I need help with a math question involving finding points on a curve where the tangent is horizontal
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I need help with a math question involving finding points on a curve where the tangent is horizontal

[From: ] [author: ] [Date: 11-10-24] [Hit: ]
Substitute x = -2 and x = 3 into the equation of the curve to find the y-coordinates of the tangent points.-Ooh, Ill have to use calculus for this one. So.y = 6x^2 - 6x - 36.6x^2 - 6x - 36 = 0.......
Q: Find the points on curve y = 2x^3 - 3x^2 - 36x + 4 where the tangent is horizontal.

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If the tangent is horizontal, the slope is zero. The slope of the tangent to a curve is the derivative of the curve at the point of tangency. If the slope of the tangent is zero, the derivative at that point is also zero.

y(x) = 2x³ - 3x² - 36x + 4
y'(x) = 6x² - 6x - 36 = 0
x² - x - 6 = 0
(x-3)(x+2) = 0

The tangents are horizontal at x = -2 and x = 3
Substitute x = -2 and x = 3 into the equation of the curve to find the y-coordinates of the tangent points.

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Ooh, I'll have to use calculus for this one. So.
y' = 6x^2 - 6x - 36.
6x^2 - 6x - 36 = 0.
6(x^2 - x - 6) = 0.
6(x-3)(x+2) = 0.
x = 3, x = -2.

EDIT: Aww, someone beat me to it. Someone else knows calculus on this site. Oh well. He explained it better anyway, go ahead and give him the points.

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First find the derivative of the equation:

y= 6x^2-6x-36

Then you set it = to 0.

6x^2-6x-36=0
6(x^2-x-6)=0
6(x+2)(x-3)=0

Therefore: x=-2 and x=3

To find the y coordinate for each point plug it back into the original equation and there's your two points.
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