Chapter 9, Problem 21SGR

To determine

To graph and describe translation of the function in relation to the graph of $f(x)=x^2$ .

Explanation of Solution

Given information :

The function that is provided: $f\left(x\right)=\frac{1}{4}{x}^2$ .

Graph :

  

Interpretation :

The graph of a quadratic equation ( $y=a{x}^2+bx+c$ ) is a parabola. The parabola is symmetric in nature and the point of symmetry is known as the vertex. This vertex lies on the line of symmetry and is symmetric about this line.

This vertex point shall be:

Highest point (if $a<0$ ) and be called the maximum. Here, the graph opens downwards.

Or, lowest point (if $a>0$ ) and can be called the minimum. Here the graph opens upwards.

In this case, ‘a’ is lesser than 0 hence the graph will have a maximum and will open downwards.

A parabola always points to infinity, either negative or positive.

In the function $f\left(x\right)=\frac{1}{4}{x}^2$ , the value of a is positive and thus it opens upwards. The transformation of the function with respect to the graph of $f(x)=x^2$ , as seen in the graph shown.

The graph of quadratic function $f(x)$ when multiplied by a positive constant a , the resulting graph $af(x)$ is a vertical dilation of $f(x)$ . The function is stretched or compressed vertically by a factor of $\left|a\right|$ .

If $\left|a\right|$ $>1$ , the graph of $f(x)$ is stretched vertically, that is, all points on the graph $f(x)$ move farther away from the x -axis.

If $\left|a\right|<1$ , the graph of $f(x)$ is compressed vertically, that is, all points on the graph $f(x)$ move closer to the x -axis.

In the function, the value of $\left|a\right|<1$ at $f\left(x\right)=\frac{1}{4}{x}^2$ . Thus, the transformation of $f(x)$ when multiplied by a , as seen in the graph, is that it is compressed vertically by a factor of $\frac{1}{4}$ .

Therefore, as seen in the graph, the transformation of the function with respect to the graph of $f(x)=x^2$ is the vertical compression by a factor of $\frac{1}{4}$ .