^2
The point(s) where the tangent to the curve is horizontal is when the derivative of the curve is equal to zero. The derivative of the curve is:
y' = 4(x+1)(x-29) + 2(x+1)^2
Setting y' = 0 and solving for x yields x = -1 and x = 29. Therefore, the point(s) where the tangent to the curve is horizontal is (x,y) = (-1,0) and (x,y) = (29,0).
Find the point(s) where the tangent to the curve is horizontal.
y =2(x-29)(x+1)
3 answers
I just knew there was no way the bot could handle such a Calculus question!!
dy/dx = 2(x-29) + 2(x+1) = 0 for a hozizontal tangent
x - 29 = -x - 1
2x = 28
x = 14
then y = 2(14-29)(14+1) = -450
The point is (14,-450)
come on Mr. bot, the function is just a weird parabola, how can it have
more than one vertices ?????
dy/dx = 2(x-29) + 2(x+1) = 0 for a hozizontal tangent
x - 29 = -x - 1
2x = 28
x = 14
then y = 2(14-29)(14+1) = -450
The point is (14,-450)
come on Mr. bot, the function is just a weird parabola, how can it have
more than one vertices ?????
one vertex