SOLUTION: the minimum value of the function y=h(x) corresponds to the point (-3,2) on the x-y plane. What is the maximum value of g(x)=6-h(x+2)?

Let's do it by finding specific quadratic functions for y=h(x) 
and y=g(x).

Let's find a quadratic function for y=h(x) that has (-3,2) as a
minimum value.

h(x) = y = A(x-H)² + K which has vertex (H,K).

So choose H=-3 and K=2 and A=1, positive so it will
open upward and the vertex will be at the bottom,
making it a minimum, so

h(x) = (x+3)² + 2

Here is the graph of h(x)



Then 

h(x+2) = (x+2+3)² + 2

h(x+2) = (x+5)² + 2

g(x) = 6-h(x+2)

g(x) = 6-[(x+5)² + 2]

g(x) = 6 - (x+5)² - 2

g(x) = -(x+5)² + 4

which has vertex (-5,4) and opens downward,
because of the negative sign before (x+5)²

so its vertex (-5,4) is a maximum.

Answer: (-5,4)

Here are both graphed on the same set of axes.
The red graph is of y=h(x) and the green
one is y=g(x)



Edwin