Answer
Standard form: $f(x)=(x+4)^2-6$
Vertex: $(-4,-6)$
Axis of symmetry: $x=-4$.
x-intercepts: $(4+\sqrt 6,0)$ and $(4-\sqrt 6,0)$
Work Step by Step
$f(x)=x^2+8x+10~~$ ($a=1,~b=8,~c=10$)
$f(x)=x^2+8x+16-16+10$
$f(x)=[x^2+2(4)x+4^2]-6$
$f(x)=(x+4)^2-6$
$-\frac{b}{2a}=-\frac{8}{2(1)}=-4$
$f(-4)=(4-4)^2-6=-6$
Vertex: $(-\frac{b}{2a},f(-\frac{b}{2a}))=(-4,-6)$
So, the axis of symmetry is $x=-\frac{b}{2a}=-4$.
The x-intercepts:
$f(x)=0$
$(x+4)^2-6=0$
$(x+4)^2=6$
$x+4=±\sqrt 6$
$x=4±\sqrt 6$
So, the x-intercept points are: $(4+\sqrt 6,0)$ and $(4-\sqrt 6,0)$
