Question

What is the equation of the tangent line of `f(x)=x^2/e^(x+2)` at `x=-2`?

Answer 1

`y=-8x-12`

Explanation:

Writing
`f(x)=x^2e^(-x-2)` so we get by the product and chain rule
`f'(x)=2xe^(-x-2)+x^2e^(-x-2)(-1)`
`f'(x)=e^(-x-2)(2x-x^2)`
and we get

`f'(-2)=e^(2-2)(-4-4)=-8`
and we have

`y=-8x+n`
since `f(-2)=4`

we obtain `n=-12`