$\{\begin{array}{c}\text{:}[x^2-x-12]\\ (x-4)(x& 3vv12)=0\\ 3\end{array}=0$

Therefore, $x=4$ and $x=-3,vv,-3.$

Step $2$

We have determined that the points of intersection occur when $x=4$ and $x=-3.$ To find the mathematically.

When $x=4$ we have the following value of $y.$

$y=x1$

$=$

Similarly, when $x=-3$ we have $y=$

Use the findings above to state the points of intersection.

smaller $x-$value $(x,y)=(\frac{x}{x})$

larger $x-$value $(x,y)=(\frac{x}{x})$