Answer
$y=-\frac{16}{3}(x+\frac{5}{2})^2$
Work Step by Step
Vertex: $(h,k)=(-\frac{5}{2},0)$
Standard form:
$y=a(x−h)^2+k$
$y=a[x-(-\frac{5}{2})]^2+0$
$y=a(x+\frac{5}{2})^2$
Now, use the point $(-\frac{7}{2},-\frac{16}{3})$ to find $a$:
$-\frac{16}{3}=a(-\frac{7}{2}+\frac{5}{2})^2$
$-\frac{16}{3}=a(-1)^2=a(1)$
$a=-\frac{16}{3}$
$y=-\frac{16}{3}(x+\frac{5}{2})^2$
