Question

The graph of the polynomial function $f\left(x\right)=ax\left(x-4\right)\left(x+1\right)$ passes through the point $\left(5,15\right)$. Sketch the graph of the function.

Short answer

The graph of the function $f\left(x\right)=2x\left(x-4\right)\left(x+1\right)$ is shown below.

Step-by-step solution

Step 1. Write down the given information.

It is given that the graph of the polynomial function $f\left(x\right)=ax\left(x-4\right)\left(x+1\right)$ passes through the point $\left(5,15\right)$.

$f\left(x\right)=ax\left(x-4\right)\left(x+1\right)....\left(1\right)$

Step 2. Calculation.

Since the graph of the polynomial function $f\left(x\right)=ax\left(x-4\right)\left(x+1\right)$ passes through the point $\left(5,15\right)$. Therefore, the point $\left(5,15\right)$will satisfy $f\left(x\right)=ax\left(x-4\right)\left(x+1\right)$.

$\begin{array}{c}15=a\left(5\right)\left(\left(5\right)-4\right)\left(\left(5\right)+1\right)\\ 15=30a\\ a=\frac{1}{2}\end{array}$

Plugging $a=\frac{1}{2}$in (1),

$f\left(x\right)=2x\left(x-4\right)\left(x+1\right)....\left(2\right)$

Now sketch the graph of the function (2).

Step 3. Interpretation from the graph.

From the graph it can be interpreted that the graph cuts x-axis at points $-1,0\text{and 4}$. Therefore, the real roots are $x=-1,x=0 \text{and}x=4$.