# Find the equation of the tangent line to the function at a given point f(x)=e^-3x+1 ; (0,e)?

Jan 21, 2018

$y - e = - 3 e \left(x - 0\right)$.

#### Explanation:

From the given point I'm assuming the function is $f \left(x\right) = {e}^{- 3 x + 1}$ which contains the point $\left(0 , e\right)$.

Using the Chain Rule on $f \left(x\right)$ we have:

$f ' \left(x\right) = {e}^{- 3 x + 1} \cdot \left(- 3\right)$

So $f ' \left(0\right) = - 3 e$.

The equation of the line can be written in point-slope form:

$y - e = - 3 e \left(x - 0\right)$.

This can be rewritten as $y = - 3 e \cdot x + e$