Q. 315 - Intermediate Algebra

Question

For the function, $f\left(x\right)=8x^2-18x+5$:

a. Find $x$, when $f\left(x\right)=-4$.

b. Use this information to find two points that lie on the graph of the function.

Step-by-Step Solution

Answer

a. The value of $x$ is $\frac{3}{2}$ or $\frac{3}{4}$ when $f\left(x\right)=-4$.

b. $(\frac{3}{2}, -4)$ and $(\frac{3}{4}, -4)$ lie on the graph.

Part a. Find x when f ( x ) = - 4 .

Substituting $f\left(x\right)=-4$ in the given equation,

$$\begin{gathered} f\left(x\right)=8x^2-18x+5 \\ -4=8x^2-18x+5 \\ 8x^2-18x+9=0 \\ 8x^2-12x-6x+9=0 \\ 4x(2x-3)-3(2x-3)=0 \\ (2x-3)(4x-3)=0 \end{gathered}$$

Using Zero Product Property,

$2x-3=0$ or

$4x-3=0$

When $2x-3=0$,

$x=\frac{3}{2}$.

When $4x-3=0$,

$x=\frac{3}{4}$.

Checking each value separately,

When $x=\frac{3}{2}$

$$\begin{gathered} f\left(x\right)=8x^2-18x+5 \\ f\left(\frac{3}{2}\right)=8{\left(\frac{3}{2}\right)}^2-18\left(\frac{3}{2}\right)+5 \\ f\left(\frac{3}{2}\right)=18-27+5 \\ f\left(\frac{3}{2}\right)=-4 \end{gathered}$$

This is true.

When $x=\frac{3}{4}$,

$$\begin{gathered} f\left(x\right)=8x^2-18x+5 \\ f\left(\frac{3}{4}\right)=8{\left(\frac{3}{4}\right)}^2-18\left(\frac{3}{4}\right)+5 \\ f\left(\frac{3}{4}\right)=\frac{9}{2}-\frac{27}{2}+5 \\ f\left(\frac{3}{4}\right)=-4 \end{gathered}$$

This is true too.

Part b. Find points that lie on the graph of the function

Since $f\left(\frac{3}{2}\right)=-4$ and $f\left(\frac{3}{4}\right)=-4$, the points $(\frac{3}{2}, -4)$ and $(\frac{3}{4}, -4)$ lie on the graph of the function.