Question 72240
In the general equation of a quadratic 
{{{y=(x-h)^2+k}}} Represents a quadratic in "vertex form", in other words the vertex is (h,k). The constant h is the horizontal shift of the vertex and k is the vertical shift of the vertex. For instance if you have {{{y=x^2}}} it looks like this
{{{graph( 300, 200, -5, 5, -5, 5, x^2)}}}The graph of {{{y=x^2}}}
Where the vertex is at (0,0). If you want to shift the graph up or down, you simply add or subtract the appropriate units. For instance, to shift 3 units up, just add 3.
{{{graph( 300, 200, -5, 5, -5, 5, x^2+3)}}}The graph of {{{y=x^2+3}}} (note: h=0 k=3)
For this problem, the shifted units are given to you and all you have to do is find the vertex. If the vertex started at (0,0) and we shifted 2 units up and 5 units to the right then the vertex would be (5,2).
{{{graph( 300, 200, 0, 10, -5, 5, 3*(x-5)^2+2)}}}The vertex is the lowest point on the graph