Thursday, October 24, 2019

College Algebra, Chapter 4, 4.2, Section 4.2, Problem 24

Sketch the graph of polynomial function P(x)=(x1)2(x+2)3 make sure the graph shows all intercepts and exhibits the proper end behaviour.
The function has an odd degree 5 and a positive leading coefficient. Thus, its end behaviour is y as x and y as x.
To solve for the y-intercept, we set y=0.


y=(01)2(0+2)3y=(1)2(2)3y=8


To solve for the x-intercept, we set x=0
0=(x1)2(x+2)3

By zero product property, we have
(x1)2 and (x+2)3=0
x=1 and x=2

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