Chapter 2.2, Problem 4Q

To determine

The rule for $g$ and identify the vertex.

Answer to Problem 4Q

The required rule for $g$ is $g\left(x\right)=-6\left(x^2+2\right)$ and its vertex is $\left(0,-12\right)$ .

Explanation of Solution

Given information : Given that $g$ be a translation $2$ units up followed by a reflection in the $x-\text{axis}$ and a vertical stretch by a factor of $6$ of the graph of $f\left(x\right)=x^2$ .

By observing the given information of the transformation of $f$ , the function $g$ will be of the form $g\left(x\right)=-k\left(x^2+h\right)$ , where the $h$ is vertical translation , $k$ is vertical stretch and negative sign is for reflection in the $x-\text{axis}$ .

According to the given information about $g$ , $h=2$ and $k=6$ .

Thus, required rule for $g$ is $g\left(x\right)=-6\left(x^2+2\right)$ .

To find the vertex of $g\left(x\right)=-6\left(x^2+2\right)$ , write it in the form $y=g\left(x\right)$ .

  $\begin{array}{c}y=-6\left(x^2+2\right)\\ y=-6x^2-12\\ -6x^2=y+12\\ x^2=-\frac{1}{6}\left(y+12\right)\end{array}$

Now, equate the above equation with the standard equation of parabola ${\left(x-\alpha \right)}^2=-a\left(y-\beta \right)$ , where $\left(\alpha ,\beta \right)$ be the vertex of the parabola.

We have, $\left(\alpha ,\beta \right)=\left(0,-12\right)$ is vertex of $g\left(x\right)=-6\left(x^2+2\right)$ .