Question

Graph the system of quadratic inequalities.
y leq −x2 + 3
y > x2 − 4x − 2
Which x−value is part of the solution?

Answer 1

\[y \le -x^2+3,~or~x^2\le -y+3 ...(1)\]
it is a downward parabola.
vertex is (0,3), cuts x-axis at \[-\sqrt{3}~and~ \sqrt{3}\]
put x=0 and y=0 in (1)
\[0\le0+3,~or~0\le3\]which is true.
Hence origin lies in the graph.
\[y>x^2-4x-2\]
\[or~y>x^2-4x+(\frac{ 4 }{ 2 })^2-\left( \frac{ 4 }{2 } \right)^2-2\]
\[or~y>\left( x-2 \right)^2-4-2\]
\[or~y+6>\left( x-2 \right)^2 ....(2)\]
\[or~\left( x-2 \right)^2<y+6\]
it is an upward parabola with vertex at(2,-6)
it cuts x-axis when y=0
\[\left| x-2 \right|<\sqrt{6},\]
\[-\sqrt{6}<x-2<\sqrt{6} ,~or~2-\sqrt{6}<x<2+\sqrt{6}\]
put x=0,y=0 in (2)
6>4
which is true.
Hence origin lies in the graph.

shade portion is the required region.