Answer
Standard form: $y=[x-(-5)]^2-11$
Vertex: $(-5,-11)$
Axis of symmetry: $x=-5$
The x-intercepts are: $(-5+\sqrt {11},0)$ and $(-5-\sqrt {11},0)$
Work Step by Step
Standard form:
$y=a(x-h)^2+k$, in which $(h,k)$ is the vertex.
$y=x^2+10x+14$
$y=x^2+2(5)x+5^2-5^2+14$
$y=(x+5)^2-11$
$y=[x-(-5)]^2-11$
Vertex: $(-5,-11)$
Axis of symmetry: $x=-5$
$y=0$
$[x-(-5)]^2-11=0$
$[x-(-5)]^2=11$
$x+5=±\sqrt {11}$
$x=-5+\sqrt {11}$ or $x=-5-\sqrt {11}$
The x-intercepts are: $(-5+\sqrt {11},0)$ and $(-5-\sqrt {11},0)$
