Answer
Standard form: $y=[x-(-4)]^2-5$
Vertex: $(-4,-5)$
Axis of symmetry: $x=-4$
The x-intercepts are: $(-4+\sqrt 5,0)$ and $(-4-\sqrt 5,0)$
Work Step by Step
Standard form:
$y=a(x-h)^2+k$, in which $(h,k)$ is the vertex.
$y=x^2+8x+11$
$y=x^2+2(4)x+4^2-4^2+11$
$y=(x+4)^2-5$
$y=[x-(-4)]^2-5$
Vertex: $(-4,-5)$
Axis of symmetry: $x=-4$
$y=0$
$[x-(-4)]^2-5=0$
$[x-(-4)]^2=5$
$x+4=±\sqrt 5$
$x=-4+\sqrt 5$ or $x=-4-\sqrt 5$
The x-intercepts are: $(-4+\sqrt 5,0)$ and $(-4-\sqrt 5,0)$
