Question

What is the equation of the line tangent to `f(x)=-(x+2)^2-5-3` at `x=5`?

Answer 1

`y=-12x+3`

Explanation:

`f(x)=-[(x+2)^2]-5-3=-[(x+2)^2]-8`

Applying the chain rule:

`f'(x)=-[2(x+2)^1xx1]`

`=-2x-2`

The tangent line is of the form:

`y=mx+c`

`:.m=-2x-2`

When `x=5` then `m=-(2xx5)-2=-12`

To get the `y` coordinate:

`f(5)=-[(5+2)^2]-8=-57`

`:.-57=-12xx5+c`

`c=60-57=3`

So the equation of the tangent line`rArr`

`y=-12x+3`

graph{(-12x+3-y)(-(x+2)^(2)-8-y)=0 [-285.2, 285.4, -142.6, 142.6]}