Q50. - Algebra 2

Question

The graph of the polynomial function $f\left(x\right)=ax\left(x-4\right)\left(x+1\right)$ passes through the point $\left(5,15\right)$. For what value of x, $f\left(x\right)=0$.

Step-by-Step Solution

Answer

The value of x is $0,-1 \text{ and }4$ for which $f\left(x\right)=0$.

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 $\left(5,15\right)$. Therefore, the point $\left(5,15\right)$ will satisfy,

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

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),

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

Substitute $f\left(x\right)=0$ in (2), to find the value of $\left(x\right)$, for which $f\left(x\right)=0$. Therefore from (2),

 $\begin{array}{c}\left(0\right)=\frac{1}{2}x\left(x-4\right)\left(x+1\right)\\ x\left(x-4\right)\left(x+1\right)=0\\ x=0,x=-1\text{ and }x=4\end{array}$

Step 3. Conclusion.

The value of x is $0,-1 \text{ and }4$ for which $f\left(x\right)=0$.