Q. 438 - Intermediate Algebra

Question

For the function, $f\left(x\right)=9x^2-18x+5$, (a) find

when $f\left(x\right)=-3$. (b) Use this information to find two

points that lie on the graph of the function.

Step-by-Step Solution

Answer

(a) The $x$ values for $f\left(x\right)=-3$ are $x=\frac{4}{3},x=\frac{2}{3}$

(b) Two points that lie on the graph of function are $\left(\frac{4}{3},-3\right),\left(\frac{2}{3},-3\right)$

Step 1. Given

The function is $f\left(x\right)=9x^2-18x+5$

(a) We have to find when $f\left(x\right)=-3$

(b) We also have to find two points that lie on the graph of the function.

Step 2. Find values of x when f ( x ) = - 3

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

So,

$x=\frac{4}{3},\frac{2}{3}$

Step 3. Check the value of x at x = 4 3 , 2 3

Substitute $x=\frac{4}{3}$ in $f\left(x\right)$,

$$\begin{gathered} f\left(x\right)=9x^2-18x+5 \\ f\left(\frac{4}{3}\right)=9{\left(\frac{4}{3}\right)}^2-18\left(\frac{4}{3}\right)+5 \\ f\left(\frac{4}{3}\right)=9\left(\frac{16}{9}\right)-18\left(\frac{4}{3}\right)+5 \\ f\left(\frac{4}{3}\right)=16-24+5 \\ f\left(\frac{4}{3}\right)=-3 \end{gathered}$$

Substitute $x=\frac{2}{3}$ in $f\left(x\right)$,

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

Step 4. Find two points that lie on the graph of the function.

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