document.write( "Question 973603: Find the equation of the tangent line to y=e^(-9t) at t=0
\n" ); document.write( "I already used t=0 to find a point- f(0)=e^(-9x0)= 1 so my point is at (0,1) but I'm unsure of what to do next. Any help is appreciated thank you! \n" ); document.write( "

Algebra.Com's Answer #595737 by Fombitz(32388) $\"\"$ $\"About$

You can put this solution on YOUR website!
The slope of the tangent line at a point is equal to the value of the derivative at that point.
\n" ); document.write( "The derivative of f is,
\n" ); document.write( " $\"df%2Fdt=-9e%5E%28-9t%29\"$
\n" ); document.write( "So the value at $\"t=0\"$ is,
\n" ); document.write( " $\"df%2Fdt=-9%281%29=-9\"$
\n" ); document.write( "So then using the point-slope form of a line,
\n" ); document.write( " $\"y-1=-9%28x-0%29\"$
\n" ); document.write( " $\"y-1=-9x\"$
\n" ); document.write( " $\"y=-9x%2B1\"$
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