Chapter 2.1, Problem 30E

To determine

To find: Describe the transformation of parent function to $f(x)=\frac{1}{2}{(x-1)}^2$ and to identify the vertex.

Answer to Problem 30E

The function $f(x)=\frac{1}{2}{(x-1)}^2$ is vertically shrunk by a factor of $\frac{1}{2}$ and followed by a translation of $1$ unit right of the parent quadratic function.

Vertex of the given function is $(1,0)$ .

Explanation of Solution

Given information: $f(x)=\frac{1}{2}{(x-1)}^2$ .

Describing the transformation:

Notice that the function is in the form of $g(x)=a{(x-h)}^2$ , where $a=\frac{1}{2}\text{and}h=1$ .

Vertical shrink occurs when $0<a<1$ .

From the given function $f(x)=\frac{1}{2}{(x-1)}^2$ , we know that $a=\frac{1}{2}$ which is smaller than $1$ and greater than $0$ .

The graph is vertically shrunk by the factor $\frac{1}{2}$ .

  $h=1$ , so the graph is translated $1$ unit right.

Therefore, we can conclude that the function $f(x)=\frac{1}{2}{(x-1)}^2$ is a vertical shrink by a factor $\frac{1}{2}$ and followed by the translation $1$ unit right.

Identifying the vertex:

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

Rewrite the given function in the general form.

  $\begin{array}{l}$ y=\frac{1}{2}{(x-1)}^2$$ \text{By}\text{using}{(a+b)}^2=a^2+b^2-2ab,$y=\frac{1}{2}(x^2+1^2-2x(1))\\ y=\frac{1}{2}(x^2-2x+1)\\ y=\frac{1}{2}{x}^2-x+\frac{1}{2}\\ a=\frac{1}{2},b=-1,c=\frac{1}{2}.\end{array}$

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

  $\begin{array}{l}x=-\frac{(-1)}{2\times \frac{1}{2}}\\ x=-\frac{(-1)}{1}\\ x=1\end{array}$

Substitute $x=1$ in $y=\frac{1}{2}{(x-1)}^2$ ,

  $\begin{array}{l}$ y=\frac{1}{2}{(1-1)}^2$y=\frac{1}{2}(0)\\ y=0\end{array}$

The vertex of the given function is $(1,0)$ .