Question

graph the solution set of the system of inequalities or indicate that the system has no solution.
$x^{2}+y^{2}\leq64$
$y - x^{2}>0$

Explanation:

Step1: Analyze the first inequality \(x^{2}+y^{2}\leq64\)

The equation \(x^{2}+y^{2}=64\) represents a circle with center \((0,0)\) and radius \(r = 8\). The inequality \(x^{2}+y^{2}\leq64\) represents the set of all points \((x,y)\) that lie on or inside the circle.

Step2: Analyze the second inequality \(y - x^{2}>0\)

Rewrite it as \(y>x^{2}\). The equation \(y = x^{2}\) is a parabola opening upwards with vertex at \((0,0)\). The inequality \(y>x^{2}\) represents the set of all points \((x,y)\) that lie above the parabola \(y = x^{2}\).

Step3: Find the intersection of the two regions

We need to find the region that is inside (or on) the circle \(x^{2}+y^{2}=64\) and above the parabola \(y=x^{2}\).

Answer:

The solution set is the region that is inside (or on) the circle \(x^{2}+y^{2}=64\) and above the parabola \(y = x^{2}\). To graph it:

1. Draw the circle \(x^{2}+y^{2}=64\) (solid line since the inequality is \(\leq\)).
2. Draw the parabola \(y=x^{2}\) (dashed line since the inequality is \(>\)).
3. Shade the region that satisfies both inequalities.