Math 135 Polynomial Functions Solutions
1. Find the leading term and use it determine the long-term behavior of each polyno-
mial function.
(a) f (x) = x2 + 3x + 1
Answer 1. The leading term of f (x) is x2 so f (x) → ∞ as x → ∞
and f (x) → −∞ as x → −∞.
(b) g(x) = −3x + 1
Answer 2. The leading term of g(x) is −3x so g(x) → −∞ as x → ∞
and g(x) → ∞ as x → −∞.
(c) p(x) = −x4 + x3 + x − 4
Answer 3. The leading term of p(x) is −x4 so p(x) → −∞ as x → ±∞.
(d) t(x) = (2x − 1)2 (3x + 2)2 (x − 1)(x + 2)
Answer 4. The leading term of t(x) is the product of the leading terms of each
factor: (2x)2 · (3x)2 · x · x = 36x6 . Thus, t(x) → ∞ as x → ±∞.
(e) h(x) = (x2 + 2x + 1)2 (2x + 3)4
Answer 5. The leading term of h(x) is the product of the leading terms of each
factor: (x2 )2 · (2x)4 = 14x8 . Thus, h(x) → ∞ as x → ±∞.
2. Find all roots and their degrees. Describe the behavior of the graph at each root.
(a) f (x) = (x − 2)2 (x + 2)(x + 4)
Answer 6. f (x) is already completely factored so we read off the roots: x = 2
(degree 2), x = −2 (degree 1), x = −4 (degree 1). The graph of f (x) crosses the
x-axis at each degree 1 root and touches the x-axis at the degree 2 root x = 2.
Moreover, since f (1.99) > 0 the graph touches the x-axis from above at x = 2.
(b) g(x) = (x + 5)4 (x − 1)
Answer 7. g(x) is completely factored so we read off the roots: x = −5 (degree
4) and x = 1 (degree 1). The graph of g(x) crosses the x-axis at x = 1 and
touches the x-axis at x = −5. Moreover, since g(−5.01) > 0 the graph touches
the x-axis from above at x = −5.
(c) p(x) = (x2 − 3x + 2)(x2 − x − 6)
Answer 8. We fist factor p(x) completely and then read off the roots. p(x) =
(x2 − 3x + 2)(x2 − x − 6) = (x − 1)(x − 2)(x − 3)(x + 2). Thus, the roots are
x = −2, 1, 2, 3 and since all are of degree 1 the graph of p(x) will cross the x-axis
at each root.
(d) h(x) = (x2 + x + 1)(x − 3)2 (x + 1)2
Answer 9. h(x) is already completely factored so we read off the roots: x = 3
(degree 2) and x = −1 (degree 2). Both roots are of even degree and since
h(2.99) > 0 and h(−0.99) > 0 the graph touches the x-axis from above at each
root.
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Math 135 Polynomial Functions Solutions
(e) r(x) = (x2 + 4)(x3 + 1)(x + 1)3
Answer 10. We first factor r(x) completely and the only factor which can be
simplified is (x3 +1) = (x+1)(x2 −x+1). Thus, r(x) = (x2 +4)(x2 −x+1)(x+1)4
and so x = −1 (degree 4) is the only root. Since it is an even degree root the
graph of r(x) will touch the x-axis at x = −1 and since r(−1.01) > 0 it follows
that the graph touch the x-axis form above.
3. Give the degree of each polynomial function. At most how many turning points
does each graph have?
(a) f (x) = (x − 2)(x + 2)(x − 1)(x + 1)(x + 3)
Answer 11. f (x) is a degree 5 polynomial so it can have at most 4 turning
points.
(b) g(x) = (x − 10)2 (x + 10)2
Answer 12. g(x) is a degree 4 polynomial so it can have at most 3 turning
points.
(c) p(x) = (x3 + x2 + x + 1)3
Answer 13. p(x) is a degree 9 polynomial so it can have at most 8 turning
points.
(d) t(x) = x(x2 − 3x + 2)2 (x − 7)2
Answer 14. t(x) is a degree 7 polynomial so it can have at most 6 turning points.
(e) h(x) = (5x2 − 2)(x2 + x + 1)3
Answer 15. h(x) is a degree 8 polynomial so it can have at most 7 turning
points.
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4. Graph each polynomial function. As usual, correctly scale and label the graph and
all axes. Label all roots with their degrees and mark all intercepts. The graph must
be smooth and continuous.
(a) p(x) = −x(x − 2)(x − 3)
(b) h(x) = (x + 4)2 (x − 1)2 (x − 5)
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(c) g(x) = −(x − 3)3 (x − 5)
(d) s(x) = (x − 2)2 (x2 + 4x + 4)(x2 + 6x + 9)
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(e) f (x) = x(2 − x2 )(x + 1)(x − 3)
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Math 135 Polynomial Functions Solutions
5. Working backwards. Find a possible polynomial function for each graph with the
given degree. The y-axis is left intentionally without scale.
(a) degree 4
Answer 16. (x + 2)2 (x − 3)2
(b) degree 2
Answer 17. (x + 2)(x − 3)
(c) degree 4 [Not the reflection of (B) about the x-axis.]
Answer 18. −(x + 2)(x − 3)(x2 + 1)
(d) degree 6
Answer 19. (x + 2)(x − 3)(x2 + 1)2
(e) degree 6 [Not the reflection of (A) about the x-axis.]
Answer 20. −(x + 2)2 (x − 3)2 (x2 + 1)
A B C
−5 −4 −3 −2 −1 1 2 3 4 5 −5 −4 −3 −2 −1 1 2 3 4 5 −5 −4 −3 −2 −1 1 2 3 4 5
D E
−5 −4 −3 −2 −1 1 2 3 4 5 −5 −4 −3 −2 −1 1 2 3 4 5
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