We want to solve the given system of inequalities by graphing. Note that both inequalities of the system are quadratic inequalities. ${y>x_{2}+2xy>x_{2}−1 (I)(II) $ Let's graph each of them, one at a time.
$x$ | $x_{2}+2x$ | $y=x_{2}+2x$ |
---|---|---|
$0$ | $0_{2}+2(0)$ | $0$ |
$-2$ | $(-2)_{2}+2(-2)$ | $0$ |
The points $(0,0)$ and $(-2,0)$ are on the parabola. Let's plot the points and connect them with a smooth curve.
Now that we have the boundary curve, we need to determine which region to shade. To do so, we will use $(-1,1)$ as a test point. If the point satisfies the inequality, we will shade the region that contains the point. If not, we will shade the opposite region.Since the substitution produced a true statement, we will shade the region that contains the point $(-1,1).$ Because we have a strict inequality, the boundary curve will be dashed.
Once again, we can write the boundary curve by replacing the greater than sign with an equals sign. $y=x_{2}−1 $ We can draw the second parabola following the same procedure as with the first.
Vertex | Axis of Symmetry | Two Points |
---|---|---|
$(0,-1)$ | $x=0$ $(y$-axis) | $(1,0)$ and $(-1,0)$ |
Any point on the plane can be used to determine the region we should shade. Because this inequality is strict, the curve will be dashed.
The solution set is the overlapping region.