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

Sketch the graph of polynomial function $\displaystyle 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 $y \rightarrow -\infty \text{ as } x \rightarrow -\infty \text{ and } y \rightarrow \infty \text{ as } x \rightarrow \infty$.
To solve for the $y$-intercept, we set $y = 0$.


$
\begin{equation}
\begin{aligned}
y &= (0 - 1)^2 (0+2)^3\\
\\
y &= (-1)^2 (2)^3\\
\\
y &= 8
\end{aligned}
\end{equation}
$


To solve for the $x$-intercept, we set $x = 0$
$0 = (x-1)^2(x+2)^3$

By zero product property, we have
$(x - 1)^2 \text{ and } (x + 2)^3 = 0$
$x = 1 \text{ and } x = -2$