Algebra 2 — Semester A
Free Practice · 10 Questions · 20 min
20:00 Exit
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Question 1 of 10
TEKS 1A-1G Medium Diagram

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

x + 6x + 4
A 2x + 10
B x² + 24
C x² + 10x + 24
D x² + 10x + 10
Explanation
(x + 4)(x + 6) = x² + 6x + 4x + 24 = x² + 10x + 24.
Question 2 of 10
TEKS 1A-1G Easy
A rectangle has length 2x + 3 and width x − 1. Which polynomial represents its area?
A 2x² + x − 3
B 2x² + 5x − 3
C 2x² + 4x − 3
D 2x² − x − 3
Explanation
(2x + 3)(x − 1) = 2x² − 2x + 3x − 3 = 2x² + x − 3.
Question 3 of 10
TEKS 2A-2I Easy Diagram

Which transformation produces this graph from y = x²?

A Horizontal shift right
B Reflection across the x-axis
C Reflection across the y-axis
D Vertical shift down
Explanation
Standard parabola opens up; this one opens down — that is y = −x², a reflection across the x-axis.
Question 4 of 10
TEKS 4A-4F Hard
Solve using the quadratic formula: 3x² − 4x − 1 = 0.
A x = (4 ± √16) / 6
B x = (4 ± √28) / 3
C x = (4 ± √28) / 6
D x = (−4 ± √28) / 6
Explanation
a=3, b=−4, c=−1. Discriminant = 16 + 12 = 28. x = (4 ± √28) / 6.
Question 5 of 10
TEKS 2A-2I Easy Diagram

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

A f(x) = √x
B f(x) = |x|
C f(x) = x
D f(x) = x²
Explanation
Sharp V-shape with a corner at the vertex is the absolute-value parent function. Quadratic would be smooth.
Question 6 of 10
TEKS 4A-4F Medium
Solve by completing the square: x² + 6x + 5 = 0.
A x = −1 or x = −5
B x = −1 or x = 5
C x = 1 or x = 5
D x = −2 or x = −4
Explanation
x² + 6x = −5 → x² + 6x + 9 = 4 → (x + 3)² = 4 → x + 3 = ±2 → x = −1 or x = −5.
Question 7 of 10
TEKS 2A-2I Hard
If f(x) = √x and g(x) = x − 2, what is the domain of (f ∘ g)(x)?
A x ≥ 2
B x > 2
C x ≥ 0
D all real numbers
Explanation
(f ∘ g)(x) = √(x − 2). Inside even root must be ≥ 0, so x − 2 ≥ 0, i.e., x ≥ 2.
Question 8 of 10
TEKS 4A-4F Medium
The discriminant of x² + 4x + 5 = 0 is:
A −16
B 0 (one real root)
C −4 (no real roots)
D 4 (two real roots)
Explanation
Discriminant = b² − 4ac = 16 − 20 = −4. Since it is negative, the equation has two complex (non-real) roots.
Question 9 of 10
TEKS 2A-2I Medium
If f(x) = 2x + 1, what is the inverse f⁻¹(x)?
A 2x − 1
B (x + 1) / 2
C 2 / (x + 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 2A-2I Medium
If f(x) = x + 3 and g(x) = x², what is (f ∘ g)(2)?
A 11
B 7
C 25
D 5
Explanation
(f ∘ g)(x) = f(g(x)) = f(x²) = x² + 3. At x = 2: 4 + 3 = 7.

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