High School Algebra 2 practice questions and exam guide

The equations 2x + 4y = 8 and x + 2y = 4 form a system with:

1. no solution
2. exactly one solution
3. the solution (0, 0)
4. infinitely many solutions

Answer: D — infinitely many solutions

D) Correct — The first equation is twice the second, so they are the same line. A) Would need parallel, different lines. B) The lines coincide. C) (0, 0) doesn't satisfy either equation.

What is the inverse of f(x) = 2x + 3?

1. f⁻¹(x) = (x + 3)/2
2. f⁻¹(x) = 2x − 3
3. f⁻¹(x) = 1/(2x + 3)
4. f⁻¹(x) = (x − 3)/2

Answer: D — f⁻¹(x) = (x − 3)/2

D) Correct — Swap x and y: x = 2y + 3, then solve for y. A) Adds 3 instead of subtracting. B) Only changes the sign of the constant. C) Takes the reciprocal, not the inverse function.

What is the sum of the roots of 2x² − 7x + 3 = 0?

1. −7/2
2. 7/2
3. 3/2
4. 7

Answer: B — 7/2

B) Correct — Sum of roots = −b/a. A) Uses b/a instead of −b/a. C) Is the product of the roots, c/a. D) Forgets to divide by a.

A ball's height is h = −16t² + 64t feet. What is its maximum height?

1. 2 ft
2. 48 ft
3. 64 ft
4. 128 ft

Answer: C — 64 ft

C) Correct — The vertex is at t = 64/32 = 2; h(2) = −64 + 128. A) Is the time of the maximum, not the height. B) Is the height at t = 1. D) Leaves out the −16t² term.

As x → ∞, what happens to f(x) = −2x³ + 5x² − 7?

1. f(x) → −∞
2. f(x) → ∞
3. f(x) → −7
4. f(x) → 0

Answer: A — f(x) → −∞

A) Correct — The leading term −2x³ dominates and becomes very negative. B) Ignores the negative leading coefficient. C) Uses the constant term. D) Polynomials don't level off.

If p(3) = 0, the Factor Theorem says:

1. (x + 3) is a factor of p(x)
2. p(x) has a remainder of 3
3. 3 is the leading coefficient
4. (x − 3) is a factor of p(x)

Answer: D — (x − 3) is a factor of p(x)

D) Correct — A zero at x = c means (x − c) is a factor. A) Gets the sign wrong. B) The remainder is 0. C) Is unrelated.

Simplify (x² − 9)/(x² − x − 6).

1. (x − 3)/(x − 2)
2. (x + 3)/(x − 2)
3. (x + 3)/(x + 2)
4. 3/2

Answer: C — (x + 3)/(x + 2)

C) Correct — (x − 3)(x + 3) / ((x − 3)(x + 2)). A) Factors the denominator incorrectly. B) Gets the sign in the denominator factor wrong. D) Cancels terms instead of factors.

Evaluate 16^(3/4).

1. 12
2. 64
3. 8
4. 4096

Answer: C — 8

C) Correct — Take the fourth root first: 16^(1/4) = 2, then 2³. A) Multiplies 16 by 3/4. B) Computes 4³ instead. D) Cubes 16 before taking the root, then stops.

Evaluate log₂(8).

1. 4
2. 3
3. 16
4. 6

Answer: B — 3

B) Correct — 2³ = 8. A) Divides 8 by 2. C) Multiplies 8 by 2. D) Subtracts 2 from 8.

An arithmetic sequence has a₁ = 3 and d = 4. What is a₁₅?

1. 63
2. 59
3. 56
4. 60

Answer: B — 59

B) Correct — aₙ = a₁ + (n − 1)d = 3 + 14 · 4. A) Uses n · d instead of (n − 1)d. C) Forgets to add a₁. D) Computes 15 · 4.

A circle has radius 4. What is the arc length for a central angle of π/3?

1. 4π/3
2. 8π/3
3. π/12
4. 240

Answer: A — 4π/3

A) Correct — s = rθ = 4 · π/3. B) Uses the diameter. C) Divides θ by r. D) Multiplies 4 by 60°.

Two cards are drawn without replacement. What is P(both hearts)?

1. 1/16
2. 1/4
3. 1/17
4. 12/51

Answer: C — 1/17

C) Correct — (13/52)(12/51) = 156/2652. A) Uses (1/4)(1/4), as if the card were replaced. B) Gives the probability for one card. D) Gives only the second-draw probability.

Worked Example: Simplify the complex expression (3 + 2i)(4 − 5i).

Step 1: Apply FOIL (distribute). (3 + 2i)(4 − 5i) = 3·4 + 3·(−5i) + 2i·4 + 2i·(−5i) = 12 − 15i + 8i − 10i² Step 2: Substitute i² = −1. −10i² = −10·(−1) = 10 Expression becomes 12 − 15i + 8i + 10 Step 3: Combine real parts and imaginary parts separately. Real: 12 + 10 = 22 Imaginary: −15i + 8i = −7i Step 4: Write in standard form a + bi. 22 − 7i Final answer: 22 − 7i

Standard equation of a circle centered at (h, k) with radius r?

(x − h)² + (y − k)² = r².

Simplify √50.

5√2. (√(25 × 2) = 5√2.)

Worked Example: Convert y = 2x² − 12x + 23 to vertex form by completing the square.

Step 1: Factor the leading coefficient out of the x-terms. y = 2(x² − 6x) + 23 Step 2: Take half of the linear coefficient (−6), square it, and add/subtract inside the parentheses. Half of −6 = −3; (−3)² = 9 y = 2(x² − 6x + 9 − 9) + 23 Step 3: Group the perfect-square trinomial. y = 2[(x − 3)² − 9] + 23 Step 4: Distribute the 2 to the −9 and combine constants. y = 2(x − 3)² − 18 + 23 y = 2(x − 3)² + 5 Step 5: Identify the vertex. Vertex form y = a(x − h)² + k → vertex (h, k) = (3, 5). Final answer: y = 2(x − 3)² + 5; vertex (3, 5).

Multiply: (x − 3)(x + 3).

x² − 9. (Difference of squares.)

What's the 'vertex form' of a quadratic?

y = a(x − h)² + k. The vertex is (h, k).