SOLUTION: Find the equation of the tangent line to y=e^(-9t) at t=0 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

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Find the equation of the tangent line to y=e^(-9t) at t=0 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       Log On


   

Question 973603: Find the equation of the tangent line to y=e^(-9t) at t=0
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!
Answer by Fombitz(32388) About Me  (Show Source):
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.
The derivative of f is,
df%2Fdt=-9e%5E%28-9t%29
So the value at t=0 is,
df%2Fdt=-9%281%29=-9
So then using the point-slope form of a line,
y-1=-9%28x-0%29
y-1=-9x
y=-9x%2B1
.
.
.
.