Answer
$ x+y \leq 2 $
$ y \geq x^{2}-4$
.
Work Step by Step
First sentence: $\quad x+y \leq 2 \quad\Rightarrow y \leq -x+2$
Second sentence: $\quad y \geq x^{2}-4 $
1st inequality ($ \leq$ inequality sign, border solid):
Graph $y=-x+2:$
y intercept=$2$, and for x=$2$, y=0 .
Draw the line through (0,$2$) and ($2$, 0)
Test point: (0,0). Is $0 \leq -(0)+2\quad ?$
Yes. shade the region containing (0,0).
2nd inequality ($ \geq $ inequality sign, border solid)::
Graph $y=x^{2}-4 $(parabola, opens up)
Vertex: (0,-4)
x-intercepts 2,$-2$
Draw the parabola.
Test point: (0,0). $\quad $Is $0 \geq 0^{2}-4\quad ?$
Yes. shade the region containing (0,0).
The solution set is the region with both shadings.
(grey on the screenshot)