Answer
a) The zeros of the function $f\left( x \right)={{x}^{3}}-5{{x}^{2}}-4x+20$ are $-2,\ 2,\ 5$.
b) The graph is shown below
Work Step by Step
(a)
Consider the given function, $f\left( x \right)={{x}^{3}}-5{{x}^{2}}-4x+20$
To find all the zeros of the function, equate the function $f\left( x \right)={{x}^{3}}-5{{x}^{2}}-4x+20$ to $0$ as,
${{x}^{3}}-5{{x}^{2}}-4x+20=0$
Factorize and simplify as follows:
$\begin{align}
& {{x}^{2}}\left( x-5 \right)-4\left( x-5 \right)=0 \\
& \left( {{x}^{2}}-4 \right)\left( x-5 \right)=0 \\
& \left( x-2 \right)\left( x+2 \right)\left( x-5 \right)=0 \\
& x=-2,2,5
\end{align}$
Thus, the zeros of $f\left( x \right)={{x}^{3}}-5{{x}^{2}}-4x+20$ are $-2,\ 2,\ 5$.
(b)
From part (a), the zeros of $f\left( x \right)={{x}^{3}}-5{{x}^{2}}-4x+20$ are $-2,\ 2,\ 5$.
Since the polynomial degree is odd and the leading coefficient is positive, the function initially rises, then falls and then rises again.
From the above obtained graph, it can be observed that the graph is not symmetrical about the y-axis, and the y-intercept of the graph is 20.
