The quadratic function f(x) = a(x - h)^2 + k, a not equal to zero, is said to be in standard form.
So the vertex is the point (h,k)
Which is (-2,-5) substitute this in the above equation
f(x)=a(x-(-2))^2 -5
f(x)=a(x+2)^2-5
And we have passing point (0,3)
F(0)=3
3=a(0+2)^2-5
3=a(4)-5
3=4a-5
3+5=4a
8=4a divide both sides by 4
a=2
F(x)=2(x+2)^2-5