Answer
$y=-2x^2+(1)x+4$
or:
$y=-2(x-\frac{1}{4})^2+\frac{33}{8}$
Work Step by Step
$y=ax^2+bx+c$
Plug in the the given points which lie on the parabola, we can write
$a=-2$ and $b=1$
Thus, $y=-2x^2+(1)x+4$
$y-4=-2(x^2-\frac{1}{2}x)$
Therefore,
$y=-2(x-\frac{1}{4})^2+\frac{33}{8}$
