Rise Placement Test Functions and Quadratic Equations 3 — Questions and Answers
Question 1: Using the quadratic formula, what are the solutions to x² + 4x + 3 = 0?
- x = 1 and x = 3
- x = −1 and x = −3 (Correct answer)
- x = 2 and x = −2
- x = −4 and x = 0
Correct answer: x = −1 and x = −3
Factoring gives (x+1)(x+3) = 0, so x = −1 or x = −3.
Question 2: What is the domain of f(x) = √(x − 4)?
- x < 4
- x ≤ 4
- x ≥ 4 (Correct answer)
- All real numbers
Correct answer: x ≥ 4
The expression under the square root must be non-negative, so x − 4 ≥ 0, giving x ≥ 4.
Question 3: If g(x) = x² − 4x + 4, what is g(x) in factored form?
- (x − 2)(x + 2)
- (x − 4)(x − 1)
- (x − 2)² (Correct answer)
- (x + 2)²
Correct answer: (x − 2)²
x² − 4x + 4 is a perfect square trinomial equal to (x − 2)².
Question 4: A ball is thrown upward with height h(t) = −16t² + 64t. At what time does it reach its maximum height?
- t = 1 sec
- t = 2 sec (Correct answer)
- t = 4 sec
- t = 8 sec
Correct answer: t = 2 sec
The vertex time is t = −b/(2a) = −64/(2·(−16)) = 2 seconds.
Question 5: Which transformation moves the graph of f(x) = x² three units to the right?
- f(x) = x² + 3
- f(x) = x² − 3
- f(x) = (x+3)²
- f(x) = (x−3)² (Correct answer)
Correct answer: f(x) = (x−3)²
Replacing x with (x − 3) shifts the graph 3 units to the right.
Question 6: What is the range of f(x) = −x² + 6?
- y ≥ 6
- y ≤ 6 (Correct answer)
- All real numbers
- y ≥ 0
Correct answer: y ≤ 6
Since the parabola opens downward with vertex at (0,6), the maximum value is 6, so y ≤ 6.
Question 7: Which of the following is NOT a quadratic function?
- y = 3x² − 2x + 1
- y = (x+1)(x−2)
- y = 5x − 7 (Correct answer)
- y = −x² + 4
Correct answer: y = 5x − 7
y = 5x − 7 is a linear function, not quadratic, because it has no x² term.
Rise Placement Test Functions and Quadratic Equations 3
Using the quadratic formula, what are the solutions to x² + 4x + 3 = 0?