College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 5 - Section 5.1 - Polynomial Functions and Models - 5.1 Assess Your Understanding - Page 340: 120

Answer

(a) The x-intercepts are (-2,0) and (4,0) (b) The x-intercepts are (0,0) and (6,0)

Work Step by Step

Real zeros (or the x-intercepts) are found by making the function equal to zero and solving for x. This is made easier if the polynomial function is in the factored form $f(x)=a(x-r_1)(x-r_2)...(x-r_n)$ since we can solve all $(x-r)$'s for zero. In this case, we solve $(x+2) \text{ and } (x-4)^3$ for zero: $x_1+2=0$ $x_1=-2$ $(x_2-4)^3=0$ $\sqrt[3]{(x_2-4)^3}=\sqrt[3]0$ $x_2-4=0$ $x_2=4$ The x-intercepts of h(x) are (-2,0) and (4,0). To solve (b), we first find G(x+3) and then do the same procedure again: $h(x-2)=(x-2+2)(x-2-4)^3$ $h(x-2)=x(x-6)^3$ $x_1=0$ $(x_2-6)^3=0$ $\sqrt[3]{(x_2-6)^3}=\sqrt[3]0$ $x_2-6=0$ $x_2=6$ The x-intercepts of h(x-2) are (0,0) and (6,0).
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