Question

Finding an Equation of a Tangent Line In Exercise, find an equation of the tangent line to the graph of the function at the given point.
y = x^2 e^-x, (1,1/e)

Answer 1

Answer:

Equation of tangent will be $y=\frac{x}{e}$

Step-by-step explanation:

We have given the function $y=x^2e^{-x}$

We have to find the equation of tangent at the point $(1,\frac{1}{e})$

Equation of tangent is equal to $\frac{dy}{dx}$

So $\frac{dy}{dx}=x^2\frac{d}{dx}e^{-x}+e^{-x}\frac{d}{dx}2x=-x^2e^{-x}+2e^{-x}$

Now we have given point $(1,\frac{1}{e})$

So putting these points in the equation of tangent

$\frac{dy}{dx}=-1^2e^{-1}+2e^{-1}$

$\frac{dy}{dx}=\frac{1}{e}$

Now equation of tangent passing through $(1,\frac{1}{e})$

$y-\frac{1}{e}=\frac{1}{e}(x-1)$

$y=\frac{x}{e}-\frac{1}{e}+\frac{1}{e}=\frac{x}{e}$