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.
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.
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.
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.