If $P(x) = x^3 - 3x^2 - 4x$, then
a.) Find all zeros of $P$, and state their multiplicities.
b.) Sketch the graph of $P$.
a.) To find the zeros of $P$, we factor $P$ to obtain
$
\begin{equation}
\begin{aligned}
P(x) =& x^3 - 3x^2 - 4x
&& \text{Given}
\\
\\
=& x (x^2 - 3x - 4)
&& \text{Factor out } x
\\
\\
=& x (x - 4)(x + 1)
&& \text{Factor the quadratic function}
\end{aligned}
\end{equation}
$
It shows that the function has zeros of $0, 4$ and $-1$. And all the zeros have multiplicity of $1$.
b.) To sketch the graph of $P$, we must know first the intercepts of the function. The values of the $x$ intercepts are the zeros of the function, that is $0, 4$ and $-1$. To determine the $y$ intercept, we set $x = 0$ so, $P(0) = 0(0 - 4)(0 + 1) = 0$
The $y$ intercept is .
Since the function has an odd degree and a positive leading coefficient, then its end behavior is $y \to \infty$ as $x \to \infty$ and $y \to - \infty$ as $x \to - \infty$. Then, the graph is
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