How do i graph the quadratic y=4(x-2)^2-1 ?

Let's find the vertex first, h=-2 <0 so the graph move the right 2 units and k=1 >0 the graph shift up 1 unit. The vertex is (2,1)
Now we find some point of the graph . ( at least 3 (include the vertex))
the normal point when the vertex is (0,0) are (1,1) and (-1,1). Now take those points plus the vertex like 
for the point ( 1,1) plus (2,1) : (1+2,1+1) = (3,2) is the first point. 
another point is (-1+2,1+1)=(1,2) . 

Number nine is f(x)= 40x : ) hoped i ya