Question 58658
graph y=2(x+5)^-3 and label the vertex and all exact interceps, not to scale on the Y axis

On the off chance that you meant to type y=2(x+5)^2-3, I'm going to give this a try.
{{{highlight(y=a(x-h)^2+k)}}} is vertex form, where the vertex is (h,k)
{{{y=2(x+5)^2-3}}}
Vertex: (h,k)=(-5,-3)
:
Y-intercepts happen when x=0, let x=0 and solve for y.
{{{y=2(0+5)^2-3}}}
{{{y=2(5)^2-3}}}
{{{y=2(25)-3}}}
{{{y=50-3}}}
{{{y=47}}}  There is a y-intercept at (0,47)
:
x-intercepts happen when y=0, let y=0 and solve for x using the square root property:
{{{0=2(x+5)^2-3}}}
{{{0+3=2(x+5)^2-3+3}}}
{{{3=2(x+5)^2}}}
{{{3/2=(2/2)(x+5)^2}}}
{{{3/2=(x+5)^2}}}
{{{+-sqrt(3/2)=sqrt((x+5)^2)}}}
{{{+-sqrt(3/2)=x+5}}}
{{{-5+-sqrt(3/2)=x+5-5}}}
{{{-5+-sqrt(3)/sqrt(2)=x}}}  Some teachers are happy here.
{{{-5+-sqrt(3)*sqrt(2)/(sqrt(2)*sqrt(2))=x}}} 
{{{-5+-sqrt(6)/2=x}}}
The x intercepts are at (-5-sqrt(6)/2,0) and (-5+sqrt(6)/2,0)
{{{graph(300,200,-10,5,-5,50,2(x+5)^2-3)}}}
Happy Calculating!!!