Chapter 9.2, Problem 45HP

To determine

To state a quadratic function that opens downwards and is wider than the parent graph

Answer to Problem 45HP

  $f(x)=(-a)x^2$ where $\left|a\right|<1$

Explanation of Solution

Given information :

Parent function:- $f(x)=x^2$

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 $f(x)=(-a)x^2$ , the value of a is negative and thus it opens downwards. Therefore, the transformation of the function with respect to the graph of $f(x)=x^2$ is that the function has been inverted and now it open downwards.

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.

Therefore, a graph $f(x)=(-a)x^2$ where $\left|a\right|<1$ would open downwards and be wider than $f(x)=x^2$