Find the vertex of the graph of the equation x - y^2 + 8y = 13 + 6y + y^2.
Need to 'massage' this around to vertex form of a parabola ( a sideways parabola due to x = y^2 form) x = a(y-k)^2 + h
x - y^2 + 8y = 13 + 6y + y^2
x = 2y^2 -2y + 13 complete the square for 'y'
x = 2 ( y^2-1) + 13
x = 2 ( y-1/2)^2 - 1/2 + 13
x = 2 (y-1/2)^2 + 12.5
vertex is (12.5 , .5)