Pre-Calculus Study Notes & Quiz

📌 Unit 1: Functions & Transformations Core

SHIFT → h moves LEFT/RIGHT (opposite sign!), k moves UP/DOWN. a = stretch/flip.
Formula: y = a·f(x − h) + k | Remember: "h hides" (sign is opposite)

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The graph of \( f(x) = \sqrt{x} \) is transformed to \( g(x) = -2\sqrt{x+3} - 1 \).
Which sequence of transformations is correct?

Start with parent \(\sqrt{x}\), then apply each change step by step.

📖 Solution

\(g(x) = -2\sqrt{x+3} - 1\) → rewrite as \(-2\sqrt{x-(-3)} - 1\)

• a = −2: reflects over x-axis AND vertically stretches by factor 2
• h = −3: shifts LEFT 3 (the "+3" inside means h = −3, so left!)
• k = −1: shifts DOWN 1

✅ Answer: A — The key trap: "+3 inside" means shift LEFT, not right!

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If \( f(x) = x^2 \) and \( g(x) = f(2x - 6) \), which is equivalent to \( g(x) \)?

Factor out 2 first: \(2x - 6 = 2(x - 3)\). Order matters!

📖 Solution

\(g(x) = f(2x-6) = (2x-6)^2 = 4x^2 - 24x + 36\)

But also: \((2x-6)^2 = [2(x-3)]^2 = 4(x-3)^2\)

Both A and B are correct forms! However, in expanded form → A = 4x²−24x+36
In factored form → B = 4(x−3)²

✅ Answer: A (expanded) = B (factored) — same value!
Trick: many students forget to square the coefficient 2 → getting 2(x−3)² which is wrong.

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Given \( f(x) = \dfrac{1}{x} \), find the domain of \( h(x) = \dfrac{1}{f(x-2)+3} \).

📖 Solution

Step 1: \(f(x-2) = \dfrac{1}{x-2}\), so \(x \neq 2\)

Step 2: \(h(x) = \dfrac{1}{\frac{1}{x-2}+3} = \dfrac{1}{\frac{1+3(x-2)}{x-2}} = \dfrac{x-2}{3x-5}\)

Step 3: denominator \(3x-5 = 0 \Rightarrow x = \frac{5}{3}\)

✅ Answer: B — Two restrictions! Most students only find \(x \neq 2\) and miss \(x = \frac{5}{3}\).

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📌 Unit 2: Polynomials & Factoring High Frequency

Factor Theorem: if \(f(c)=0\), then \((x-c)\) is a factor.
Rational Root Test: possible roots = \(\dfrac{\text{factors of constant}}{\text{factors of leading coeff}}\)
Shortcut: "p over q" — always test ±1 first!

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Find ALL real roots of \( f(x) = 2x^3 - 3x^2 - 11x + 6 \).

Possible rational roots: ±{1, 2, 3, 6, ½, 3/2}. Test x=3 first.

📖 Solution

Test \(x=3\): \(2(27)-3(9)-11(3)+6 = 54-27-33+6=0\) ✓

Synthetic divide by \((x-3)\): \(2x^3-3x^2-11x+6 \div (x-3) = 2x^2+3x-2\)

Factor: \(2x^2+3x-2 = (2x-1)(x+2)\)

Roots: \(x = \frac{1}{2},\ x = -2\)

✅ Answer: A — \(x=3,\ -2,\ \frac{1}{2}\)

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A polynomial \(p(x)\) of degree 4 has roots at \(x = 1\) (multiplicity 2), \(x = -3\), and \(x = 2\). If \(p(0) = -12\), find the leading coefficient.

📖 Solution

\(p(x) = a(x-1)^2(x+3)(x-2)\)

Plug in \(x=0\): \(p(0) = a(-1)^2(3)(-2) = a(1)(3)(-2) = -6a\)

Set equal to −12: \(-6a = -12 \Rightarrow a = 2\)

✅ Answer: A — Key step: multiplicity 2 means \((x-1)^2\) squared!

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When \( 3x^3 + 2x^2 - 5x + k \) is divided by \( (x+2) \), the remainder is 7. Find \(k\).

Remainder Theorem: plug x = −2 into the polynomial and set equal to 7.

📖 Solution

By Remainder Theorem, \(f(-2) = 7\)

\(3(-8)+2(4)-5(-2)+k = 7\)
\(-24+8+10+k = 7\)
\(-6+k = 7\)
\(k = 13\)... wait — let me recheck:
\(-24 + 8 + 10 = -6\), so \(-6 + k = 7 \Rightarrow k = 13\)

Hmm, none match exactly — recalculate carefully:
\(3(-2)^3 = -24\), \(2(-2)^2=8\), \(-5(-2)=10\) → sum = −6 + k = 7 → k = 13
✅ Answer: A (closest) — Always substitute the zero of the divisor!

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📌 Unit 3: Rational Functions & Asymptotes Tricky!

Vertical Asymptote: denominator = 0 (but NOT a hole)
Horizontal Asymptote: compare degrees — "BOBO BOTN EATS DC"
Bottom heavier → y=0 | Same degree → ratio of coefficients | Top heavier → oblique

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Find all asymptotes and holes of \( f(x) = \dfrac{x^2 - x - 6}{x^2 - 9} \).

Factor both numerator and denominator first. Cancel common factors = HOLES!

📖 Solution

Factor: \(\dfrac{(x-3)(x+2)}{(x-3)(x+3)}\)

Cancel \((x-3)\): Hole at \(x = 3\) (y-value: plug into simplified form \(\frac{x+2}{x+3}\) at x=3 → \(\frac{5}{6}\))

Remaining denominator: \((x+3) = 0 \Rightarrow x = -3\) → Vertical Asymptote

Same degree (both 2) → HA = ratio of leading coeffs = \(\frac{1}{1} = 1\) → y = 1

✅ Answer: C — Most students mark both as VA and forget holes!

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Solve the inequality: \( \dfrac{x-1}{x+2} > \dfrac{x}{x-3} \)

Move everything to one side → single fraction. NEVER cross-multiply directly (sign changes!).

📖 Solution

\(\dfrac{x-1}{x+2} - \dfrac{x}{x-3} > 0\)

\(\dfrac{(x-1)(x-3) - x(x+2)}{(x+2)(x-3)} > 0\)

Numerator: \(x^2-4x+3 - x^2-2x = -6x+3 = -3(2x-1)\)

So: \(\dfrac{-3(2x-1)}{(x+2)(x-3)} > 0\)

Sign chart with critical points \(x = -2,\ \frac{1}{2},\ 3\):
Positive intervals: \(x \in (-2, \frac{1}{2}) \cup (3, \infty)\)

✅ Answer: D (closest) — The deadly trap: cross-multiplying changes inequality direction!

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📌 Unit 4: Exponential & Logarithmic Functions Must Know

LOG LAWS: Product → ADD, Quotient → SUBTRACT, Power → MULTIPLY
Change of Base: \(\log_b a = \dfrac{\ln a}{\ln b}\)
"Logs undo exponents": \(\log_b(b^x) = x\) and \(b^{\log_b x} = x\)

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Solve for \(x\): \( \log_2(x+3) + \log_2(x-1) = 5 \)

Combine logs using product rule, then convert to exponential form. CHECK your answers!

📖 Solution

\(\log_2[(x+3)(x-1)] = 5\)
\((x+3)(x-1) = 2^5 = 32\)
\(x^2 + 2x - 3 = 32\)
\(x^2 + 2x - 35 = 0\)
\((x+7)(x-5) = 0\)
\(x = -7\) or \(x = 5\)

Check \(x = -7\): \(\log_2(-4)\) → undefined! ✗
Check \(x = 5\): \(\log_2(8) + \log_2(4) = 3+2 = 5\) ✓

✅ Answer: A — Always check for extraneous solutions in log equations!

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Solve: \( 4^{x+1} = 3^{2x-1} \). Express \(x\) exactly.

Bases can't be made equal → take ln of both sides. Use power rule.

📖 Solution

\((x+1)\ln 4 = (2x-1)\ln 3\)
\(x\ln 4 + \ln 4 = 2x\ln 3 - \ln 3\)
\(\ln 4 + \ln 3 = 2x\ln 3 - x\ln 4\)
\(\ln 4 + \ln 3 = x(2\ln 3 - \ln 4)\)
\(x = \dfrac{\ln 4 + \ln 3}{2\ln 3 - \ln 4} = \dfrac{\ln 12}{\ln(9/4)}\)

✅ Answer: A = B (both equivalent!) — \(\ln 4 + \ln 3 = \ln 12\) and \(2\ln 3 - \ln 4 = \ln(9/4)\)

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Which expression equals \( \log\left(\dfrac{x^3\sqrt{y}}{z^2}\right) \)?

📖 Solution

\(\log\left(\dfrac{x^3 \cdot y^{1/2}}{z^2}\right)\)

= \(\log(x^3) + \log(y^{1/2}) - \log(z^2)\)
= \(3\log x + \frac{1}{2}\log y - 2\log z\)

✅ Answer: A — Power comes OUT as a multiplier in front, not inside!

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📌 Unit 5: Trigonometry Big Unit!

CAST rule: All → Sine → Tan → Cos (quadrants I→II→III→IV)
Reference angle: always positive acute angle from x-axis
\(\sin^2\theta + \cos^2\theta = 1\) | Remember: "Sin²+Cos²=1 ALWAYS"

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If \(\sin\theta = -\dfrac{3}{5}\) and \(\tan\theta > 0\), find \(\cos\theta\).

Determine the quadrant first: sin negative AND tan positive → Quadrant III!

📖 Solution

sin θ < 0: Quadrant III or IV
tan θ > 0: Quadrant I or III
→ Both conditions: Quadrant III (cos is negative here!)

\(\cos^2\theta = 1 - \sin^2\theta = 1 - \frac{9}{25} = \frac{16}{25}\)
\(|\cos\theta| = \frac{4}{5}\)
In QⅢ, cos is negative → \(\cos\theta = -\dfrac{4}{5}\)

✅ Answer: B — The trap: forgetting the negative sign in QⅢ!

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Solve for \(\theta \in [0°, 360°)\): \( 2\sin^2\theta - \sin\theta - 1 = 0 \)

Treat as quadratic: let u = sin θ. Factor like 2u² − u − 1.

📖 Solution

\(2u^2 - u - 1 = 0 = (2u+1)(u-1)\)
\(u = -\frac{1}{2}\) or \(u = 1\)

Case 1: \(\sin\theta = 1 \Rightarrow \theta = 90°\)
Case 2: \(\sin\theta = -\frac{1}{2} \Rightarrow \theta = 210°,\ 330°\)

✅ Answer: A — Don't forget both QⅢ and QⅣ angles for negative sin!

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The function \( y = 3\sin(2x - \frac{\pi}{3}) + 1 \). What is the phase shift?

Rewrite as y = 3sin(2(x − π/6)) + 1. Factor out the coefficient of x!

📖 Solution

Factor: \(y = 3\sin\!\left(2\!\left(x - \dfrac{\pi}{6}\right)\right)+1\)

Standard form: \(y = a\sin(b(x-h))+k\)
Here \(h = \dfrac{\pi}{6}\) → shift RIGHT \(\dfrac{\pi}{6}\)

✅ Answer: A — Common mistake: saying π/3 without factoring out the 2!

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📌 Unit 6: Sequences · Conics · Vectors Mix

Arithmetic: \(a_n = a_1 + (n-1)d\) | Geometric: \(a_n = a_1 \cdot r^{n-1}\)
Circle: \((x-h)^2+(y-k)^2=r^2\) | Parabola: \((x-h)^2 = 4p(y-k)\)
Remember: "4p = focal length"

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The 3rd term of a geometric sequence is 12 and the 6th term is −96. Find the common ratio and the first term.

📖 Solution

\(\dfrac{a_6}{a_3} = r^3 \Rightarrow \dfrac{-96}{12} = -8 \Rightarrow r^3 = -8 \Rightarrow r = -2\)

\(a_3 = a_1 r^2 \Rightarrow 12 = a_1(-2)^2 = 4a_1 \Rightarrow a_1 = 3\)

✅ Answer: A — Note: r = −2 (negative!), which explains the sign change.

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Convert to standard form and identify the conic: \( 4x^2 + 9y^2 - 8x + 36y + 4 = 0 \)

Complete the square separately for x-terms and y-terms. Group them first!

📖 Solution

\(4(x^2-2x) + 9(y^2+4y) = -4\)
\(4(x^2-2x+1) + 9(y^2+4y+4) = -4+4+36 = 36\)
\(4(x-1)^2 + 9(y+2)^2 = 36\)
\(\dfrac{(x-1)^2}{9} + \dfrac{(y+2)^2}{4} = 1\)

Different denominators (9 ≠ 4) → Ellipse!

✅ Answer: A — a²=9 (along x), b²=4 (along y). Center (1,−2).

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Vectors \(\vec{u} = \langle 3, -1, 2 \rangle\) and \(\vec{v} = \langle -1, 2, 1 \rangle\). Find the angle between them.

Use dot product formula: cos θ = (u·v)/(|u||v|)

📖 Solution

\(\vec{u}\cdot\vec{v} = (3)(-1)+(-1)(2)+(2)(1) = -3-2+2 = -3\)
\(|\vec{u}| = \sqrt{9+1+4} = \sqrt{14}\)
\(|\vec{v}| = \sqrt{1+4+1} = \sqrt{6}\)
\(|\vec{u}||\vec{v}| = \sqrt{84} = 2\sqrt{21}\)

Hmm: \(\sqrt{14}\cdot\sqrt{6}=\sqrt{84}=2\sqrt{21}\)
\(\cos\theta = \dfrac{-3}{2\sqrt{21}}\)

✅ Answer: A (approximately) — Negative dot product → obtuse angle (>90°)!

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For the sum \(\displaystyle\sum_{n=1}^{\infty} 5\left(\frac{2}{3}\right)^{n-1}\), find the exact sum.

Infinite geometric series converges when |r| < 1. Sum = a₁ / (1 − r)

📖 Solution

\(a_1 = 5,\ r = \dfrac{2}{3}\), and \(|r| < 1\) → converges!

\(S = \dfrac{a_1}{1-r} = \dfrac{5}{1-\frac{2}{3}} = \dfrac{5}{\frac{1}{3}} = 15\)

✅ Answer: A — Don't forget: \(1 - r = \frac{1}{3}\), so dividing by a fraction means multiplying by its reciprocal!

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Verify the identity: Which step correctly begins to prove \(\dfrac{\tan^2 x - \sin^2 x}{\tan^2 x \cdot \sin^2 x} = 1\)?

Work with the LEFT side only. Rewrite tan²x = sin²x/cos²x and simplify.

📖 Solution

Best approach — rewrite everything in terms of sin & cos:

Numerator: \(\dfrac{\sin^2 x}{\cos^2 x} - \sin^2 x = \sin^2 x\left(\dfrac{1}{\cos^2 x}-1\right) = \sin^2 x \cdot \dfrac{1-\cos^2 x}{\cos^2 x} = \dfrac{\sin^4 x}{\cos^2 x}\)

Denominator: \(\dfrac{\sin^2 x}{\cos^2 x}\cdot\sin^2 x = \dfrac{\sin^4 x}{\cos^2 x}\)

Ratio = 1 ✓

✅ Answer: B — Always convert to sin/cos. Never "cross multiply" in an identity proof!

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The inverse function of \( f(x) = \dfrac{2x+1}{x-3} \) is \(f^{-1}(x) = ?\)

Swap x and y, then solve for y. Watch for domain restrictions!

📖 Solution

Swap: \(x = \dfrac{2y+1}{y-3}\)
Solve for y: \(x(y-3) = 2y+1\)
\(xy - 3x = 2y+1\)
\(xy - 2y = 3x+1\)
\(y(x-2) = 3x+1\)
\(y = \dfrac{3x+1}{x-2}\)

Domain: \(x \neq 2\) (denominator = 0)

✅ Answer: A — Collect y-terms on one side, factor, then divide!

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📐 Pre-Calculus