Answer
$(-\sqrt 2,2), (\sqrt 2,2), (-2,4), (2,4)$
Work Step by Step
We are given the equations:
$\begin{cases}
y=x^2\\
x^2+y^2-7y+8=0
\end{cases}$
Substitute $y$ for $x^2$ in the second equation and solve it:
$y+y^2-7y+8=0$
$y^2-6y+8=0$
$y^2-2y-4y+8=0$
$y(y-2)-4(y-2)=0$
$(y-2)(y-4)=0$
$y-2=0\Rightarrow y=2$
$y-4=0\Rightarrow y=4$
Determine $x$:
$y=2\Rightarrow x^2=2\Rightarrow x=\pm\sqrt 2$
$y=4\Rightarrow x^2=4\Rightarrow x=\pm\sqrt 4=\pm 2$
There are 4 intersection points:
$(-\sqrt 2,2), (\sqrt 2,2), (-2,4), (2,4)$
Draw the curves: