document.write( "Question 1155282: Find the slope of the line tangent to the following at point, where x=2
\n" ); document.write( "f(x)=e^(-0.1x^2+2x-3).
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Algebra.Com's Answer #777859 by Alan3354(69443) $\"\"$ $\"About$

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Find the slope of the line tangent to the following at point, where x=2
\n" ); document.write( "f(x)=e^(-0.1x^2+2x-3).
\n" ); document.write( "--------------
\n" ); document.write( "f(x)=e^(-0.1x^2+2x-3)
\n" ); document.write( "---
\n" ); document.write( "f'(x) = $\"e%5E%28-0.1x%5E2%2B2x-3%29%2A%28-0.2x+%2B+2%29\"$
\n" ); document.write( "Sub 2 for x
\n" ); document.write( "===================
\n" ); document.write( "---
\n" ); document.write( "f'(2) = $\"e%5E%28-0.4+%2B+4+-3%29%2A%28-0.8+%2B+2%29\"$
\n" ); document.write( "f'(2) = $\"1.2e%5E%280.6%29\"$ \n" ); document.write( "

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