Chapter 2.2, Problem 103E

To determine

To identify: The symmetry with respect to t -axis,g -axis, and the origin andt -intercept (s) of polynomial function $g\left(t\right)$ using graphing utility to graph the function.

Answer to Problem 103E

The function $g\left(t\right)=-\frac{1}{2}{\left(t-4\right)}^2{\left(t+4\right)}^2$ is symmetric about g -axis but not with respect tot -axis and the origin, and t -intercepts $g\left(t\right)$ of are $\left(-4,0\right)$ and $\left(4,0\right)$ .

Explanation of Solution

Given data:

The given polynomial function $g\left(t\right)=-\frac{1}{2}{\left(t-4\right)}^2{\left(t+4\right)}^2$ .

Method/ concept used:

The graph of the function is plotted using graphing calculator. The x -intercept of the function $f\left(x\right)$ is the point $\left(a,0\right)$ , where $f\left(a\right)=0$ .

Graph:

The graph of the polynomial function $g\left(t\right)=-\frac{1}{2}{\left(t-4\right)}^2{\left(t+4\right)}^2$ is shown in Figure-1.

  

Interpretation of the Graph:

Since the graph of $g\left(t\right)=-\frac{1}{2}{\left(t-4\right)}^2{\left(t+4\right)}^2$ is not identical about t- axis (both sides of x -axis) that implies the function is not symmetric about t -axis.

[Note: if replacing y by -y function $y=f\left(x\right)$ does not change then the function is symmetrical about x − axis, e.g., $y^2=ax$ ]

Since the graph of $g\left(t\right)=-\frac{1}{2}{\left(t-4\right)}^2{\left(t+4\right)}^2$ is identical about g -axis (both sides of g -axis) that implies the function is not symmetric about g -axis.

[Note: if replacing xby −x function $y=f\left(x\right)$ does not change then the function is symmetrical about y - axis, e.g., $y=a{x}^2$ .]

Here, $g\left(-t\right)=-\frac{1}{2}{\left(-t-4\right)}^2{\left(-t+4\right)}^2=-\frac{1}{2}{\left(t+4\right)}^2{\left(t-4\right)}^2=g\left(t\right)$ . Therefore, $g\left(t\right)$ is symmetric about vertical axis.

Since the graph of $g\left(t\right)=-\frac{1}{2}{\left(t-4\right)}^2{\left(t+4\right)}^2$ is not identical about t -axis and g -axis simultaneously) that implies the function is not symmetrical about g -axis.

[Note: A function is said to be symmetric about the origin if $\left(x,y\right)$ is on the graph then $\left(-x,-y\right)$ is also, e.g., $y^2=x^2$ .]

Thus, the function $g\left(t\right)=-\frac{1}{2}{\left(t-4\right)}^2{\left(t+4\right)}^2$ is symmetric with respect tog -axis, but not with respect to t -axis and the origin.

The graph of $g\left(t\right)=-\frac{1}{2}{\left(t-4\right)}^2{\left(t+4\right)}^2$ touches t -axis at points $\left(-4,0\right)$ and $\left(4,0\right)$ .

Hence, $\left(-4,0\right)$ and $\left(4,0\right)$ are t -interceptsof $g\left(t\right)$ .

Conclusion:

The function $g\left(t\right)=-\frac{1}{2}{\left(t-4\right)}^2{\left(t+4\right)}^2$ is symmetric about g -axis but not with respect tot -axis and the origin, and t -interceptsof $g\left(t\right)$ are $\left(-4,0\right)$ and $\left(4,0\right)$ .