Question 795063
The vertex is right there, in front of your eyes.
I'll show you how to look for it.
1. {{{h(x) = (x + 2)^2 – 1 }}} is {{{-1}}} when {{{x=-2}}},
because then {{{(x+2)^2=0^2=0}}}.
For all other values of {{{x}}},
{{{(x+2)^2>0}}} and {{{h(x) = (x + 2)^2 – 1>-2 }}}.
So (-2,-1), with {{{x=-2}}} and {{{y=(-2+2)^-1=-1}}} is the minimum and the vertex.
 
2. {{{h(x) = (x + 1)^2 – 2}}} has a minimum and vertex at (-1,-2), because 
when {{{x=-1}}} {{{h(x)=(x + 1)^2 – 2}}} takes the minimum value
{{{h(x)=(-1 + 1)^2 – 2=0^2-2=-2}}}
 
3. {{{h(x) = (x - 2)^2 - 1}}} has a maximum and vertex at (2,-1).
 
4. {{{h(x) = (x - 1)^2 – 2}}} has a maximum and vertex at (1,-2)
 
5. {{{h(x) = (x + 2)^2 + 1}}} has a maximum and vertex at (-2,1)
 
6. {{{h(x) = (x + 1)^2 + 2}}} has a maximum and vertex at (-1,2)
 
NOTES:
If you had a minus sign in front of that square, as in
{{{f(x)=-2(x-3)^2+4}}} the vertex, at (3,4) in this case, would be a maximum, because for {{{x<>3}}} {{{-2(x-3)^2<0}}} and {{{h(x)=-2(x-3)^3+4<4=h(3)}}}
 
When they give you the function as
{{{h(x)=x^2+2x+3}}} you will have to transform it to the form given above
{{{h(x)=x^2+2x+3=x^2+2x+1+2=(x + 1)^2 + 2}}} to find the vertex.
 
You an write your exponents with a "^" in front, as in
h(x) = (x + 1)^2 + 2 , and everyone will know you mean {{{h(x) = (x + 1)^2 + 2}}}