Question

$\begin{cases} y \le -x^2 \\ y \ge x^2 - 2x - 8 \end{cases}$

          $\begin{cases} y \le -x^2 \\ y \ge x^2 - 2x - 8 \end{cases}$

y ≤ -x^2
y ≥ x^2 - 2x - 8

Added by Emilio Z.

Geometry A Common Core Curriculum

Ron Larson, Laurie Boswell 1st Edition

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Solved by Expert Lashaumbe Jernigan

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