Chapter 2.2, Problem 6Q

To determine

The rule for $g$ and identify the vertex.

Answer to Problem 6Q

The required rule for $g$ is $g\left(x\right)=\frac{1}{4}{\left(x-1\right)}^2-10$ and its vertex is $\left(1,-10\right)$ .

Explanation of Solution

Given information : Given that $g$ be a horizontal shrink by a factor of $\frac{1}{4}$ , followed by a translation of $1$ unit up and $3$ units right of the graph of $f\left(x\right)={\left(2x+1\right)}^2-11$ .

By observing the given information of the transformation of $f$ , the function $g$ will be of the form $g\left(x\right)=f\left(a\left(x-h\right)\right)+k$ , where the $h$ is horizontal translation, $k$ is vertical translation and $a$ is horizontal shrink.

According to the given information about $g$ , $h=3,a=\frac{1}{4}$ and $k=1$ .

Thus, required rule for $g$ is

  $\begin{array}{c}g\left(x\right)=f\left(\frac{1}{4}\left(x-3\right)\right)+1\\ ={\left(2\times \frac{1}{4}\left(x-3\right)+1\right)}^2-11+1\\ ={\left(\frac{\left(x-3\right)}{2}+1\right)}^2-10\\ ={\left(\frac{x-3+2}{2}\right)}^2-10\\ =\frac{1}{4}{\left(x-1\right)}^2-10\end{array}$ .

Thus, the rule for $g$ is $g\left(x\right)=\frac{1}{4}{\left(x-1\right)}^2-10$ .

To find the vertex of $g\left(x\right)=\frac{1}{4}{\left(x-1\right)}^2-10$ , write it in the form $y=g\left(x\right)$ .

  $\begin{array}{c}y=\frac{1}{4}{\left(x-1\right)}^2-10\\ y+10=\frac{1}{4}{\left(x-1\right)}^2\\ {\left(x-1\right)}^2=4\left(y+10\right)\end{array}$

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, $\left(\alpha ,\beta \right)=\left(1,-10\right)$ is vertex of $g\left(x\right)=\frac{1}{4}{\left(x-1\right)}^2-10$ .