Exercise 6 Page 539

In case you want to see the steps we took to graph the system, we have included the process. First we graphed the function $y=x^2+2x-8.$ We began by determining the values of $a,$ $b,$ and $c.$ Therefore, $a=1,$ $b=2,$ and $c=\text{-}8.$ Then we identified the axis of symmetry. The axis of symmetry for this function is $x=\text{-}1.$ Next we determined the vertex. The axis of symmetry always intersects the parabola at its vertex. Therefore, $x=\text{-}1$ is the $x\text{-}$coordinate of the vertex. We used this to determine the corresponding $y\text{-}$coordinate. The vertex of the function is $(\text{-}1,\text{-}9).$ We then found one more point that lies on the parabola. To do so we used $x=4.$ The point $(4,16)$ lies on the parabola. Finally, we plotted the axis of symmetry $x=\text{-}1,$ the vertex $(\text{-}1,\text{-}9),$ and the point $(4,16).$ We also reflected this point across the axis of symmetry.

Finally we connected the three points with a smooth curve.

Next we graphed the linear function $y=5x+2.$ We identified the values of the slope $m$ and the $y\text{-}$intercept $b.$ We plotted the $y\text{-}$intercept and used the slope to plot another point. Then we connected the points with a straight edge.

The last step was connecting the graphs.