We analyze the chart and observe that the linear function is [tex]y=x[/tex], since this relation holds for all values in the table. Drawing this line over the quadratic function shows that they intersect twice, at both the positive and negative x-coordinates.
This is by far the easiest way to solve this problem, but if you're interested in learning how to do it algebraically, read on! To prove this more rigorously, we can find that the equation of the parabola is [tex]y=\frac13 x^2 - 2.[/tex] Substituting in [tex]y=x[/tex], we find that the intersection points occur where [tex]\frac13 x^2 - 2 = x[/tex], or [tex]\frac13 x^2 - x - 2 = 0,[/tex] or [tex]x^2 - 3x - 6 = 0.[/tex]
This equation doesn't factor nicely, so we use the quadratic formula to learn that [tex]x = \frac{3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-6)}}{2} = \frac{3 \pm \sqrt{33}}{2}.[/tex] Hence, the x-coordinates of the intersection points are [tex]\frac{3 + \sqrt{33}}{2}[/tex], which is positive, and [tex]\frac{3 - \sqrt{33}}{2}[/tex], which is negative. This proves that there are intersection points on both ends of the axis.
