Question

Find the point on the curve x=4y-y^2 where the tangent to the curve is a vertical line.

My work: Finding the derivative.

1=4(dy/dx)-2y(dy/dx)
1=dy/dx(4-2y)
dy/dx=1/4-2y
Therefore, y cannot equal +2 or -2
Right?

Answer 1

take the derivative respect to y not x
if you graph it its a sideway parabola.
x=4y-y^2
x'=4-2y
when derivative = 0 you have the min of the function.
0=4-2y
-4=-2y
2=y
plug 2 in to the equation
x=4y-y^2
x=4(2)-(2)^2=4
vertical line at point(4,2)
x=4 is vertical tangent to the curve.