We do this by setting $y=4-x^{2}$ equal to $y=x^{2}$ and solving for $x$:
\begin{align*}
4-x^{2} &= x^{2} \\
4 &= 2x^{2} \\
x^{2} &= 2 \\
x &= \pm \sqrt{2}
\end{align*}
Substituting $x=\sqrt{2}$ and $x=-\sqrt{2}$ into either of the original equations gives $y=2$.
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