91Ó°ÊÓ

We have:

$f\left(x\right)=-2x^2-12x-5$

The function is $f\left(x\right)=-2x^2-12x-5$.

Add the some value to both sides:

$f\left(x\right)=-2\left(x^2-6\right)-5$

Subtract $18$to both sides:

$$\begin{gathered} f\left(x\right)-18=-2\left(x^2+6x+9\right)-5 \\ f\left(x\right)=-2\left(x^2+6x+9\right)-5+18 \\ f\left(x\right)=-2{\left(x+3\right)}^2+13 \end{gathered}$$

The parent function of the function is:

$f\left(x\right)=x^2$

At $x=0:$

$f\left(0\right)={\left(0\right)}^2=0$

By the same procedure find some points:

$\left(-3,9\right),\left(-2,4\right),\left(0,0\right),\left(2,4\right),\left(3,9\right)$

Plot the calculated points. Draw a smooth curve through the plotted points. The graph of $f\left(x\right)=x^2$is shown below. The graph of $f\left(x\right)=x^2$is horizontally translated left by $3$unit. The transform equation will be $f\left(x\right)={\left(x+3\right)}^2$. The graph of $f\left(x\right)={\left(x+3\right)}^2$is a vertical reflection over $x$-axis by a factor of $2.$The transform equation will be $f\left(x\right)=2{\left(x+3\right)}^2$. The graph of $f\left(x\right)=2{\left(x+3\right)}^2$is vertically translated up by $13$units. The transform equation will by:$f\left(x\right)=2{\left(x+3\right)}^2+13$