The given problem is already in the vertex form.

#color(blue)"The vertex form"#

#color(maroon)(y=a(x-h)^2+k)#

Where #(h,k)# is the vertex.

Our problem

#y=-2(x+1)^2+7#

#y=-2(x-(-1))^2+7#

#(h,k) = (-1,7)#

**The vertex is #(-1,7)#**

Intercepts on #x# and #y# axes occur where the curve crosses them.

To find #y# intercept we need to plug in #x=0#

#y=-2(0+1)^2+7#

#y=-2(1)+7#

#y=-2+7#

#y=5#

**The #y-#intercept is #(0,5)#**

For finding #x-#intercepts, we need to plug in #y=0#

#0=-2(x+1)^2+7#

Subtract #7# from both ends and isolating the term containing #x#

#-7 = -2(x+1)^2#

Let us rewrite it as #-2(x+1)^2=-7# It looks better to when the variable is kept of the left side of the equation.

#-2(x+1)^2=-7# dividing by #-2# on both sides isolates #(x+1)^2#

We get

#(x+1)^2=7/2#

Take square root on both the sides we get

#sqrt((x+1)^2) = +-sqrt(7/2)#

#x+1 = +-sqrt(7/2)#

Subtract #1# from both sides to solve for #x#

#x=-1+-sqrt(7/2)#

**The #x-#intercepts are #(-1+sqrt(7/2))# and #(-1-sqrt(7/2))#**