Question

choose the graph of y > x² - 9. done

Explanation:

Step1: Analyze the parabola equation

The inequality is \( y > x^2 - 9 \). The related equation is \( y = x^2 - 9 \), which is a parabola. For a quadratic function \( y = ax^2 + bx + c \), here \( a = 1>0 \), so the parabola opens upward. The vertex of \( y = x^2 - 9 \) is at \( (0, - 9) \) (since in \( y=x^2 - 9 \), \( h = 0,k=-9 \) for vertex form \( y=a(x - h)^2+k \)).

Step2: Determine the line type and region

Since the inequality is \( y > x^2 - 9 \) (not \( \geq \)), the parabola should be dashed. Then, we test a point not on the parabola, say \( (0,0) \). Substitute into the inequality: \( 0>0^2 - 9\), which is \( 0 > - 9 \), true. So the region above the parabola \( y = x^2 - 9 \) should be shaded.
Looking at the graphs:

• The first graph: Dashed parabola opening upward, shaded above (since the shaded region is above the vertex at \( (0, - 9) \)) and the parabola is dashed (correct for \( > \)).
• The second graph: Parabola opening downward (since \( a\) would be negative), so not matching \( y=x^2 - 9 \) (which opens upward).
• The third graph: Solid parabola (but we need dashed for \( > \)), so incorrect.
• The fourth graph: Parabola opening downward, incorrect.

Answer:

The first graph (the one with the dashed upward - opening parabola and the region above the parabola shaded)