SLCC MATH 1050 — College Algebra
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1. What is the domain of the function f(x) = 1/(x – 3)?
a) All real numbers
b) All real numbers except 0
c) All real numbers except 3
d) x > 3
Rationale: The denominator cannot be zero. Since x – 3 = 0 when x =
3, the domain excludes 3.
2. Solve for x: 2x – 5 = 9
a) 3
b) 7
, c) –3
d) 14
Rationale: Add 5 to both sides → 2x = 14 → x = 7.
3. Which of the following is the graph of y = –x² + 4?
a) Parabola opening upward, vertex (0,4)
b) Parabola opening downward, vertex (0,4)
c) Parabola opening upward, vertex (0,–4)
d) Line with slope –1
Rationale: The negative coefficient of x² means it opens downward;
constant term gives vertex at (0,4).
4. Which is equivalent to log₅(125)?
a) 2
b) 4
c) 3
d) 1/3
Rationale: 5³ = 125, so log₅(125) = 3.
5. Solve for x: (x – 4)(x + 2) = 0
a) –2
b) 4
c) –2 and 4
d) No solution
Rationale: Zero product property → x – 4 = 0 or x + 2 = 0 → x = 4 or –
2.
, 6. Which of the following is a horizontal asymptote of f(x) = (3x² – 1)/(x² +
2)?
a) y = 0
b) y = 3
c) y = 2
d) y = 1
Rationale: Degree of numerator = denominator, so horizontal
asymptote is ratio of leading coefficients: 3/1 = 3.
7. Solve: |x – 5| = 2
a) x = 3 only
b) x = 7 only
c) x = 3 or 7
d) No solution
Rationale: Absolute value equation splits into x – 5 = 2 → x = 7, and x
– 5 = –2 → x = 3.
8. The graph of y = (x – 2)² + 1 has its vertex at:
a) (0,0)
b) (0,1)
c) (2,1)
d) (–2,1)
Rationale: In vertex form (x – h)² + k, vertex is (h,k) = (2,1).
9. Solve for x: log₂(x) = 5
a) 10
Practice Exam Questions And Correct
Answers (Verified Answers) Plus
Rationales 2025|2026 Q&A | Instant
Download Pdf
1. What is the domain of the function f(x) = 1/(x – 3)?
a) All real numbers
b) All real numbers except 0
c) All real numbers except 3
d) x > 3
Rationale: The denominator cannot be zero. Since x – 3 = 0 when x =
3, the domain excludes 3.
2. Solve for x: 2x – 5 = 9
a) 3
b) 7
, c) –3
d) 14
Rationale: Add 5 to both sides → 2x = 14 → x = 7.
3. Which of the following is the graph of y = –x² + 4?
a) Parabola opening upward, vertex (0,4)
b) Parabola opening downward, vertex (0,4)
c) Parabola opening upward, vertex (0,–4)
d) Line with slope –1
Rationale: The negative coefficient of x² means it opens downward;
constant term gives vertex at (0,4).
4. Which is equivalent to log₅(125)?
a) 2
b) 4
c) 3
d) 1/3
Rationale: 5³ = 125, so log₅(125) = 3.
5. Solve for x: (x – 4)(x + 2) = 0
a) –2
b) 4
c) –2 and 4
d) No solution
Rationale: Zero product property → x – 4 = 0 or x + 2 = 0 → x = 4 or –
2.
, 6. Which of the following is a horizontal asymptote of f(x) = (3x² – 1)/(x² +
2)?
a) y = 0
b) y = 3
c) y = 2
d) y = 1
Rationale: Degree of numerator = denominator, so horizontal
asymptote is ratio of leading coefficients: 3/1 = 3.
7. Solve: |x – 5| = 2
a) x = 3 only
b) x = 7 only
c) x = 3 or 7
d) No solution
Rationale: Absolute value equation splits into x – 5 = 2 → x = 7, and x
– 5 = –2 → x = 3.
8. The graph of y = (x – 2)² + 1 has its vertex at:
a) (0,0)
b) (0,1)
c) (2,1)
d) (–2,1)
Rationale: In vertex form (x – h)² + k, vertex is (h,k) = (2,1).
9. Solve for x: log₂(x) = 5
a) 10
