Question

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

Answer 1

`y=435.8x-655.2`. A near-origin graph and a contracted high-y graph are inserted to show that the tangent at ( 2, 218.4) is nearly vertical. .

Explanation:

At (x = 2, y = 4e^4=218.4, nearly.

So, the point of contact is P( 2, 218.4)

`y' =xe^(x+2)(2+x)=8e^2=436.8`, at P.

The tangent is inclined at `tan^(-1)436.8=89.9^o`, nearly to the x-

axis.

The equation of the tangent is

`y-218.4=436.8(x-2)`. Rearranging,

`y=436.8x-655.2`

graph{x^2e(x+2) [-606, 606, -303, 303]}
graph{x^2e(x+2) [-10, 10, -5, 5]}