x=36y-3y^2-64 what is the vertex of this equation?
to solve for the vertex, there are actually many ways, but i think the fastest is to get the derivative (since the derivative is either a maximum or a minimum --- like the vertex, which can be a maximum if parabola is concave downward, or a minimum if it is concave upward)
getting the derivative of x with respect to y, or dx/dy:
x = -3y^2 + 36y - 64
dx/dy = -6y + 36
we equate this to zero since at the max or min, the slope is zero:
0 = -6y + 36
6y = 36
y = 6
this is the y-coordinate of the vertex,, to get the x-coordinate, we substitute this to the original equation:
x = -3y^2 + 36y - 64
x = -3*6^2 + 36*6 - 64
x = 44
thus vertex is at (44,6)
hope this helps~ :)