document.write( "Question 1136901: find the equation of the normal to the curve with the equation y=e^3x-2 at the point (1,e)
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Algebra.Com's Answer #754741 by Alan3354(69443) $\"\"$ $\"About$

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y=e^3x-2
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\n" ); document.write( "Is the exponent 3x ?
\n" ); document.write( "Or 3x - 2 ?
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\n" ); document.write( "Parentheses are free. Use some.
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\n" ); document.write( " $\"f%28x%29+=+e%5E%283x-2%29\"$
\n" ); document.write( "f'(x) = $\"%283x-2%29%2Ae%5E%283x-2%29%2A3+=+%289x-6%29%2Ae%5E%283x-2%29\"$ = slope
\n" ); document.write( "f'(1) = 3e
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\n" ); document.write( "slope of normal = -1/3e
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\n" ); document.write( "y-e = (-1/3e)*(x-1)
\n" ); document.write( " $\"y+=+e+%2B+%281-x%29%2F3e\"$ \r
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