Polynomial Functions: ACT Math Study Guide

A function's rate of change between two points is the same slope formula used for lines: the change in output divided by the change in input. For a linear function, this rate is constant everywhere; for a quadratic or higher-degree function, it describes only the average rate between those two specific points.

The end behavior of a polynomial is controlled entirely by its degree and leading coefficient. An even degree means both ends point in the same direction, up if the leading coefficient is positive and down if negative. An odd degree means the ends point in opposite directions, rising on the right and falling on the left if the leading coefficient is positive, and the reverse if negative.

An arithmetic sequence changes by the same constant difference d at every step: aₙ = a₁ + (n − 1)d. Finding a term far down the sequence, or finding the common difference from two known terms that aren't adjacent, both use this same formula, just solved for a different piece of it.

Where this appears as a model: a box built by folding a flat sheet. Cutting a 2-inch square from each corner of a 12-by-8 sheet and folding up the sides gives a box with volume 2(12−4)(8−4)=64 cubic inches, using the same polynomial expression x(12−2x)(8−2x) for any cut size x.

▸ End behavior from degree and leading coefficient

Worked Example Describe the end behavior of f(x) = 3x⁵ − 2x² + 1.

Degree 5 is odd; the leading coefficient is positive.

Odd degree: the ends point in opposite directions.

Positive leading coefficient: up on the right, down on the left.

The next section previews one arithmetic-sequence skill early, since it uses the same algebraic habit as the polynomial work just covered: setting up and solving for an unknown from a general formula. Sequences and series receive full, dedicated treatment later in this Part.

▸ Arithmetic sequences from non-adjacent terms

Worked Example An arithmetic sequence has a₂ = 9 and a₅ = 21. Find a₁₀.

Find the common difference first, then extend to the requested term.

From a₂ to a₅ is 3 steps: d = (21−9)/3 = 4

a₁ = 9 − 4 = 5

a₁₀ = 5 + (10−1)(4) = 41

▸ Practice

Question 1

Find the rate of change of a function passing through (2, 5) and (6, 17).

A. 4
B. 3
C. 1/3
D. −3

Question 2

Find the zeros of f(x) = x² + x − 12.

F. x = 4, −3
G. x = −4, 3
H. x = −4, −3
J. x = 4, 3

Question 3

Describe the end behavior of f(x) = x⁴ − 3x² + 1.

A. Up on both ends
B. Down on both ends
C. Up on the left, down on the right
D. Down on the left, up on the right

Question 4

Find the x-intercepts of f(x) = x² − 9.

F. x = 9, −9
G. x = 3 only
H. x = −3, 3
J. x = −3 only

Question 5

An arithmetic sequence has a₁ = 5 and d = −3. Find a₁₂.

A. −25
B. −31
C. 28
D. −28

Question 6

An arithmetic sequence has a₃ = 11 and a₇ = 27. Find a₂₀.

F. 79
G. 75
H. 83
J. 71

Question 7

A function passes through (2, 9) and (6, 25). Find its rate of change.

A. 8
B. 4
C. 2
D. 16

Question 8

Find the zeros of f(x) = x² − 2x − 15.

F. x = 5, −3
G. x = −5, 3
H. x = 5, 3
J. x = −5, −3

Question 9

Describe the end behavior of f(x) = −2x³ + x − 4.

A. Down on the right, up on the left
B. Up on both ends
C. Down on both ends
D. Up on the right, down on the left

Question 10

An arithmetic sequence has a₁ = 7 and d = −2. Find a₉.

F. −11
G. −7
H. 9
J. −9

Question 11

Find the zeros of f(x) = x² + 2x − 24.

A. x = 6, −4
B. x = 6, 4
C. x = −6, −4
D. x = −6, 4

Question 12

Describe the end behavior of f(x) = −2x⁴ + 3x² − 1.

F. Up on both ends
G. Up on the right, down on the left
H. Down on the right, up on the left
J. Down on both ends

Question 13

A line passes through (2, 11) and (5, 20). Find its y-intercept.

A. 3
B. 11
C. 5
D. 2

Question 14

How many real zeros does f(x) = x³ − 4x have?

F. 3
G. 1
H. 2
J. 0

Question 15

A function passes through (3, 7) and (9, 25). Find its rate of change.

A. 6
B. 9
C. 3
D. 18

Question 16

Find the zeros of f(x) = x² − 6x + 8.

F. −4, 2
G. 4, −2
H. 4, 2
J. −4, −2

Question 17

Describe the end behavior of f(x) = −x⁵ + 2x.

A. Up on the right, down on the left
B. Down on the right, up on the left
C. Up on both ends
D. Down on both ends

Question 18

An arithmetic sequence has a₁ = 12 and d = −5. Find a₈.

F. −18
G. −23
H. −13
J. −28

Question 19

Find the zeros of f(x) = x² + 3x − 10.

A. −5, 2
B. 5, −2
C. 5, 2
D. −5, −2

Question 20

A line passes through (1, 6) and (4, 15). Find its y-intercept.

F. 6
G. 3
H. 9
J. −3

▸ Answers and Explanations

Show the answers and explanations

1: B. (17 − 5) ⁄ (6 − 2) = 12 ⁄ 4 = 3. Choice A reports the denominator of the slope calculation (6−2=4) instead of completing the division.

2: G. Factor: (x + 4)(x − 3) = 0, giving x = −4 or x = 3. Choice F reverses the sign of both roots.

3: A. Degree 4 is even, so both ends point the same direction; the leading coefficient is positive, so both ends rise. Choice B treats the leading coefficient as negative instead of positive, reversing the conclusion.

4: H. Factor as a difference of squares: (x−3)(x+3)=0, giving x=−3 or x=3. Choice F uses the constant 9 directly as a root instead of taking its square root.

5: D. a₁₂ = 5 + (12 − 1)(−3) = 5 − 33 = −28. Choice A uses one fewer step than required, an off-by-one error in the sequence formula.

6: F. From a₃=11 to a₇=27 is 4 steps, so d=(27−11)⁄4=4; a₁=11−2(4)=3; a₂₀=3+(19)(4)=79. Choice G uses one fewer step than required when computing from a₁ to a₂₀, an off-by-one error.

7: B. (25−9)⁄(6−2)=16⁄4=4. Choice A doubles the correct rate of change, a common arithmetic slip when simplifying the fraction.

8: F. (x−5)(x+3)=0, giving x=5 or x=−3. Choice G reverses the sign of both roots.

9: A. Degree 3 is odd, and the leading coefficient is negative, so as x→∞, f→−∞, and as x→−∞, f→∞: down on the right, up on the left. Choice D treats the leading coefficient as positive instead of negative, reversing the conclusion.

10: J. a₉=7+(9−1)(−2)=7−16=−9. Choice F uses one extra step beyond what's required, an off-by-one error in the sequence formula.

11: D. (x+6)(x−4)=0, giving x=−6 or x=4. Choice A reverses the sign of both roots.

12: J. Degree 4 is even, and the leading coefficient is negative, so both ends fall. Choice G misapplies the even-degree rule as if the polynomial had an odd degree instead.

13: C. Slope=(20−11)⁄(5−2)=3; using point (2,11): 11=3(2)+b, so b=5. Choice B reports the y-coordinate of the given point (2, 11) instead of solving for the actual y-intercept.

14: F. Factor out x: x(x²−4)=x(x−2)(x+2)=0, giving three real zeros: 0, 2, and −2. Choice H drops the zero at x=0 from the factored form, reporting only two of the three real zeros.

15: C. (25−7)⁄(9−3)=18⁄6=3. Choice A doubles the correct rate of change, a common arithmetic slip when simplifying the fraction.

16: H. (x−4)(x−2)=0, giving x=4 or x=2. Choice F drops the sign on one root only.

17: B. Degree 5 is odd, and the leading coefficient is negative, so as x→∞, f→−∞, and as x→−∞, f→∞: down on the right, up on the left. Choice A treats the leading coefficient as positive instead of negative, reversing the conclusion.

18: G. a₈=12+(8−1)(−5)=12−35=−23. Choice F uses one fewer step than required, an off-by-one error in the sequence formula.

19: A. (x+5)(x−2)=0, giving x=−5 or x=2. Choice C drops the negative sign on one root.

20: G. Slope=(15−6)⁄(4−1)=3; using point (1,6): 6=3(1)+b, so b=3. Choice F reports the y-coordinate of the given point (1, 6) instead of solving for the actual y-intercept.

Polynomial Functions

▸ End behavior from degree and leading coefficient

▸ Arithmetic sequences from non-adjacent terms

▸ Practice

▸ Answers and Explanations

PART III · FUNCTIONS