Answer
Standard form: $y=[x-(-\frac{1}{2})]^2-\frac{121}{4}$
Vertex: $(-\frac{1}{2},-\frac{121}{4})$
Axis of symmetry: $x=-\frac{1}{2}$
The x-intercepts are: $(5,0)$ and $(-6,0)$
Work Step by Step
Standard form:
$y=a(x-h)^2+k$, in which $(h,k)$ is the vertex.
$y=-(x^2+x-30)$
$y=-[x^2+2(\frac{1}{2})x+(\frac{1}{2})^2-(\frac{1}{2})^2-30]$
$y=-[(x+\frac{1}{2})^2-\frac{121}{4}]$
$y=[x-(-\frac{1}{2})]^2-\frac{121}{4}$
Vertex: $(-\frac{1}{2},-\frac{121}{4})$
Axis of symmetry: $x=-\frac{1}{2}$
$y=0$
$[x-(-\frac{1}{2})]^2-\frac{121}{4}=0$
$[x-(-\frac{1}{2})]^2=\frac{121}{4}$
$x+\frac{1}{2}=±\frac{11}{2}$
$x=5$ or $x=-6$
The x-intercepts are: $(5,0)$ and $(-6,0)$
