document.write( "Question 183055: Find the point(s) where the tangent to the curve is horizontal for:
\n" ); document.write( "y=2(x-29)(x+1)
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Algebra.Com's Answer #137429 by stanbon(75887)\"\" \"About 
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Find the point(s) where the tangent to the curve is horizontal for:
\n" ); document.write( "y=2(x-29)(x+1)
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\n" ); document.write( "Take the derivative to get:
\n" ); document.write( "y' = 2[(x-29)*1 + (x+1)*1]\r
\n" ); document.write( "\n" ); document.write( "y' = 2[2x-28)
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\n" ); document.write( "Let y' = 0 and solve for x:
\n" ); document.write( "2[2x-28]= 0
\n" ); document.write( "x = 14
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\n" ); document.write( "The slope is zero when x = 14
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\n" ); document.write( "f(14) = 2(14-29)(14+1) = 2(-15)(15) = -450
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\n" ); document.write( "Ans: (14,-450)
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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