Question

What is the equation of the line tangent to `f(x)=4x^2 + 4x - 4` at `x=-2`?

Answer 1

`y=-12x-20`

Explanation:

The tangent line will pass through the point `(-2,f(-2))`.

`f(-2)=4(-2)^2+4(-2)-4=4`

Thus, the tangent line will pass through the point `(-2,4)` and have a slope of `f'(-2)`.

`f'(x)=8x+4`

`f'(-2)=8(-2)+4=-12`

The slope of the tangent line is `-12` and it passes through `(-2,4)`.

Write this in point-slope form.

`y-4=-12(x+2)`

In slope-intercept form:

`y=-12x-20`

graph{(y+12x+20)(4x^2+4x-4-y)=0 [-23.14, 17.41, -8.89, 11.37]}