throw at you.

▶3.7Rational Functionsfull lesson →

Formulas

General Form

f(x) = \frac{p(x)}{q(x)}

where p(x) and q(x) are polynomials and q(x) ≠ 0

Vertical Asymptotes

\text{Set } q(x) = 0 \text{ (after canceling common factors)}

The x-values where the denominator is zero (and doesn't cancel) give vertical asymptotes

Horizontal Asymptote (degrees equal)

y = \frac{a_n}{b_m}

When deg(p) = deg(q), the HA is the ratio of leading coefficients

Horizontal Asymptote (numerator smaller)

y = 0

When deg(p) < deg(q)

Hole Location

\text{If } (x - c) \text{ cancels from both } p(x) \text{ and } q(x), \text{ hole at } x = c

Key Points

Factor the numerator and denominator completely — this single step reveals holes, vertical asymptotes, and x-intercepts
Hole vs. vertical asymptote: if a factor cancels from top and bottom, it's a hole; if it stays in the denominator, it's a VA
Horizontal asymptote rule: $\deg(p) < \deg(q) \to y = 0$ ; equal degrees $\to y = a_n/b_m$ ; $\deg(p) > \deg(q) \to$ no HA
Domain of a rational function = all real numbers except where the original denominator is zero

Don't do this

Calling every denominator zero a vertical asymptote — if the factor ALSO cancels from the numerator, it's a hole, not a VA
Forgetting to find the $y$ -coordinate of a hole — plug the $x$ -value into the SIMPLIFIED function (after canceling) to get the actual point
Comparing degrees after canceling common factors for the horizontal asymptote — always compare the degrees of the ORIGINAL numerator and denominator

Strategy

Factor the numerator and denominator completely — this single step reveals everything. Common factors that cancel give holes; remaining denominator factors give vertical asymptotes; remaining numerator factors give ** $x$ -intercepts. Compare degrees for the horizontal asymptote**: smaller numerator degree → $y = 0$ , equal degrees → ratio of leading coefficients.

▶sz.ineqPolynomial & Rational Inequalitiesfull lesson →

Formulas

Sign Chart Method

\begin{aligned} &1.\;\text{Move everything to one side} \\ &2.\;\text{Factor completely} \\ &3.\;\text{Find critical points} \\ &4.\;\text{Test intervals} \end{aligned}

Critical points = zeros of numerator + zeros of denominator

Polynomial Inequality Setup

p(x) > 0 \;\text{ or }\; p(x) < 0 \;\text{ or }\; p(x) \geq 0 \;\text{ or }\; p(x) \leq 0

Always move everything to one side first so you're comparing to 0

Rational Inequality Setup

\frac{p(x)}{q(x)} \leq 0 \implies \text{critical points at } p(x)=0 \text{ and } q(x)=0

Never multiply both sides by the denominator — you don't know its sign

Interval Notation Reminders

[a, b] \text{ includes endpoints}, \quad (a, b) \text{ excludes endpoints}

Use ( ) at ±∞ and at values where the function is undefined

Sign Change Rule

\text{Sign changes at single roots; sign stays the same at double (even) roots}

A factor like (x - 2)² touches zero but doesn't change sign

Key Points

Always use a sign chart: find critical points, test one value per interval, determine the sign
Critical points include both numerator zeros and denominator zeros — the sign can change at either
Use brackets $[\;]$ at zeros where the expression equals 0 (if the inequality allows equality); always use parentheses $(\;)$ at undefined points and at $\pm\infty$
Never multiply both sides of a rational inequality by the denominator — you don't know its sign

Don't do this

Multiplying both sides of a rational inequality by the denominator — you don't know its sign, so the inequality might flip; always use a sign chart instead
Using a bracket at an undefined point — if $x = 3$ makes the denominator zero, ALWAYS use a parenthesis there, even with $\leq$ or $\geq$
Forgetting that even-multiplicity roots don't change the sign — $(x - 2)^2$ is always $\geq 0$ , so the expression doesn't flip sign at $x = 2$

Strategy

Factor the expression completely, find all zeros (numerator and denominator), then build a sign chart: place critical points on a number line, test one value per interval, and determine the sign. Use brackets $[\;]$ at zeros where equality is allowed and parentheses $(\;)$ at undefined points. Never multiply both sides by the denominator.

▶4.1Exponential Functionsfull lesson →

Formulas

Exponential Function (General)

f(x) = b^x, \quad b > 0, \; b \neq 1

b > 1 means growth; 0 < b < 1 means decay

Natural Exponential Function

f(x) = e^x, \quad e \approx 2.71828

The most important base — it makes calculus cleanest

Compound Interest

A = P\left(1 + \frac{r}{n}\right)^{nt}

P = principal, r = annual rate (decimal), n = compounds per year, t = years

Continuous Compounding

A = Pe^{rt}

The limit of compound interest as n → ∞

Key Properties of b^x

b^0 = 1, \quad b^x > 0 \text{ for all } x, \quad \text{HA: } y = 0

Key Points

Exponential functions: the variable is in the exponent ( $b^x$ ), not the base ( $x^b$ ) — this is what makes growth so fast
$b > 1$ means growth; $0 < b < 1$ means decay; every exponential passes through $(0, 1)$ with HA at $y = 0$
Compound interest: use $A = P(1 + r/n)^{nt}$ for periodic compounding and $A = Pe^{rt}$ for continuous
Always convert the interest rate to a decimal before plugging in: $6\% \to r = 0.06$

Don't do this

Forgetting to convert the interest rate from a percentage to a decimal — $6\%$ should be $r = 0.06$ , not $r = 6$
Confusing $2^{-3}$ with $-2^3$ : a negative exponent gives a reciprocal ( $\frac{1}{8}$ ), not a negative number ( $-8$ )
Using the wrong compounding formula: "continuously" means $A = Pe^{rt}$ , not $A = P(1 + r/n)^{nt}$

Strategy

For compound interest, identify $P$ , $r$ (as a decimal!), $n$ , and $t$ , then plug into $A = P(1 + r/n)^{nt}$ . If it says "continuously," use $A = Pe^{rt}$ instead. For solving equations like $2^{2x+1} = 32$ , rewrite both sides with the same base and set the exponents equal.

▶4.3Logarithmic Functionsfull lesson →

Formulas

Definition of logarithm

\log_b(x) = y \iff b^y = x

b > 0, b ≠ 1, and x > 0

Common logarithm

\log(x) = \log_{10}(x)

Natural logarithm

\ln(x) = \log_e(x)

e ≈ 2.718

Inverse relationships

\log_b(b^x) = x \quad \text{and} \quad b^{\log_b(x)} = x

Logs and exponentials cancel each other

Change-of-base formula

\log_b(x) = \frac{\ln(x)}{\ln(b)} = \frac{\log(x)}{\log(b)}

Use this to evaluate any log on a calculator

Key Points

Logarithms answer the question: what exponent gives me this number? $\log_b(x) = y$ means $b^y = x$
The natural log $\ln$ uses base $e \approx 2.718$ ; the common log $\log$ uses base 10
You can only take the log of a positive number — the domain of $\log_b(x)$ is $x > 0$
Use the change-of-base formula $\log_b(x) = \frac{\ln x}{\ln b}$ to evaluate any log on a calculator

Don't do this

Thinking $\log(x + y) = \log(x) + \log(y)$ — there is NO log rule for sums; the product rule is $\log(xy) = \log x + \log y$
Forgetting that you can only take the log of a positive number — $\log_b(0)$ and $\log_b(-5)$ are undefined
Confusing $\log$ (base 10) with $\ln$ (base $e$ ) — when no base is written, it's base 10

Strategy

When asked to evaluate a log, convert to exponential form: $\log_b(x) = y$ means $b^y = x$ , then figure out the exponent. For unusual bases, use change-of-base: $\log_b(x) = \frac{\ln x}{\ln b}$ . If asked to convert between forms, remember the base stays the base, the exponent becomes the answer, and the result goes inside the log.

Your turn — work on paper, then check

Try it· easy

Evaluate

\log_3(81)

▶4.4Graphs of Logarithmic Functions★ top pickfull lesson →

Formulas

Parent logarithmic function

f(x) = \log_b(x)

Passes through (1, 0); vertical asymptote at x = 0

Horizontal shift

f(x) = \log_b(x - h)

Asymptote moves to x = h; domain is (h, ∞)

Vertical shift

f(x) = \log_b(x) + k

Shifts graph up/down; asymptote and domain unchanged

General transformed log

f(x) = a \cdot \log_b(x - h) + k

a = vertical stretch/reflect, h = horizontal shift, k = vertical shift

Exponential-log reflection relationship

y = b^x \;\longleftrightarrow\; y = \log_b(x) \text{ reflected over } y = x

Key Points

Log graphs are the reflection of exponential graphs over the line $y = x$ — swap the roles of $x$ and $y$
The parent $y = \log_b(x)$ has a vertical asymptote at $x = 0$ and passes through $(1, 0)$
Horizontal shifts move the vertical asymptote: $\log_b(x - h)$ has VA at $x = h$ and domain $(h, \infty)$
To find the domain of any log function, set the argument $> 0$ and solve

Don't do this

Confusing which function has which asymptote: exponentials have horizontal asymptotes; logs have vertical asymptotes
Thinking a vertical shift ( $+k$ ) moves the vertical asymptote — only horizontal shifts change where the asymptote is
Forgetting to flip the inequality when dividing by a negative while finding the domain (e.g., $-2x > -6$ becomes $x < 3$ )

Strategy

For domain questions, set the argument of the log $> 0$ and solve the inequality — the boundary is also where the vertical asymptote lives. For transformations, identify $h$ (horizontal shift = asymptote location) and $k$ (vertical shift). Remember: only horizontal shifts move the asymptote; vertical shifts ( $+k$ ) do not.

Worked example — read this carefully

Graph

f(x) = \log_2(x - 3) + 1

and state the domain, range, and asymptote.

01
Identify transformations from the parent $y = \log_2(x)$ .
The $(x - 3)$ inside shifts the graph 3 units right. The $+1$ outside shifts it 1 unit up.
02
Find the vertical asymptote.
$x = 3$
The parent asymptote $x = 0$ shifts right by 3. Set the argument equal to zero: $x - 3 = 0 \implies x = 3$ .
03
Determine the domain and range.
$\text{Domain: } (3, \infty), \quad \text{Range: } (-\infty, \infty)$
The domain is everything to the right of the asymptote. The range of a log function is always all real numbers — vertical shifts don't change that.
04
Plot key points.
$(1, 0) \to (4, 1), \quad (2, 1) \to (5, 2)$
Shift each parent point right 3 and up 1 to sketch the curve.

Your turn — work on paper, then check

Try it· easy

Find the domain and vertical asymptote of

f(x) = \log_5(x - 7)

Try it· easy

State the vertical asymptote, domain, and range of

f(x) = \ln(x + 3) - 4

▶4.6Exponential & Logarithmic Equationsfull lesson →

Formulas

Same-Base Strategy

b^{f(x)} = b^{g(x)} \implies f(x) = g(x)

Only works when both sides share the same base

Take-Log-of-Both-Sides Strategy

a^x = c \implies x = \frac{\ln c}{\ln a}

Use when bases cannot be matched

Log-to-Exponential Conversion

\log_b(x) = y \iff b^y = x

The key move for solving logarithmic equations

One-to-One Property of Logarithms

\log_b(M) = \log_b(N) \implies M = N

If two logs with the same base are equal, their arguments are equal

Extraneous Solution Check

\text{Domain requirement: arguments of all logs must be } > 0

Always verify solutions in the original equation

Key Points

For exponential equations: try to match bases first ( $b^{f(x)} = b^{g(x)} \implies f(x) = g(x)$ ); if you can't, take $\ln$ of both sides
For log equations: condense to a single log, convert to exponential form, then solve
Always check for extraneous solutions in log equations — reject any answer that makes a log argument $\leq 0$
When you see $e^{2x}$ , think substitution: let $u = e^x$ and solve the resulting quadratic

Don't do this

Forgetting to check for extraneous solutions in log equations — always plug answers back in and reject any that make a log argument $\leq 0$
Trying to take $\ln$ of both sides when the bases can be matched — check for same-base first, it's faster and avoids decimals
Distributing a log across addition: $\log(x + 3) \neq \log(x) + \log(3)$ — there is no rule for the log of a sum

Strategy

For an exponential equation, first ask: can I rewrite both sides with the same base? If yes, set exponents equal. If not, take $\ln$ of both sides. For a log equation, condense into a single log, convert to exponential form, solve, then always check that every log argument is $> 0$ .

Your turn — work on paper, then check

Try it· easy

Solve:

9^x = 27

▶4.7Exponential & Logarithmic Modelsfull lesson →

Formulas

Exponential Growth/Decay Model

N(t) = N_0 \, e^{kt}

k > 0 for growth, k < 0 for decay

Doubling Time

t_{\text{double}} = \frac{\ln 2}{k}

Time for a quantity to double (assumes k > 0)

Half-Life

t_{1/2} = \frac{\ln 2}{|k|}

Time for a quantity to halve (assumes k < 0)

Newton's Law of Cooling

T(t) = T_s + (T_0 - T_s)\,e^{kt}

T_s = surrounding temp, T_0 = initial temp, k < 0

Logistic Growth Model

P(t) = \frac{c}{1 + a\,e^{-bt}}

c = carrying capacity; growth slows as P approaches c

Key Points

The continuous growth/decay model $N = N_0 e^{kt}$ : use $k > 0$ for growth, $k < 0$ for decay
Doubling time = $\ln 2 / k$ ; half-life = $\ln 2 / |k|$ — both derived from the same model
Word problem strategy: use the given data point to find $k$ first, then answer the question with $k$
In Newton's law of cooling, $T_s$ is the surrounding temperature — the object's temp approaches $T_s$ over time

Don't do this

Using a positive $k$ for decay problems — if something is shrinking, $k$ must be negative
Plugging in the percentage directly instead of a decimal: $6\%$ growth means $k = 0.06$ , not $k = 6$
Confusing the half-life shortcut $t = \frac{\ln 2}{|k|}$ with the general model — the shortcut only gives you the half-life itself, not the amount remaining

Strategy

Write $N = N_0 e^{kt}$ and identify what's given. **Use the data point to find $k$ first** — plug in the known measurement and solve. Then plug $k$ back in to answer the actual question. For half-life or doubling time, use the shortcut $t = \frac{\ln 2}{|k|}$ directly.

Block 4· 20 min

Functions, inverses, polynomials (quick refresh)

You've seen this material twice already. Skim the key takeaways and formulas — don't re-learn it, just refresh.

▶1.fnFunctions, Domain & Rangefull lesson →

Formulas

Domain of a Rational Function

f(x) = \frac{p(x)}{q(x)} \implies \text{exclude all } x \text{ where } q(x) = 0

Set the denominator equal to zero and solve — those x-values are NOT in the domain

Domain of a Square Root Function

f(x) = \sqrt{g(x)} \implies g(x) \geq 0

The expression under the radical must be zero or positive

Interval Notation Quick Reference

(a, b) \;\text{open} \quad [a, b] \;\text{closed} \quad [a, b) \;\text{half-open} \quad (-\infty, \infty) \;\text{all reals}

Infinity always gets a parenthesis, never a bracket

Key Points

A function assigns exactly one output to each input — the vertical line test checks this visually
Domain = all valid inputs; Range = all outputs that actually occur
Two domain killers: division by zero (denominator $= 0$ ) and square root of a negative (radicand $< 0$ )
Interval notation: parentheses $(\,)$ for excluded endpoints, brackets $[\,]$ for included — infinity always gets a parenthesis

Don't do this

Writing the domain of $\sqrt{x - 3}$ as $x > 3$ instead of $x \geq 3$ — the square root of zero is perfectly fine, it's only negatives that break things
Forgetting to exclude BOTH values when the denominator factors into two pieces — e.g., $x^2 - 9 = 0$ gives $x = 3$ AND $x = -3$ , not just one of them
Confusing domain (valid inputs / $x$ -values) with range (outputs / $y$ -values) — domain is horizontal, range is vertical

Strategy

First, check what type of function you have. If there's a fraction, set the denominator $\neq 0$ . If there's a square root, set the radicand $\geq 0$ . If you have both, combine the restrictions. For evaluation problems like $f(a+h)$ , just replace every $x$ with the entire expression — use parentheses to avoid sign errors.

▶1.compComposition & Transformationsfull lesson →

Formulas

Average Rate of Change

\frac{f(b) - f(a)}{b - a}

Slope of the secant line between (a, f(a)) and (b, f(b))

Composition of Functions

(f \circ g)(x) = f(g(x))

Apply g first, then f — read right to left

Vertical & Horizontal Shifts

f(x) + k \;\text{(up/down)} \qquad f(x - h) \;\text{(right } h \text{)}

Inside changes move opposite: x - h shifts RIGHT h units

Stretches & Reflections

a \cdot f(x) \;\text{(vert. stretch)} \qquad f(bx) \;\text{(horiz. compress by } \tfrac{1}{b}\text{)}

-f(x) reflects over x-axis; f(-x) reflects over y-axis

Absolute Value (Vertex Form)

f(x) = a|x - h| + k, \quad \text{vertex } (h, k)

a > 0 opens up (V), a < 0 opens down (∧)

Key Points

$(f \circ g)(x) = f(g(x))$ — apply $g$ first, then $f$ ; order matters
Outside changes ( $+k$ , $a \cdot f$ ) affect $y$ directly; inside changes ( $x - h$ , $bx$ ) affect $x$ in the opposite way
$f(x - h)$ shifts right $h$ ; $-f(x)$ reflects over the $x$ -axis; $f(-x)$ reflects over the $y$ -axis
Average rate of change $= \frac{f(b) - f(a)}{b - a}$ is just the slope between two points

Don't do this

Applying composition in the wrong order — $(f \circ g)(x) = f(g(x))$ means apply $g$ first, then $f$ ; mixing this up gives a completely different answer
Shifting the graph the wrong direction — $f(x - 3)$ shifts RIGHT 3, not left; inside changes always go the opposite way from what you'd expect
Forgetting to distribute the negative in transformations — in $-2(x + 1)^2 + 5$ , the $-$ reflects over the $x$ -axis AND the $2$ stretches vertically; don't miss either piece

Strategy

For composition, work inside-out: in $(f \circ g)(x)$ , evaluate $g(x)$ first, then plug the result into $f$ . For transformations, match the equation to the template $a \cdot f(x - h) + k$ — read $h$ and $k$ for shifts, check the sign of $a$ for reflections. Remember: inside changes ( $x - h$ , $bx$ ) do the opposite of what they look like.

▶1.invInverse Functions★ top pickfull lesson →

Formulas

Inverse Verification

f(f^{-1}(x)) = x \quad \text{and} \quad f^{-1}(f(x)) = x

Both compositions must equal x — checking only one isn't enough

Finding an Inverse Algebraically

y = f(x) \;\longrightarrow\; \text{swap } x \text{ and } y \;\longrightarrow\; \text{solve for } y = f^{-1}(x)

Replace f(x) with y, swap every x and y, then isolate y

Domain–Range Relationship

\text{Domain of } f = \text{Range of } f^{-1}, \quad \text{Range of } f = \text{Domain of } f^{-1}

Inputs and outputs swap roles when you invert

Key Points

To find $f^{-1}$ : replace $f(x)$ with $y$ , swap $x$ and $y$ , then solve for $y$
A function has an inverse only if it's one-to-one — use the horizontal line test to check
Domain of $f$ = Range of $f^{-1}$ , and Range of $f$ = Domain of $f^{-1}$
To verify inverses: both $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$ must hold

Don't do this

Solving for $x$ without swapping first — if you skip the swap step, you'll just get back the original function instead of the inverse
Confusing $f^{-1}(x)$ with $\frac{1}{f(x)}$ — the $-1$ is NOT an exponent; $f^{-1}$ means the inverse function, not the reciprocal
Checking only one composition direction when verifying — you need BOTH $f(f^{-1}(x)) = x$ AND $f^{-1}(f(x)) = x$

Strategy

If asked to find an inverse, follow "swap and solve": replace $f(x)$ with $y$ , swap $x$ and $y$ , then solve for $y$ . If asked to verify two functions are inverses, compose both ways — $f(g(x))$ and $g(f(x))$ must both simplify to $x$ . For rational functions, multiply to clear fractions after swapping, then collect all $y$ -terms on one side and factor.

Worked example — read this carefully

Find the inverse of

f(x) = 3x - 7

01
Replace f(x) with y
$y = 3x - 7$
This just makes the algebra easier to work with.
02
Swap x and y
$x = 3y - 7$
This is the key step — we're switching inputs and outputs.
03
Solve for y
$x + 7 = 3y \implies y = \frac{x + 7}{3}$
Add 7 to both sides, then divide by 3. So $f^{-1}(x) = \frac{x+7}{3}$ .

Your turn — work on paper, then check

Try it· easy

f(x) = x^3 + 2

a one-to-one function? Explain briefly.

▶2.linLinear Functionsfull lesson →

Formulas

Slope

m = \frac{y_2 - y_1}{x_2 - x_1}

Rise over run — the rate of change between any two points

Slope-Intercept Form

y = mx + b

m = slope, b = y-intercept

Point-Slope Form

y - y_1 = m(x - x_1)

Use when you know the slope and one point $(x_1, y_1)$

Parallel & Perpendicular Slopes

\text{Parallel: } m_1 = m_2 \qquad \text{Perpendicular: } m_1 \cdot m_2 = -1

Perpendicular slopes are negative reciprocals (e.g. $\frac{2}{3}$ and $-\frac{3}{2}$ )

Key Points

Slope $m = \frac{\text{rise}}{\text{run}} = \frac{y_2 - y_1}{x_2 - x_1}$ — positive means uphill, negative means downhill, zero means horizontal
Point-slope form $y - y_1 = m(x - x_1)$ is the fastest way to write a line equation when you have a point and slope
Parallel lines have equal slopes ( $m_1 = m_2$ ); perpendicular lines have negative reciprocal slopes ( $m_1 \cdot m_2 = -1$ )

Don't do this

Finding the perpendicular slope by flipping the fraction but forgetting to change the sign — perpendicular to $\frac{2}{3}$ is $-\frac{3}{2}$ , not $\frac{3}{2}$
Putting the slope formula upside down — it's $\frac{y_2 - y_1}{x_2 - x_1}$ (rise over run), not $\frac{x_2 - x_1}{y_2 - y_1}$
Distributing incorrectly in point-slope form — in $y - y_1 = m(x - x_1)$ , multiply $m$ by BOTH the $x$ and the $-x_1$ inside the parentheses

Strategy

Almost every line problem starts the same way: find the slope first. If given two points, use $m = \frac{y_2 - y_1}{x_2 - x_1}$ . If asked for parallel/perpendicular, grab the slope from the given line (same slope for parallel, flip-and-switch for perpendicular). Then plug the slope and a point into point-slope form $y - y_1 = m(x - x_1)$ and simplify.

▶3.cqComplex Numbers & Quadraticsfull lesson →

Formulas

Standard Form of a Complex Number

z = a + bi, \quad i^2 = -1

Complex Conjugate Product

(a + bi)(a - bi) = a^2 + b^2

Multiply by the conjugate to clear i from a denominator

Vertex Form of a Quadratic

f(x) = a(x - h)^2 + k

Vertex at (h, k); axis of symmetry x = h

Quadratic Formula

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Discriminant

\Delta = b^2 - 4ac

Δ > 0 → two real roots; Δ = 0 → one repeated root; Δ < 0 → two complex roots

Key Points

Complex numbers: $i^2 = -1$ ; add/subtract by combining like terms; multiply with FOIL and replace $i^2$ with $-1$
To divide complex numbers, multiply top and bottom by the conjugate of the denominator
Vertex form $a(x - h)^2 + k$ gives vertex $(h, k)$ — complete the square to convert from standard form
Discriminant $b^2 - 4ac$ : positive → 2 real roots, zero → 1 repeated root, negative → 2 complex roots

Don't do this

Forgetting the negative sign on $b$ in the quadratic formula — when $b = -5$ , you get $-(-5) = 5$ , not $-5$
Writing $i^2 = 1$ instead of $i^2 = -1$ — the whole point of $i$ is that its square is negative one
When completing the square, adding a number inside the parentheses but forgetting to balance the other side — if you add $9$ inside, add $9$ (times the leading coefficient) to the other side too

Strategy

For complex arithmetic, treat $i$ like a variable and FOIL — just replace $i^2$ with $-1$ at the end. For division, multiply top and bottom by the conjugate. For quadratics, check the discriminant $b^2 - 4ac$ first to know what type of answers to expect, then use the quadratic formula or complete the square to convert to vertex form.

Your turn — work on paper, then check

Try it· easy

Simplify

(5 + 3i) + (2 - 7i)

▶3.polyPolynomialsfull lesson →

Formulas

End Behavior (Leading Term Test)

f(x) = a_n x^n + \cdots \implies \text{end behavior matches } a_n x^n

Even degree → both ends same direction; odd degree → opposite ends

Division Algorithm

f(x) = d(x) \cdot q(x) + r(x)

dividend = divisor × quotient + remainder (degree of r < degree of d)

Remainder Theorem

f(c) = \text{remainder when } f(x) \div (x - c)

Evaluate f(c) without plugging in — just read the remainder from synthetic division

Factor Theorem

(x - c) \text{ is a factor of } f(x) \iff f(c) = 0

Rational Zero Theorem

\text{Possible rational zeros} = \pm\frac{p}{q}

p = factors of the constant term, q = factors of the leading coefficient

Key Points

End behavior depends only on the leading term: even degree → both ends same direction, odd degree → opposite ends
Zeros with odd multiplicity cross the $x$ -axis; zeros with even multiplicity touch and bounce
Remainder Theorem: $f(c)$ equals the remainder when dividing $f(x)$ by $(x - c)$ ; if the remainder is $0$ , then $(x - c)$ is a factor
Rational Zero Theorem: possible rational zeros are $\pm\frac{\text{factors of constant}}{\text{factors of leading coeff}}$

Don't do this

Forgetting to use $0$ as a placeholder for missing terms in synthetic division — if there's no $x^3$ term, you still need a $0$ in that slot or everything after shifts
Confusing end behavior with local behavior — end behavior depends ONLY on the leading term $a_n x^n$ , no matter what the middle of the graph looks like
Saying a zero with even multiplicity "crosses" the axis — even multiplicity means the graph TOUCHES and turns around; odd multiplicity is what crosses

Strategy

For end behavior, look only at the leading term — even degree means same direction on both ends, odd means opposite. For finding zeros, list candidates with the Rational Zero Theorem ( $\pm p/q$ ), test with synthetic division until one works, then factor the quotient. Don't forget $0$ placeholders for missing terms in synthetic division.

Memorize Before Bed

If you only remember these, you'll bank a huge chunk of points. Write each one out on paper from memory at least once.

Unit circle key values (first quadrant)

$\sin: 0, \tfrac{1}{2}, \tfrac{\sqrt{2}}{2}, \tfrac{\sqrt{3}}{2}, 1$ for $0, \tfrac{\pi}{6}, \tfrac{\pi}{4}, \tfrac{\pi}{3}, \tfrac{\pi}{2}$
$\cos$ uses the same values in reverse order
ASTC: QI All positive, QII Sine, QIII Tangent, QIV Cosine

Pythagorean identities (memorize all three)

$\sin^2\theta + \cos^2\theta = 1$
$1 + \tan^2\theta = \sec^2\theta$
$1 + \cot^2\theta = \csc^2\theta$

Log rules (these are the three you need)

$\log_b(MN) = \log_b M + \log_b N$
$\log_b(M/N) = \log_b M - \log_b N$
$\log_b(M^p) = p \log_b M$
Change of base: $\log_b x = \dfrac{\ln x}{\ln b}$

Conics — standard forms (Exam 4!)

Circle: $(x-h)^2 + (y-k)^2 = r^2$
Ellipse: $\dfrac{(x-h)^2}{a^2} + \dfrac{(y-k)^2}{b^2} = 1$
Hyperbola: $\dfrac{(x-h)^2}{a^2} - \dfrac{(y-k)^2}{b^2} = 1$

Sequences & Series (Exam 4!)

Arithmetic $n$ th term: $a_n = a_1 + (n-1)d$
Arithmetic sum: $S_n = \dfrac{n(a_1 + a_n)}{2}$
Geometric $n$ th term: $a_n = a_1 r^{n-1}$
Geometric sum: $S_n = a_1 \cdot \dfrac{1 - r^n}{1 - r}$
Infinite geometric (only if $|r|<1$ ): $S = \dfrac{a_1}{1-r}$

Binomial Theorem (Exam 4!)

$(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$
The $(k+1)$ th term is $\binom{n}{k} a^{n-k} b^k$ — 1st term uses $k=0$
If $b$ is negative (like $x-3$ ), set $b=-3$ ; signs handle themselves via powers
$\binom{n}{k} = \dfrac{n!}{k!(n-k)!}$ — cancel factorials before multiplying!

Quadratic formula (never lose these points)

$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
Discriminant $b^2 - 4ac$ : positive → 2 real, zero → 1 real, negative → complex
Vertex of $ax^2+bx+c$ : $x = -\dfrac{b}{2a}$

Don't Lose Points On These

The top mistakes across the whole semester. Read these out loud once. Read them again right before you walk into the exam.

01Reference angles are measured to the $x$ -axis, never the $y$ -axis. QII: $\pi - \theta$ . QIII: $\theta - \pi$ . QIV: $2\pi - \theta$ .
02When $b$ is negative in $(a+b)^n$ — like $(x-3)^4$ — set $b = -3$ , not $3$ . Signs alternate automatically.
03 $(2x)^3 = 8x^3$ , not $2x^3$ . Raise the entire term, including the coefficient.
04The $(k+1)$ th term uses $k$ , not $k+1$ . The 4th term uses $k=3$ .
05Don't lose the $\pm$ when you take a square root in a quadratic or Pythagorean problem.
06Domain of $\log(x)$ is $x > 0$ . Check your solutions — extraneous roots are common.
07When solving $\sin x = c$ on $[0, 2\pi)$ , find ALL angles — usually two, sometimes four.
08Common ratio $r$ is each term divided by the previous one, not subtracted.
09Infinite geometric series only has a sum when $|r| < 1$ . If $|r| \ge 1$ , the answer is "no sum / diverges."
10Inverse functions: swap $x$ and $y$ , then solve for $y$ . The domain of $f^{-1}$ is the range of $f$ .
11Log equations: condense first using log rules, then convert to exponential form. Check for extraneous solutions at the end.
12Read each problem TWICE. Underline what they're asking. Many wrong answers come from solving the wrong thing.

Morning Warm-Up

One problem per major topic, pulled from your lessons with full step-by-step solutions. Do these on paper before the exam — if you can crank through all 10, you're ready. Tap to reveal each solution.

01Unit circle5.2 →

Find the exact value of

\cos\frac{5\pi}{4}

02Trig equation5.2 →

Find all angles

\theta

[0, 2\pi)

where

\cos\theta = -\frac{1}{2}

03Log equation4.4 →

Find the domain and vertical asymptote of

f(x) = \log_5(x - 7)

04Exponential4.4 →

State the vertical asymptote, domain, and range of

f(x) = \ln(x + 3) - 4

05Circlecircles →

Write the equation of the circle with center

(4, -1)

and radius

3

06Arithmetic sequence11.2 →

Determine whether the sequence

7, 3, -1, -5, -9, \ldots

is arithmetic. If so, find the common difference.

07Geometric series11.3 →

Write the explicit formula for the geometric sequence

2, 10, 50, 250, \ldots

and find

a_6

08Binomial expansion11.6 →

Expand

(x - 3)^4

09Binomial — find a term11.6 →

Find the 3rd term of

(2x + y)^6

10Inverse function1.inv →

f(x) = x^3 + 2

a one-to-one function? Explain briefly.

One more thing.

You don't need a perfect score. You need every point you can grab. Read each problem twice. Write the formula down before you start plugging in numbers. If you get stuck, skip and come back. The easy points are worth the same as the hard ones — collect them first. You can do this.