document.write( "Question 330494: please find the equation of the tangent to the curve given by
\n" ); document.write( "y= x^3 - 5x + 15
\n" ); document.write( "at the point where x= -1
\n" ); document.write( "I got the answer to be
\n" ); document.write( "y= -2x +17
\n" ); document.write( "but im not sure if this is right.
\n" ); document.write( "could you show working if im incorrect please.thanks
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Algebra.Com's Answer #236895 by Fombitz(32388)\"\" \"About 
You can put this solution on YOUR website!
The slope of the tangent line equals the value of the derivative at x=-1.
\n" ); document.write( "\"y=x%5E3-5x%2B15\"
\n" ); document.write( "\"dy%2Fdx=3x%5E2-5\"
\n" ); document.write( "At \"x=-1\"
\n" ); document.write( "\"y=%28-1%29%5E3-5%28-1%29%2B15=-1%2B5%2B15=-19\"
\n" ); document.write( "\"dy%2Fdx=3%28-1%29%5E2-5=-2\"
\n" ); document.write( "Use the point-slope form of a line, \"y-yp=m%28x-xp%29\"
\n" ); document.write( "\"y-%28-19%29=-2%28x-%28-1%29%29\"
\n" ); document.write( "\"y%2B19=-2%28x%2B1%29\"
\n" ); document.write( "\"y=-2x-2-19\"
\n" ); document.write( "\"y=-2x%2B17\"
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\n" ); document.write( "That's what I get too.
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