AP Precalculus Polynomial Functions — Questions and Answers
Question 1: What is the end behavior of f(x) = −2x⁵ + 3x² − 1?
- f(x) → ∞ as x → ∞ and f(x) → ∞ as x → −∞
- f(x) → −∞ as x → ∞ and f(x) → −∞ as x → −∞
- f(x) → −∞ as x → ∞ and f(x) → ∞ as x → −∞ (Correct answer)
- f(x) → ∞ as x → ∞ and f(x) → −∞ as x → −∞
Correct answer: f(x) → −∞ as x → ∞ and f(x) → ∞ as x → −∞
The leading term −2x⁵ has odd degree and a negative coefficient, so the graph falls to the right and rises to the left.
For large |x|, a polynomial behaves like its leading term. Here that term is −2x⁵. As x → ∞, x⁵ is large and positive, so −2x⁵ → −∞. As x → −∞, x⁵ is large and negative, so −2x⁵ → +∞. Odd-degree polynomials have opposite end behaviors, and the negative leading coefficient makes the graph fall on the right.
Question 2: What is the average rate of change of f(x) = x² − 4x + 1 over the interval [1, 5]?
Correct answer: 2
f(5) = 6 and f(1) = −2, so the average rate of change is (6 − (−2))/(5 − 1) = 2.
The average rate of change of f on [a, b] is (f(b) − f(a))/(b − a), which is the slope of the secant line. Here f(5) = 25 − 20 + 1 = 6 and f(1) = 1 − 4 + 1 = −2. The change in output is 8 and the change in input is 4, giving 2. Note that this is not f(5) or the value at the midpoint; it is the secant slope over the whole interval.
Question 3: A quadratic function has average rates of change of 3 on [0, 1] and 7 on [1, 2]. If this pattern continues, what is its average rate of change on [2, 3]?
Correct answer: 11
For a quadratic, average rates of change over consecutive equal-length intervals change by a constant amount; here it increases by 4, so the next value is 11.
A quadratic function's average rates of change over consecutive equal-length intervals form a linear (arithmetic) pattern. From 3 to 7 the rate increases by 4, so the next increase is also 4, giving 7 + 4 = 11. This constant change in the rates of change reflects the constant second difference of a quadratic. Predicting that the rates would be multiplied by a constant would instead suggest exponential behavior.
Question 4: Which is a polynomial of least degree with real coefficients that has 2 and 1 + i as zeros?
- x³ − 4x² + 4x − 4
- x³ − 4x² + 6x − 4 (Correct answer)
- x³ + 4x² + 6x + 4
- x³ − 2x − 4
Correct answer: x³ − 4x² + 6x − 4
Complex zeros of real-coefficient polynomials come in conjugate pairs, so 1 − i is also a zero, and (x − 2)(x² − 2x + 2) = x³ − 4x² + 6x − 4.
Since the coefficients are real, 1 − i must be a zero along with 1 + i. The product (x − (1 + i))(x − (1 − i)) equals x² − 2x + 2. Multiplying by (x − 2) gives x³ − 2x² + 2x − 2x² + 4x − 4 = x³ − 4x² + 6x − 4. This is the lowest-degree choice because all three zeros are required.
Question 5: The point (4, 5) lies on the graph of y = f(x). Which point must lie on the graph of g(x) = −f(x − 3) + 2?
- (7, 7)
- (1, 7)
- (7, −3) (Correct answer)
- (1, −3)
Correct answer: (7, −3)
Replacing x with x − 3 shifts the graph right 3, giving x = 7; the negative reflects the y-value to −5 and +2 raises it to −3.
For g(x) = −f(x − 3) + 2, g(7) = −f(4) + 2 because 7 − 3 = 4. Since f(4) = 5, g(7) = −5 + 2 = −3. The input shift moves the point right 3 units (opposite the sign inside the parentheses), the negative reflects over the x-axis, and +2 shifts the graph up 2. So the image point is (7, −3).
Question 6: What is the greatest possible number of relative extrema (turning points) of a polynomial function of degree 6?
Correct answer: 5
A polynomial of degree n has at most n − 1 turning points, so a degree 6 polynomial has at most 5.
The derivative of a degree-6 polynomial has degree 5, so it can have at most 5 real zeros, and each relative extremum corresponds to one of these sign changes. Thus the graph can turn at most 5 times. A particular polynomial may have fewer turning points, but 5 is the greatest possible. An even-degree polynomial always has an odd number of turning points, which agrees with this.
Question 7: Let f(x) = (x + 1)²(x − 3)³. Which describes the behavior of the graph of f at x = 3?
- It touches the x-axis and turns around
- It crosses the x-axis like a straight line
- It touches the x-axis with a corner and turns around
- It crosses the x-axis and flattens near the zero (Correct answer)
Correct answer: It crosses the x-axis and flattens near the zero
The zero x = 3 has odd multiplicity 3, so the graph crosses the axis, and because the multiplicity is greater than 1 it flattens near the intercept.
The multiplicity of a zero determines local behavior. An even multiplicity (like the 2 at x = −1) makes the graph touch the axis and turn around. An odd multiplicity makes the graph cross the axis, and when the multiplicity is 3 or more, the graph flattens out near the zero in a cubic-like shape. At x = 3 the factor (x − 3)³ gives a cross-with-flattening pattern.
Question 8: On an interval, a function's average rates of change over consecutive equal-length subintervals are all positive and increasing. What does this indicate about the graph of the function on that interval?
- Increasing and concave up (Correct answer)
- Increasing and concave down
- Decreasing and concave up
- Decreasing and concave down
Correct answer: Increasing and concave up
Positive rates of change mean the function is increasing, and rates that grow larger mean the graph is concave up.
Positive average rates of change mean the outputs rise as inputs increase, so the function is increasing. When those rates of change themselves get larger, the graph gets steeper, which is the definition of concave up. If the positive rates were decreasing, the function would still increase but would be concave down. Sign of the rate controls increasing versus decreasing, and the change in the rate controls concavity.
Question 9: What is the remainder when f(x) = x³ − 2x² + 5x − 1 is divided by x − 2?
Correct answer: 9
By the Remainder Theorem, the remainder equals f(2) = 8 − 8 + 10 − 1 = 9.
The Remainder Theorem says that dividing f(x) by x − c leaves remainder f(c). With c = 2, f(2) = 2³ − 2(2²) + 5(2) − 1 = 8 − 8 + 10 − 1 = 9. If the remainder had been 0, x − 2 would be a factor and 2 would be a zero. Since it is 9, 2 is not a zero of f.
Question 10: What is the range of the quadratic function g(x) = −(x + 2)² + 5?
- (−∞, −2]
- (−∞, 5] (Correct answer)
- [5, ∞)
- (−∞, ∞)
Correct answer: (−∞, 5]
The graph of y = x² is shifted left 2, reflected over the x-axis, and shifted up 5, so it opens downward with vertex (−2, 5) and maximum value 5.
In vertex form a(x − h)² + k, this function has a = −1, h = −2, and k = 5, so the vertex is (−2, 5). Because a is negative, the parabola opens downward and the vertex is the highest point. Every output is therefore at most 5, giving a range of (−∞, 5]. The value −2 describes the axis of symmetry and the domain-related shift, not the range.
