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

1 Answer
Dec 29, 2015

y = x+8

Explanation:

To find the gradient, we need to differentiate, here, using the product rule, as we have two functions multiplied together:

(uv)' = u'v + v'u
u = x^2- x
u' = 2x - 1
v = e^(x+2)
v' = e^(x+2)

Hence (uv)' = u'v + v'u

= (2x-1)e^(x+2) + (e^(x+2))(x^2-x)
=(e^(x+2))(2x-1+x^2-x)

f'(x)=(e^(x+2))(x^2+x-1)

Now, we fill in -2 for x:

f'(-2) = (e^((-2)+2))((-2)^2+(-2)-1)
= e^0(4-2-1)

As e^0 = 1:
f'(-2) = 1*1 = 1

Now you know the gradient, you need to sub into y-b = m(x-a); to get point b, you need to sub x =-2 into the original equation.

f(x) = ((-2)^2 - (-2))*(e^((-2)+2))
= ((4+2)*1)

=6

Now we have our gradient m = 1 and the point (-2,6), so the equation of the tangent is:

y-b = m(x-a)

y-6 = 1(x-(-2))

y = x+2+6

y = x+8