Find the point(s) where the tangent to the curve is horizontal.
y =2(x-29)(x+1)
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 ?????
one vertex
To find the point(s) where the tangent to the curve is horizontal, we need to find the values of x where the slope of the curve is zero.
The slope of the curve y = 2(x-29)(x+1) can be found by taking its derivative with respect to x. Let's differentiate the equation using the product rule:
dy/dx = 2(x+1)(d/dx)(x-29) + (x-29)(d/dx)(x+1)
Now, let's find the derivatives of (x+1) and (x-29):
(d/dx)(x+1) = 1
(d/dx)(x-29) = 1
Substituting these derivatives back into the equation:
dy/dx = 2(x+1)(1) + (x-29)(1)
dy/dx = 2x + 2 + x - 29
dy/dx = 3x - 27
To find the point(s) where the tangent is horizontal, we need to set the slope equal to zero and solve for x:
3x - 27 = 0
Adding 27 to both sides:
3x = 27
Dividing both sides by 3:
x = 9
Therefore, the tangent to the curve is horizontal at the point x = 9. To find the corresponding y-coordinate, we substitute x = 9 back into the original equation:
y = 2(9-29)(9+1)
y = 2(-20)(10)
y = -400
Therefore, the point where the tangent is horizontal is (9, -400).
^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).