Given the polynomial function h(x) defined as:
[tex]h(x)=(x+3)^2(x-2)[/tex]
The y-intercept is the value of the function at x = 0. Then, evaluating h(0):
[tex]\begin{gathered} h(0)=(0+3)^2(0-2)=3^2\cdot(-2)=-9\cdot2 \\ \Rightarrow h(0)=-18 \end{gathered}[/tex]
The y-intercept is a unique value, so its multiplicity is 1. On the other hand, the x-intercepts are those x-values such that h(x) = 0. Then, solving the polynomial equation for x:
[tex](x+3)^2(x-2)=0[/tex]
This equation is 0 for:
[tex]\begin{gathered} (x+3)^2=0\Rightarrow x+3=0\Rightarrow x=-3 \\ (x-2)=0\Rightarrow x=2 \end{gathered}[/tex]
The first equation has a square exponent, so the multiplicity is 2. The multiplicity of the second equation is 1 because it is linear.
Summarizing:
x-intercepts:
i) -3, multiplicity 2
ii) 2, multiplicity 1
y-intercept:
i) -18, multiplicity 1
And the graph of the function looks like this:
For the end behavior, we need to analyze the limits for +∞ and -∞:
[tex]\begin{gathered} \lim _{x\to+\infty}(x+3)^2(x-2)=\infty^2\cdot\infty=+\infty \\ \lim _{x\to-\infty}(x+3)^2(x-2)=(-\infty)^2\cdot(-\infty)=\infty\cdot(-\infty)=-\infty \end{gathered}[/tex]
So the function tends to infinite when x tends to infinite, and to minus infinite when x tends to minus infinite.
