Chapter 2.1, Problem 27E

To determine

To find: Describe the transformation of parent function to $f(x)=3{(x+2)}^2+1$ and to identify the vertex.

Answer to Problem 27E

The function $f(x)=3{(x+2)}^2+1$ is vertically stretched by a factor of $3$ and followed by a translation of $4$ units left and $1$ unit up of the parent quadratic function.

Vertex of the given function is $(-2,1)$ .

Explanation of Solution

Given information: $f(x)=3{(x+2)}^2+1$ .

Describing the transformation:

Notice that the function is in the form of $g(x)=a{(x-h)}^2+k$ , where $a=3,h=-2andk=1$ .

Vertical stretch occurs when $a>1$ .

From the given function $f(x)=3{(x+2)}^2+1$ , we know that $a=3$ which greater than $1$ .

The graph is vertically stretched by the factor $3$ .

  $h=-2$ and $k=1$ so the graph is $2$ units to its left and $1$ unit up.

Therefore, we can conclude that the function $f(x)=3{(x+2)}^2+1$ is a vertical stretch by a factor $3$ and followed by the translation $2$ units left and $1$ unit up.

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=3{(x+2)}^2+1$$ \text{By}\text{using}{(a+b)}^2=a^2+b^2+2ab,$y=3(x^2+2^2+2x(2))+1\\ y=3(x^2+4+4x)+1\\ y=3x^2+12x+13\\ a=3,b=12,c=13.\end{array}$

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

  $\begin{array}{l}x=-\frac{12}{2\times 3}\\ x=-\frac{12}{6}\\ x=-2\end{array}$

Substitute $x=-2$ in $y=3x^2+12x+13$ ,

  $\begin{array}{l}$ y=3{(-2)}^2+12(-2)+13$y=3(4)-24+13\\ y=12-24+13\\ y=1\end{array}$

The vertex of the given function is $(-2,1)$ .