Question 973603
The slope of the tangent line at a point is equal to the value of the derivative at that point.
The derivative of f is,
{{{df/dt=-9e^(-9t)}}}
So the value at {{{t=0}}} is,
{{{df/dt=-9(1)=-9}}}
So then using the point-slope form of  a line,
{{{y-1=-9(x-0)}}}
{{{y-1=-9x}}}
{{{y=-9x+1}}}
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  *[illustration va1.JPG].