Chapter 1, Problem 4CT

In Problems $1-3$, use $P_1=\left(-1,3\right)$ and $P_2=\left(5,-1\right)$.

Graph $y=x^2-9$ by plotting points.

To determine

To graph: The equation $y=x^2-9$ by plotting points.

Explanation of Solution

Given Information:

The equation $y=x^2-9$.

Graph:

The graph of $y=x^2-9$ consists of points $\left(x,y\right)$.

To graph the equation $y=x^2-9$, plot some points $\left(x,y\right)$ and join them. To find some points $\left(x,y\right)$, choose random $x$ values, and find the corresponding $y$ values by using the equation $y=x^2-9$.

If $x=0$, then $y={\left(0\right)}^2-9=-9$. Therefore, the point $\left(0,-9\right)$ is on the graph .

If $x=1$, then $y={\left(1\right)}^2-9=-8$. Therefore, the point $\left(1,-8\right)$ is on the graph .

If $x=-1$, then $y={\left(-1\right)}^2-9=-8$. Therefore, the point $\left(-1,-8\right)$ is on the graph .

If $x=2$, then $y={\left(2\right)}^2-9=-5$. Therefore, the point $\left(2,-5\right)$ is on the graph .

If $x=-2$, then $y={\left(-2\right)}^2-9=-5$. Therefore, the point $\left(-2,-5\right)$ is on the graph .

If $x=3$, then $y={\left(3\right)}^2-9=0$. Therefore, the point $\left(3,0\right)$ is on the graph .

If $x=-3$, then $y={\left(-3\right)}^2-9=0$. Therefore, the point $\left(-3,0\right)$ is on the graph .

If $x=-5$, then $y={\left(-5\right)}^2-9=16$. Therefore, the point $\left(-5,16\right)$ is on the graph .

If $x=5$, then $y={\left(5\right)}^2-9=16$. Therefore, the point $\left(5,16\right)$ is on the graph .

Draw the graph by plotting these points and connecting them.

The graph of the equation $y=x^2-9$ represents the equation of a parabola.

Interpretation:

The graph of the equation $y=x^2-9$ represents the equation of a parabola.