Answer:
[tex]\begin{gathered} x-\text{intercept}=-3\text{ and 1} \\ y-\text{intercept}=\text{ -9} \end{gathered}[/tex][tex]\begin{gathered} x=1\text{ multiplicity 3} \\ x=-3\text{ multiplicity 2} \\ \lim _{x\rightarrow\infty}(x-1)^3(x+3)^2=\infty \\ \lim _{x\rightarrow-\infty}(x-1)^3(x+3)^2=-\infty \end{gathered}[/tex]
Step-by-step explanation:
To find the x-intercepts factor the function to the simplest form:
[tex]h(x)=(x-1)^3(x+3)^2[/tex]
As we can see the zeros to the function would be 1 and -3, then its:
[tex]\begin{gathered} x-\text{intercept}=-3\text{ and 1} \\ y-\text{intercept}=\text{ -9} \end{gathered}[/tex]
Zero has a "multiplicity", which refers to the number of times that its associated factor appears in the polynomial. Therefore, this function has multiplicity:
[tex]\begin{gathered} x=1\text{ multiplicity 3} \\ x=-3\text{ multiplicity 2} \end{gathered}[/tex]
For the end behavior:
down/up
As x approaches infinity f(x) approaches infinity
As x approaches -infinity f(x) approaches -infinity
[tex]\begin{gathered} \lim _{x\rightarrow\infty}(x-1)^3(x+3)^2=\infty \\ \lim _{x\rightarrow-\infty}(x-1)^3(x+3)^2=-\infty \end{gathered}[/tex]
