ACT Math: Functions (Drill 3)

Question 1. Which of the following sets of ordered pairs represents a function?

• A) {(1, 2), (1, 3), (2, 4)}
• B) {(2, 5), (3, 5), (4, 5)} ✓
• C) {(0, 1), (0, −1), (1, 2)}
• D) {(3, 4), (3, 7), (3, 9)}

Explanation: A relation is a function if and only if each input (x-value) maps to exactly one output (y-value). Choice B: the x-values are 2, 3, and 4, all different, so each input has exactly one output. This is a function, even though all three outputs are the same value (5). Choice A: the input x = 1 maps to both 2 and 3, so it is not a function. Choice C: the input x = 0 maps to both 1 and −1, so it is not a function. Choice D: the input x = 3 maps to 4, 7, and 9, three different outputs, so it is not a function.

Question 2. What is the range of the function f(x) = −2(x − 1)² + 8?

• A) All real numbers
• B) y ≥ 8
• C) y ≤ 8 ✓
• D) y ≤ −2

Explanation: The function is in vertex form f(x) = a(x − h)² + k with a = −2, h = 1, and k = 8. The vertex is (1, 8). Since a = −2 is negative, the parabola opens downward, meaning the vertex is the maximum. The function reaches its highest value of 8 at x = 1 and decreases without bound as x moves away from 1. So the range is y ≤ 8. Choice A is the range of a linear function, not a quadratic. Choice B confuses the direction; it would be correct if the parabola opened upward (a > 0), making the vertex a minimum. Choice D uses the coefficient −2 rather than the vertex y-value 8 as the boundary.

Question 3. The value of a car is modeled by V(t) = 24,000 · (0.88)^t, where t is the number of years since purchase. By approximately what percentage does the car's value decrease each year?

• A) 8%
• B) 12% ✓
• C) 88%
• D) 24%

Explanation: In an exponential decay model V(t) = a · b^t, the base b represents the fraction of value retained each year. Here b = 0.88, meaning the car retains 88% of its value annually. The percentage decrease is 1 − 0.88 = 0.12 = 12%. Choice A (8%) confuses the decay rate with 1 − 0.92 = 0.08, misreading 0.88 as 0.92. Choice C (88%) reports the retention rate rather than the decay rate. Choice D (24%) has no direct basis in the model and results from an arithmetic error.

Question 4. If f(x) = (x + 5) / 3, which of the following is f⁻¹(x), the inverse of f?

• A) f⁻¹(x) = 3x − 5 ✓
• B) f⁻¹(x) = 3x + 5
• C) f⁻¹(x) = (x − 5) / 3
• D) f⁻¹(x) = 3 / (x + 5)

Explanation: To find the inverse, replace f(x) with y, swap x and y, then solve for y. Start: y = (x + 5) / 3. Swap: x = (y + 5) / 3. Multiply both sides by 3: 3x = y + 5. Subtract 5: y = 3x − 5. So f⁻¹(x) = 3x − 5. Verify: f(f⁻¹(x)) = f(3x − 5) = (3x − 5 + 5) / 3 = 3x / 3 = x ✓. Choice B results from adding 5 instead of subtracting when solving for y: 3x = y + 5 → y = 3x + 5. Choice C results from failing to multiply by 3 first, simply moving the 5 and leaving the division: (x − 5) / 3. Choice D results from taking the reciprocal of f(x) rather than finding the true algebraic inverse.

Question 5. The function p(x) = (x + 2)²(x − 1)(x − 4) has how many distinct real zeros, and what is the degree of p?

• A) 3 distinct real zeros; degree 3
• B) 4 distinct real zeros; degree 4
• C) 3 distinct real zeros; degree 4 ✓
• D) 2 distinct real zeros; degree 4

Explanation: The zeros are the values of x that make p(x) = 0: x = −2 (from (x + 2)²), x = 1, and x = 4. That is 3 distinct real zeros. Note that x = −2 is a repeated zero (multiplicity 2), but it is still only one distinct zero. The degree of p is found by adding the exponents of all factors: 2 + 1 + 1 = 4. So p has 3 distinct real zeros and degree 4. Choice A counts the correct number of distinct zeros but gets the degree wrong by counting factors rather than their exponents. Choice B counts 4 zeros by treating the repeated zero at x = −2 as two separate distinct zeros. Choice D counts only the factors with exponent 1, ignoring x = −2 entirely.