posted by anonymous on .
find all points on the curve where x=2. Show there is a horizontal tangent to the curve at one of the points.
sub x = 2 into your equation to get
4 - 4y = y^2 - 1
0 = y^2 + 4y - 5
(y+5)(y-1) = 0
so when x=2, y =-5 or when x=2, y = 1
so the points are (2,-5) and (2,1)
check them by subbing them back into the original equation, they work
I found the derivative of your equation implicitly and got
dy/dx = (2x+2xy)/(2y = x^2)
for a horizontal tangent this has to be zero, which means the numerator should be zero
subbing in (2,-5) makes the top zero, so there is a horizontal tangent at the point (2,_5)