Algebra 2 — Semester A Practice

Question 1 of 10

TEKS 1A-1GMedium Diagram

A poster has dimensions (x + 4) by (x + 6). Which expression represents its area?

Ax² + 24

B2x + 10

Cx² + 10x + 24

Dx² + 10x + 10

Explanation

(x + 4)(x + 6) = x² + 6x + 4x + 24 = x² + 10x + 24.

Question 2 of 10

TEKS 4A-4FHard

Solve using the quadratic formula: 3x² − 4x − 1 = 0.

Ax = (4 ± √28) / 6

Bx = (4 ± √28) / 3

Cx = (−4 ± √28) / 6

Dx = (4 ± √16) / 6

Explanation

a=3, b=−4, c=−1. Discriminant = 16 + 12 = 28. x = (4 ± √28) / 6.

Question 3 of 10

TEKS 2A-2IHard

If f(x) = √x and g(x) = x − 2, what is the domain of (f ∘ g)(x)?

Ax ≥ 0

Bx > 2

Call real numbers

Dx ≥ 2

Explanation

(f ∘ g)(x) = √(x − 2). Inside even root must be ≥ 0, so x − 2 ≥ 0, i.e., x ≥ 2.

Question 4 of 10

TEKS 2A-2IMedium

If f(x) = x + 3 and g(x) = x², what is (f ∘ g)(2)?

C25

D11

Explanation

(f ∘ g)(x) = f(g(x)) = f(x²) = x² + 3. At x = 2: 4 + 3 = 7.

Question 5 of 10

TEKS 2A-2IEasy Diagram

Which transformation produces this graph from y = x²?

AHorizontal shift right

BVertical shift down

CReflection across the y-axis

DReflection across the x-axis

Explanation

Standard parabola opens up; this one opens down — that is y = −x², a reflection across the x-axis.

Question 6 of 10

TEKS 1A-1GEasy

A rectangle has length 2x + 3 and width x − 1. Which polynomial represents its area?

A2x² + x − 3

B2x² + 4x − 3

C2x² − x − 3

D2x² + 5x − 3

Explanation

(2x + 3)(x − 1) = 2x² − 2x + 3x − 3 = 2x² + x − 3.

Question 7 of 10

TEKS 4A-4FMedium

The discriminant of x² + 4x + 5 = 0 is:

A−16

B0 (one real root)

C−4 (no real roots)

D4 (two real roots)

Explanation

Discriminant = b² − 4ac = 16 − 20 = −4. Since it is negative, the equation has two complex (non-real) roots.

Question 8 of 10

TEKS 2A-2IEasy Diagram

The graph below shows g(x). What is the parent function?

Af(x) = x

Bf(x) = x²

Cf(x) = |x|

Df(x) = √x

Explanation

Sharp V-shape with a corner at the vertex is the absolute-value parent function. Quadratic would be smooth.

Question 9 of 10

TEKS 2A-2IMedium

If f(x) = 2x + 1, what is the inverse f⁻¹(x)?

A(x + 1) / 2

B2 / (x + 1)

C2x − 1

D(x − 1) / 2

Explanation

Swap x and y: x = 2y + 1. Solve for y: y = (x − 1)/2. So f⁻¹(x) = (x − 1)/2.

Question 10 of 10

TEKS 4A-4FMedium

Solve by completing the square: x² + 6x + 5 = 0.

Ax = −1 or x = −5

Bx = −1 or x = 5

Cx = −2 or x = −4

Dx = 1 or x = 5

Explanation

x² + 6x = −5 → x² + 6x + 9 = 4 → (x + 3)² = 4 → x + 3 = ±2 → x = −1 or x = −5.

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