Answer
vertex: $(-1/2,-49/4)$ x-intercepts: $(-4,0)$, $(3,0)$ y-intercept: $(0,-12)$
Work Step by Step
$f(x)=x^2+x-12$
$a=1$, $b=1$, $c=-12$
Vertex of a parabola is found at $x=-b/2a$
$x=-b/2a$
$x=-1/2*1$
$x=-1/2$
$x=-1/2$
$f(x)=x^2+x-12$
$f(-1/2)=(-1/2)^2+(-1/2)-12$
$f(-1/2)=1/4-1/2-12$
$f(-1/2)=-1/4 -12$
$f(-1/2)=-49/4$
$x=0$
$f(x)=x^2+x-12$
$f(0)=0^2+0-12$
$f(0)=0+0-12$
$f(0)=-12$
$f(x)=0$
$f(x)=x^2+x-12$
$0=x^2+x-12$
$0=(x+4)(x-3)$
$x+4=0$
$x+4-4=0-4$
$x=-4$
$x-3=0$
$x-3+3=0+3$
$x=3$
