## Algebra 2 (1st Edition)

$y=-\frac{1}{2}(x-3)^{2}+5$
We already know $y$, $x$, $k$ and $h$ from the given vertex and point, so in order to get the quadratic function we need to find $a$... $y=a(x-h)^{2}+k$ $-3=a(7-3)^{2}+5$ $-8=a(4)^{2}$ $16a=-8$ $a=-\frac{8}{16}$ $a=-\frac{1}{2}$ Therefore the function is... $y=-\frac{1}{2}(x-3)^{2}+5$