Chapter 2.1, Problem 33E

To determine

To find: Describe the transformations of graph of $f$ and identify the vertex.

Answer to Problem 33E

  $f(x)=8x^2-6$ , with a horizontal stretch by a factor $2$ and a reflection in the $y-\text{axis}$ , followed by a translation $2$ units up is $g(x)=2x^2-4$ .

Vertex of the given function is $(0,-4)$ .

Explanation of Solution

Given information: $f(x)=8x^2-6$ with a horizontal stretch by a factor $2$ and a reflection in the $y-\text{axis}$ , followed by a translation $2$ units up.

Describing the transformation:

Horizontal stretch occurs when $0<a<1$ , here $a=\frac{1}{2}$ .

Translation of $2$ units up occurs when $k$ is on the positive side.

Reflection in $y-\text{axis}$ is $y=x^2$ .

  $\begin{array}{c}$ g(\frac{1}{2}x)=8{(\frac{1}{2}x)}^2-6+2$y=8(\frac{1}{4}{x}^2)-4\\ y=2x^2-4\end{array}$

Therefore, the above function is the rule of $g$ described by the transformations of the graph $f$ .

Identifying the vertex:

General form of a quadratic function is $y=a{x}^2+bx+c$ .

  $\begin{array}{l}y=2x^2-4\\ a=2,b=0,c=-4,\end{array}$

The $x-$ coordinate of the vertex of the given parabola is $x=-\frac{b}{2a}$ ,

  $\begin{array}{l}x=-\frac{0}{2\times 2}\\ x=0\end{array}$

Substitute $x=0$ in $y=2x^2-4$ ,

  $\begin{array}{l}$ y=2{(0)}^2-4$y=0=-4\\ y=-4\end{array}$

The vertex of the given function is $(0,-4)$ .