document.write( "Question 183055: Find the point(s) where the tangent to the curve is horizontal for:
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document.write( "y=2(x-29)(x+1)
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Algebra.Com's Answer #137429 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! Find the point(s) where the tangent to the curve is horizontal for: \n" ); document.write( "y=2(x-29)(x+1) \n" ); document.write( "--------------------- \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) \n" ); document.write( "--------- \n" ); document.write( "Let y' = 0 and solve for x: \n" ); document.write( "2[2x-28]= 0 \n" ); document.write( "x = 14 \n" ); document.write( "------------------- \n" ); document.write( "The slope is zero when x = 14 \n" ); document.write( "------------------------------ \n" ); document.write( "f(14) = 2(14-29)(14+1) = 2(-15)(15) = -450 \n" ); document.write( "================================= \n" ); document.write( "Ans: (14,-450) \n" ); document.write( "====================== \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( " |