Self-Study Guide

Algebra 2 & Geometry

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Algebra 2
10 Core Questions Β· Key Units Β· Tricky Traps
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Quick Memory Points

Algebra 2 Power Words

FOIL DISCRIMINANT VERTEX = -b/2a LOG↔EXP REMAINDER THM ASYMPTOTE CONJUGATE PAIR f⁻¹ swap x,y
Question 01 Quadratic
Which of the following is the vertex form of the quadratic \( f(x) = x^2 - 6x + 5 \)?
πŸ’‘ Complete the square β€” don't just guess!
πŸ“– Explanation
Complete the square: \(x^2 - 6x + 5 = (x^2 - 6x + 9) - 9 + 5 = (x-3)^2 - 4\).
VERTEX FORMULA: \(h = -b/2a = -(-6)/2 = 3\), then \(k = f(3) = 9-18+5 = -4\). Vertex \((3, -4)\).
Trap: Option A has \(+4\) instead of \(-4\) β€” a sign error students often make!
Question 02 Discriminant
How many real solutions does \( 2x^2 - 4x + 5 = 0 \) have?
πŸ’‘ Check the discriminant \(b^2 - 4ac\) before solving!
πŸ“– Explanation
\(\Delta = b^2 - 4ac = (-4)^2 - 4(2)(5) = 16 - 40 = -24\)
Since \(\Delta < 0\), there are no real solutions β€” only two complex conjugate roots.
Key rule: \(\Delta > 0\) β†’ 2 real roots Β· \(\Delta = 0\) β†’ 1 repeated root Β· \(\Delta < 0\) β†’ no real roots.
Question 03 Polynomials
When \( p(x) = x^3 - 4x^2 + x + 6 \) is divided by \( (x - 3) \), what is the remainder?
πŸ’‘ REMAINDER THEOREM: just plug in \(x = 3\) !
πŸ“– Explanation
Remainder Theorem: \(p(3) = 27 - 36 + 3 + 6 = 0\)
Remainder \(= 0\), so \((x-3)\) is actually a factor of \(p(x)\)!
Factor Theorem: if \(p(a) = 0\), then \((x-a)\) is a factor.
Question 04 Logarithms
Solve: \( \log_2(x + 3) = 4 \)
πŸ’‘ LOG↔EXP: rewrite as \(2^4 = x + 3\)
πŸ“– Explanation
Convert: \(\log_2(x+3) = 4\) β†’ \(2^4 = x+3\) β†’ \(16 = x + 3\) β†’ \(x = 13\)
LOG ↔ EXP rule: \(\log_b(y) = x \Leftrightarrow b^x = y\)
Always check domain: \(x + 3 > 0\) β†’ \(x > -3\). βœ“
Question 05 Functions
If \( f(x) = 3x - 1 \), find the inverse function \( f^{-1}(x) \).
πŸ’‘ INVERSE: swap \(x\) and \(y\), then solve for \(y\)
πŸ“– Explanation
Let \(y = 3x-1\). Swap: \(x = 3y-1\). Solve: \(3y = x+1\) β†’ \(y = \dfrac{x+1}{3}\)
Verify: \(f(f^{-1}(x)) = 3\cdot\dfrac{x+1}{3}-1 = x+1-1 = x\) βœ“
Trap: Option B uses \(x-1\) β€” sign error from forgetting to move \(-1\) to the other side as \(+1\).
Question 06 Rational Functions
What is the vertical asymptote of \( f(x) = \dfrac{2x+1}{x-4} \)?
πŸ’‘ ASYMPTOTE: set the denominator = 0
πŸ“– Explanation
Set denominator \(= 0\): \(x - 4 = 0\) β†’ \(x = 4\) is the vertical asymptote.
Option A (\(x = -\tfrac{1}{2}\)) is the x-intercept (numerator = 0), not the asymptote.
Option B (\(y = 2\)) is the horizontal asymptote (ratio of leading coefficients).
Question 07 Complex Numbers
Simplify: \( (3 + 2i)(1 - i) \)
πŸ’‘ FOIL it, then remember \(i^2 = -1\)
πŸ“– Explanation
FOIL: \(3(1) + 3(-i) + 2i(1) + 2i(-i) = 3 - 3i + 2i - 2i^2\)
Since \(i^2 = -1\): \(= 3 - i - 2(-1) = 3 - i + 2 = \mathbf{5 - i}\)
Most common trap: forgetting \(i^2 = -1\) and writing \(3-i+2i^2 = 3-i-2 = 1-i\) (Option A).
Question 08 Sequences
Find the 10th term of the geometric sequence: \(2, 6, 18, 54, \ldots\)
πŸ’‘ GEO SEQ formula: \(a_n = a_1 \cdot r^{n-1}\)
πŸ“– Explanation
\(a_1 = 2\), \(r = 3\), \(n = 10\)
\(a_{10} = 2 \cdot 3^{10-1} = 2 \cdot 3^9 = 2 \cdot 19683 = 39366\)
Off-by-one trap: Using \(r^n\) instead of \(r^{n-1}\) is the #1 mistake here.
Question 09 Systems
Solve the system: \(\begin{cases} y = x^2 - 2 \\ y = x + 4 \end{cases}\)
πŸ’‘ Set equal: \(x^2 - 2 = x + 4\), then solve the quadratic
πŸ“– Explanation
Set equal: \(x^2-2 = x+4\) β†’ \(x^2 - x - 6 = 0\) β†’ \((x-3)(x+2)=0\)
So \(x = 3\) or \(x = -2\). Intersection points: \((3,7)\) and \((-2,2)\).
Trap: Option D flips the signs β€” always check factoring carefully!
Question 10 Exponential
Solve for \(x\): \( 5^{2x} = 125 \)
πŸ’‘ SAME BASE trick: express 125 as a power of 5
πŸ“– Explanation
\(125 = 5^3\), so \(5^{2x} = 5^3\) β†’ same base β†’ \(2x = 3\) β†’ \(x = \dfrac{3}{2}\)
Same Base Rule: if \(b^m = b^n\), then \(m = n\) (as long as \(b > 0,\ b \neq 1\))
Geometry
Trigonometry: Sin Β· Cos Β· Tan Β· From Basics to Application
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Quick Memory Points

Trig Power Words

SOH-CAH-TOA HYPOTENUSE OPPOSITE / ADJACENT sinΒ²+cosΒ²=1 tan=sin/cos 30-60-90 45-45-90 UNIT CIRCLE
Question 01 Basic Definition
In a right triangle, the side opposite angle \(\theta\) has length 5 and the hypotenuse has length 13. What is \(\sin\theta\)?
πŸ’‘ SOH: Sin = Opposite / Hypotenuse
πŸ“– Explanation
SOH: \(\sin\theta = \dfrac{\text{Opposite}}{\text{Hypotenuse}} = \dfrac{5}{13}\)
Using Pythagorean theorem, adjacent = \(\sqrt{13^2 - 5^2} = \sqrt{144} = 12\)
So \(\cos\theta = \dfrac{12}{13}\) and \(\tan\theta = \dfrac{5}{12}\). Classic 5-12-13 triangle!
Question 02 SOH-CAH-TOA
In a right triangle with adjacent = 8 and hypotenuse = 10, what is \(\tan\theta\)?
πŸ’‘ First find the opposite side using Pythagorean theorem!
πŸ“– Explanation
Step 1 β€” find opposite: \(\text{opp} = \sqrt{10^2 - 8^2} = \sqrt{36} = 6\)
TOA: \(\tan\theta = \dfrac{\text{Opposite}}{\text{Adjacent}} = \dfrac{6}{8} = \dfrac{3}{4}\)
Trap: Option A is \(\cos\theta\), Option B is \(\sin\theta\). Know which ratio is which!
Question 03 Special Angles
What is the exact value of \(\sin 30Β°\)?
πŸ’‘ Memorize: 30-60-90 triangle sides are 1 : √3 : 2
πŸ“– Explanation
In a 30-60-90 triangle: opp(30Β°)=1, hyp=2 β†’ \(\sin 30Β° = \dfrac{1}{2}\)
Full table to memorize:
\(\sin 30Β° = \tfrac{1}{2}\), \(\cos 30Β° = \tfrac{\sqrt{3}}{2}\), \(\tan 30Β° = \tfrac{1}{\sqrt{3}}\)
Trap: Option A is \(\cos 30Β°\), not \(\sin 30Β°\). Students often swap them!
Question 04 Special Angles
What is \(\tan 45Β°\)?
πŸ’‘ 45-45-90 triangle: sides are 1 : 1 : √2
πŸ“– Explanation
45-45-90 triangle: both legs = 1. \(\tan 45Β° = \dfrac{1}{1} = 1\)
\(\sin 45Β° = \cos 45Β° = \dfrac{\sqrt{2}}{2}\) β€” but \(\tan = \dfrac{\sin}{\cos} = \dfrac{\sqrt{2}/2}{\sqrt{2}/2} = 1\)
Trap: Options A & B are \(\sin\)/\(\cos\) of 45Β°. \(\tan 45Β°\) is the clean answer: just \(1\).
Question 05 Pythagorean Identity
If \(\sin\theta = \dfrac{3}{5}\), what is \(\cos\theta\)? (Assume \(\theta\) is in the first quadrant.)
πŸ’‘ IDENTITY: \(\sin^2\theta + \cos^2\theta = 1\)
πŸ“– Explanation
\(\cos^2\theta = 1 - \sin^2\theta = 1 - \dfrac{9}{25} = \dfrac{16}{25}\) β†’ \(\cos\theta = \dfrac{4}{5}\) (Q1, positive)
Alternatively: 3-4-5 right triangle. opp=3, hyp=5, adj=4. \(\cos = \dfrac{4}{5}\)
Identity shortcut: memorize 3-4-5, 5-12-13, 8-15-17 Pythagorean triples!
Question 06 Finding Sides
A right triangle has angle \(\theta = 60Β°\) and hypotenuse = 10. What is the length of the side opposite to \(\theta\)?
πŸ’‘ SOH: opposite = hypotenuse Γ— sin(60Β°)
πŸ“– Explanation
\(\sin 60Β° = \dfrac{\text{opp}}{\text{hyp}}\) β†’ \(\text{opp} = 10 \times \sin 60Β° = 10 \times \dfrac{\sqrt{3}}{2} = 5\sqrt{3}\)
Trap: Option A (= 5) comes from \(\sin 30Β°\), not \(\sin 60Β°\)! Make sure you use the right angle.
Question 07 Angle of Elevation
A ladder leans against a wall. The ladder is 10 m long and makes a 60Β° angle with the ground. How high up the wall does the ladder reach?
πŸ’‘ Draw a right triangle! The ground is adjacent, the wall is opposite.
πŸ“– Explanation
The angle 60Β° is between the ladder (hyp = 10m) and the ground. The wall is the opposite side.
\(\sin 60Β° = \dfrac{\text{height}}{10}\) β†’ height \(= 10\sin 60Β° = 10 \cdot \dfrac{\sqrt{3}}{2} = 5\sqrt{3}\)
Real-world tip: always identify what is the hypotenuse first (the ladder)!
Question 08 Tan Identity
Which expression is equivalent to \(\tan\theta\)?
πŸ’‘ TAN = SIN / COS β€” this is a fundamental identity!
πŸ“– Explanation
Quotient Identity: \(\tan\theta = \dfrac{\sin\theta}{\cos\theta}\)
Option A is \(\cot\theta\) (cotangent). Option D is \(\csc\theta\) (cosecant).
Memorize all 6 trig functions: sin, cos, tan, csc, sec, cot.
Question 09 Quadrant Signs
In which quadrant is \(\sin\theta > 0\) and \(\cos\theta < 0\)?
πŸ’‘ Mnemonic: "All Students Take Calculus" β€” which functions are positive in each quadrant?
πŸ“– Explanation
"All Students Take Calculus":
Q1 (All positive) Β· Q2 (Sin positive) Β· Q3 (Tan positive) Β· Q4 (Cos positive)
\(\sin > 0\) β†’ Q1 or Q2. \(\cos < 0\) β†’ Q2 or Q3. Intersection = Quadrant II.
Question 10 Finding Angles
In a right triangle, opposite = 7 and adjacent = 7. What is angle \(\theta\)?
πŸ’‘ Find tan ΞΈ first, then use inverse tan (arctan)
πŸ“– Explanation
\(\tan\theta = \dfrac{7}{7} = 1\) β†’ \(\theta = \arctan(1) = 45Β°\)
When opposite = adjacent, the triangle is a 45-45-90 isoceles right triangle.
Inverse trig: use \(\sin^{-1}\), \(\cos^{-1}\), \(\tan^{-1}\) to find angle from ratio.
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