Sketch the graph of polynomial function P(x)=(x−1)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=(0−1)2(0+2)3y=(−1)2(2)3y=8
To solve for the x-intercept, we set x=0
0=(x−1)2(x+2)3
By zero product property, we have
(x−1)2 and (x+2)3=0
x=1 and x=−2
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