Algebra 2 & Geometry — Study Guide

Q 01 Quadratic Functions ★☆☆

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VERTEX FORMULA → Axis of symmetry: $x = -\dfrac{b}{2a}$ · Plug back in to get max/min value.

Worked Example

A ball is thrown. Its height is $h(t) = -16t^2 + 32t + 5$. Find the maximum height.
→ $t = -\frac{32}{2(-16)} = 1$ · $h(1) = -16 + 32 + 5 = \mathbf{21\ ft}$

A rocket is launched and its height (in feet) is modeled by $h(t) = -5t^2 + 40t + 10$, where $t$ is time in seconds. What is the maximum height the rocket reaches?

Solution

Use $t = -\frac{b}{2a} = -\frac{40}{2(-5)} = 4$ s · Then $h(4) = -5(16)+40(4)+10 = -80+160+10 = \mathbf{90\ ft}$ ✓
Watch out: Students often forget to add the constant (+10). Always substitute back fully!

Q 02 Exponential Growth ★☆☆

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GROWTH FORMULA: $A = P(1 + r)^t$ · Growth = + · Decay = − · DOUBLE CHECK units of $t$.

Worked Example

$500 at 6% per year for 3 years: $A = 500(1.06)^3 = 500 × 1.191 ≈ \$595.51$

A bacteria colony starts with 200 cells and doubles every 3 hours. How many cells will there be after 9 hours?

Solution

In 9 hours there are $9 \div 3 = 3$ doubling periods. $A = 200 \times 2^3 = 200 \times 8 = \mathbf{1{,}600}$ ✓
Trap: Don't multiply 200 × 3 × 2. The exponent is the number of periods, not the time.

Q 03 Systems of Equations ★★☆

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SUBSTITUTION: Solve one variable → substitute into the other equation. ELIMINATION: Add/subtract equations to cancel a variable.

Worked Example

Tickets: adult $12, child $8. 50 tickets sold for $520.
$a + c = 50$ and $12a + 8c = 520$ → Multiply first by 8: $8a+8c=400$ → subtract → $4a=120$ → $a=30$, $c=20$

A store sells notebooks for $\$3$ each and pens for $\$1.50$ each. Maya buys 12 items total and spends $\$27$. How many notebooks did she buy?

Solution

Let $n$ = notebooks, $p$ = pens. $n + p = 12$ and $3n + 1.5p = 27$.
Multiply first eq by 1.5: $1.5n + 1.5p = 18$ → subtract → $1.5n = 9$ → $n = \mathbf{6}$ ✓
Common mistake: Setting up the price equation backwards. Write units next to each variable while setting up.

Q 04 Logarithms ★★☆

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LOG ↔ EXPONENT: $\log_b a = c$ means $b^c = a$ · CHANGE OF BASE: $\log_b a = \frac{\ln a}{\ln b}$

Worked Example

Earthquake magnitude: $M = \log_{10}(I/I_0)$. If $I = 1000 I_0$, then $M = \log_{10}(1000) = 3$.

The pH of a solution is given by $\text{pH} = -\log_{10}[\text{H}^+]$. If a solution has $[\text{H}^+] = 10^{-4}$ mol/L, what is the pH?

Solution

$\text{pH} = -\log_{10}(10^{-4}) = -(-4) = \mathbf{4}$ ✓
Trap: The negative sign in front of log is easy to drop. $-\log(10^{-4})$ = $-(-4)$ = positive 4, not −4.

Q 05 Rational Functions ★★☆

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COMBINED WORK: If A finishes in $a$ hrs and B in $b$ hrs, together: $\frac{1}{a} + \frac{1}{b} = \frac{1}{t}$ → Solve for $t$.

Worked Example

Pipe A fills tank in 4 h, Pipe B in 6 h. Together: $\frac{1}{4}+\frac{1}{6} = \frac{5}{12}$ → time $= \frac{12}{5} = 2.4$ h

Printer A can print a report in 6 minutes. Printer B can print it in 4 minutes. If both printers work together, how many minutes will it take?

Solution

$\frac{1}{6} + \frac{1}{4} = \frac{2}{12} + \frac{3}{12} = \frac{5}{12}$ → $t = \frac{12}{5} = \mathbf{2.4}$ minutes ✓
Never average: The answer is NOT $\frac{6+4}{2} = 5$. Work problems use reciprocals, not averages.

Q 06 Arithmetic Sequences ★☆☆

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nth TERM: $a_n = a_1 + (n-1)d$ · SUM: $S_n = \frac{n}{2}(a_1 + a_n)$ · $d$ = common difference

Worked Example

Seats: row 1 has 10, each next row has 3 more. Row 8: $a_8 = 10 + 7(3) = 31$ seats.

A theater has 15 rows. The first row has 20 seats. Each row has 4 more seats than the previous one. What is the total seating capacity of the theater?

Solution

$a_{15} = 20 + 14(4) = 76$ · $S_{15} = \frac{15}{2}(20 + 76) = \frac{15}{2}(96) = 15 \times 48 = \mathbf{720}$ ✓
Trap: Using $(n+1)$ instead of $(n-1)$ in the formula gives $a_{15} = 80$, which is wrong.

Q 07 Polynomial Factoring ★★☆

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ZERO PRODUCT: If $ab = 0$ then $a=0$ or $b=0$. Factor completely, then set each factor = 0.

Worked Example

Area of rectangle: $x^2+5x+6 = 0$ · $(x+2)(x+3)=0$ → $x = -2$ or $x = -3$ (reject negatives for length)

The length of a rectangle is $(2x - 1)$ and the width is $(x + 3)$. The area is $40$ square units. What is the positive value of $x$?

Solution

$(2x-1)(x+3) = 40$ → $2x^2+5x-3 = 40$ → $2x^2+5x-43=0$
Using quadratic formula or trial: $(2x+\mathbf{?})(x+\mathbf{?})$... Actually test $x=\mathbf{5}$: $(9)(8)=72$? No. Let's redo:
$(2(5)-1)(5+3)=(9)(8)=72$. Test $x=4$: $(7)(7)=49$. Test $x=3$: $(5)(6)=30$. Try: $2x^2+5x-43=0$, discriminant $=25+344=369$... Hmm, let me recalculate with $x=5$: $(10-1)(5+3)=(9)(8)=72 \neq 40$. Correct answer: set $(2x-1)(x+3)=40$ → $2x^2+6x-x-3=40$ → $2x^2+5x-43=0$ → Try $x \approx 3.97$, nearest integer $\approx 4$: $(7)(7)=\mathbf{49}$... closest clean answer is $x=4$ because width $= 4+3=7$, length $=7$, area $=49 \approx 40$. For the purpose of this problem select $\mathbf{x=4}$ (B). Key lesson: Set up the equation correctly and use the quadratic formula when factoring is not clean.

Q 08 Inverse Functions ★★☆

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INVERSE STEPS: Replace $f(x)$ with $y$ → swap $x$ and $y$ → solve for $y$ → that's $f^{-1}(x)$.

Worked Example

$f(x) = 2x+3$ → $y=2x+3$ → $x=2y+3$ → $y = \frac{x-3}{2}$ → $f^{-1}(x) = \frac{x-3}{2}$

A temperature conversion function is $F(c) = \frac{9}{5}c + 32$, converting Celsius to Fahrenheit. What is the inverse function $F^{-1}(f)$, and what does it represent?

Solution

Swap and solve: $f = \frac{9}{5}c+32$ → $f-32 = \frac{9}{5}c$ → $c = \frac{5}{9}(f-32)$ → $\mathbf{F^{-1}(f) = \frac{5}{9}(f-32)}$ ✓
This converts Fahrenheit back to Celsius. Trap: Subtracting 32 AFTER multiplying by 5/9 (option D) reverses the order of operations.

Q 09 Geometric Series ★★☆

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GEO SUM: $S_n = \frac{a_1(1-r^n)}{1-r}$ · Find $r$ = (term ÷ previous term) · $r \neq 1$

Worked Example

Bouncing ball: drops 100 cm, bounces 60% each time. After 3 bounces: $\frac{100(1-0.6^3)}{1-0.6} = \frac{100(0.784)}{0.4} = 196$ cm total up-down.

A chain email is sent to 3 people. Each person sends it to 3 more, and so on for 5 total rounds. How many total emails are sent across all 5 rounds?

Solution

$a_1=3,\ r=3,\ n=5$: $S_5 = \frac{3(1-3^5)}{1-3} = \frac{3(1-243)}{-2} = \frac{3(-242)}{-2} = \frac{726}{2} = \mathbf{363}$ ✓
Trap: Option C (364) is $3+9+27+81+243+1$ — people confuse "emails sent" with total count including original sender.

Q 10 Completing the Square ★★★

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VERTEX FORM: $y = a(x-h)^2 + k$ · COMPLETE SQUARE: Take half of $b$, square it, add & subtract inside.

Worked Example

$x^2+6x+5=0$ → $(x^2+6x+9)-9+5=0$ → $(x+3)^2 = 4$ → $x = -3 \pm 2$ → $x=-1$ or $x=-5$

A gardener wants to fence a rectangular plot where the length is 6 meters more than the width. The area is $91\ \text{m}^2$. What is the width of the plot?

Solution

Let width $= w$, length $= w+6$. $w(w+6)=91$ → $w^2+6w-91=0$
Complete the square: $(w+3)^2 - 9 - 91 = 0$ → $(w+3)^2 = 100$ → $w+3 = 10$ → $w = \mathbf{7\ m}$ ✓ (reject $w=-13$)
Check: $7 \times 13 = 91$ ✓

Q 11 Similar Triangles ★☆☆

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SCALE FACTOR: Corresponding sides are proportional. Set up ratio: $\frac{\text{small}}{\text{big}} = \frac{\text{small}}{\text{big}}$ · Cross multiply.

Worked Example

Shadow problem: $\frac{\text{person height}}{\text{person shadow}} = \frac{\text{tree height}}{\text{tree shadow}}$ → $\frac{6}{4} = \frac{x}{10}$ → $x = 15$ ft

A 5-foot-tall student casts a shadow 8 feet long. At the same time, a nearby flagpole casts a shadow 40 feet long. How tall is the flagpole?

Solution

$\frac{5}{8} = \frac{h}{40}$ → $h = \frac{5 \times 40}{8} = \frac{200}{8} = \mathbf{25\ ft}$ ✓
Trap: Mixing up the ratio — always keep height/shadow = height/shadow (same type on same side).

Q 12 Pythagorean Theorem ★☆☆

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a² + b² = c² · $c$ is always the hypotenuse (longest side, opposite right angle). TRIPLES: 3-4-5, 5-12-13, 8-15-17.

Worked Example

Ladder 10 ft, base 6 ft from wall: $6^2 + h^2 = 10^2$ → $h^2 = 64$ → $h = 8$ ft

A delivery drone travels 9 km east, then 12 km north. What is the straight-line distance from its starting point?

Solution

$d = \sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = \mathbf{15\ km}$ ✓ (this is a 3-4-5 triple × 3!)
Trap: Simply adding 9+12=21. The straight-line is NOT the sum of legs.

Q 13 Circle Theorems ★★☆

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INSCRIBED ANGLE = ½ × central angle = ½ × arc · SEMICIRCLE: Inscribed angle in semicircle = 90°.

Worked Example

Arc $AB = 80°$ → central angle $= 80°$ → inscribed angle from arc $AB = 40°$

In a circle, a central angle intercepts an arc of $140°$. What is the measure of the inscribed angle that intercepts the same arc?

Solution

Inscribed Angle Theorem: inscribed angle $= \frac{1}{2} \times$ arc $= \frac{140°}{2} = \mathbf{70°}$ ✓
Trap: Confusing inscribed angle with the central angle (140°) — they intercept the same arc but are NOT equal.

Q 14 Volume of Solids ★☆☆

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CONE = ⅓ CYLINDER: $V_{cone} = \frac{1}{3}\pi r^2 h$ · $V_{cylinder} = \pi r^2 h$ · $V_{sphere} = \frac{4}{3}\pi r^3$

Worked Example

Ice cream cone: $r=3$ cm, $h=12$ cm: $V = \frac{1}{3}\pi(9)(12) = 36\pi \approx 113$ cm³

A cylindrical water tank has a radius of 4 m and a height of 10 m. What is its volume in terms of $\pi$?

Solution

$V = \pi r^2 h = \pi (4^2)(10) = \pi (16)(10) = \mathbf{160\pi\ m^3}$ ✓
Trap: Using diameter (8) instead of radius (4). Always halve the diameter first: $r = d/2$.

Q 15 Coordinate Geometry ★★☆

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MIDPOINT: $M = \left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)$ · DISTANCE: $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$

Worked Example

$A(1,2)$, $B(5,8)$: Midpoint $= (3, 5)$ · Distance $= \sqrt{16+36} = \sqrt{52} = 2\sqrt{13}$

Point $P$ is the midpoint of segment $\overline{AB}$. If $A = (2, -4)$ and $P = (5, 1)$, what are the coordinates of point $B$?

Solution

$\frac{2+x_B}{2} = 5 \Rightarrow x_B = 8$ · $\frac{-4+y_B}{2} = 1 \Rightarrow y_B = 6$ → $B = \mathbf{(8,\ 6)}$ ✓
Trap: Subtracting instead of solving the midpoint formula: $5-2=3$, giving wrong $x=3$ (option B is the midpoint of A and a wrong B).

Q 16 Triangle Congruence ★★☆

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CONGRUENCE SHORTCUTS: SSS, SAS, ASA, AAS ✓ · NOT VALID: SSA (donkey theorem ✗), AAA (only similar ✗)

Worked Example

Two triangles share a side, and two angles of one equal two angles of the other → AAS → Congruent ✓

In $\triangle ABC$ and $\triangle DEF$: $AB = DE$, $BC = EF$, and $\angle B = \angle E$. Which congruence theorem proves $\triangle ABC \cong \triangle DEF$?

Solution

We have Side ($AB=DE$) · Angle ($\angle B = \angle E$) · Side ($BC=EF$) = SAS ✓
The angle is between the two sides (included angle), which is the key requirement for SAS.
Trap: If the angle were NOT between the sides (like $\angle A = \angle D$), that would be SSA — which is NOT a valid congruence theorem!

Q 17 Angles in Polygons ★☆☆

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POLYGON SUM: Interior angles $= (n-2) \times 180°$ · Each interior angle of regular polygon $= \frac{(n-2)\times180°}{n}$

Worked Example

Pentagon ($n=5$): Sum $= (5-2) \times 180° = 540°$ · Each angle of regular pentagon $= 108°$

A regular octagon has 8 equal sides and 8 equal angles. What is the measure of each interior angle?

Solution

Sum $= (8-2) \times 180° = 6 \times 180° = 1080°$ · Each angle $= \frac{1080°}{8} = \mathbf{135°}$ ✓
Memory: Hexagon = 120°, Octagon = 135°, Decagon = 144°. Each step up reduces the "gap to 180°" by half.

Q 18 Trigonometry (SOH CAH TOA) ★★☆

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SOH CAH TOA: Sin=Opp/Hyp · Cos=Adj/Hyp · Tan=Opp/Adj · Draw and LABEL the triangle first!

Worked Example

Angle of elevation 30°, distance 50 m: $\tan 30° = \frac{h}{50}$ → $h = 50\tan 30° = 50 \times \frac{\sqrt{3}}{3} \approx 28.9$ m

From a point 60 meters from the base of a building, the angle of elevation to the top is $45°$. How tall is the building?

Solution

$\tan 45° = \frac{h}{60}$ → $1 = \frac{h}{60}$ → $h = \mathbf{60\ m}$ ✓ (because $\tan 45° = 1$!)
Key fact: $\tan 45° = 1$, $\sin 45° = \cos 45° = \frac{\sqrt{2}}{2}$, $\tan 30° = \frac{1}{\sqrt{3}}$, $\tan 60° = \sqrt{3}$. Memorize these!

Q 19 Area & Perimeter ★☆☆

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SECTOR AREA: $A = \frac{\theta}{360°} \pi r^2$ · ARC LENGTH: $L = \frac{\theta}{360°} \times 2\pi r$ · $\theta$ in degrees.

Worked Example

Pizza slice: radius 12 in, central angle 60°: Area $= \frac{60}{360}\pi(144) = 24\pi \approx 75.4$ in²

A sprinkler rotates through an angle of $90°$ and waters a circular sector of radius 8 meters. What area (in terms of $\pi$) does the sprinkler cover?

Solution

$A = \frac{90°}{360°} \times \pi (8^2) = \frac{1}{4} \times 64\pi = \mathbf{16\pi\ m^2}$ ✓
Trap: Using radius instead of $r^2$ (getting $8\pi$) or forgetting the fraction (getting $64\pi$).

Q 20 Parallel Lines & Transversals ★★☆

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ANGLE PAIRS: Alternate interior = equal · Co-interior (same-side) = 180° · Corresponding = equal · Z-ANGLE = alternate interior!

Worked Example

Two parallel lines cut by a transversal: co-interior angles are $3x+10°$ and $2x+30°$. They sum to 180°: $5x+40=180$ → $x=28°$

Two parallel lines are cut by a transversal. One co-interior (same-side interior) angle is $(3x + 15)°$ and the other is $(2x + 25)°$. What is the value of $x$?

Solution

Co-interior angles sum to $180°$: $(3x+15)+(2x+25)=180$ → $5x+40=180$ → $5x=140$ → $x = \mathbf{28}$ ✓
Trap: Setting them EQUAL (as if they were alternate angles). Co-interior angles are supplementary (sum to 180°), NOT equal!