SOLUTION: Find the point(s) where the tangent to the curve is horizontal for: y=2(x-29)(x+1)

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Question 183055: Find the point(s) where the tangent to the curve is horizontal for:
y=2(x-29)(x+1)
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Find the point(s) where the tangent to the curve is horizontal for:
y=2(x-29)(x+1)
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Take the derivative to get:
y' = 2[(x-29)*1 + (x+1)*1]
y' = 2[2x-28)
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Let y' = 0 and solve for x:
2[2x-28]= 0
x = 14
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The slope is zero when x = 14
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f(14) = 2(14-29)(14+1) = 2(-15)(15) = -450
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Ans: (14,-450)
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Cheers,
Stan H.