Figure

For the polynomial function $f ( x ) =x^{2} ( x-4 ) ( x+1 )$ :

Find the $x$ - and $y$ -intercepts of the graph of $f$ .
Determine whether the graph crosses or touches the $x$ -axis at each $x$ -intercept.
Plot at least one point to the left and right of each $x$ -intercept and connect the points to obtain the graph.

Solution

The $y$ -intercept is $f ( 0 ) =0.$ The $x$ -intercepts are the zeros of the function: $0$ , $4$ , and $-1$ .
$0$ is a zero of multiplicity $2$ ; the graph of $f$ will touch the $x$ -axis at $0$ . The numbers $4$ and $-1$ are zeros of multiplicity $1$ ; the graph of $f$ will cross the $x$ -axis at $4$ and $-1$ .
Since $f ( -2 ) =24$ , $f \!\left( -\dfrac{1}{2} \right) =-\dfrac{9}{16}$ , $f ( 2 ) =-24$ , and $f ( 5 ) =150,$ the points $( -2,24 )$ , $\left( -\dfrac{1}{2},-\dfrac{9}{16} \right)$ , $( 2,-24 )$ , and $( 5,150 )$ are on the graph. See Figure 31.