Question

Find the point(s) where the tangent to the curve is horizontal.

y =2(x-29)(x+1)

Answer 1

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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).

Answer 2

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 ?????

Answer 3

one vertex