Question

How do you find the linearization at a=3 of `f(x) = 2x³ + 4x² + 6`?

Answer 1

`y = (6a^2+8a)x - (4a^3+4a^2-6)`

Explanation:

`f'(a)=6a^2+8a`

The tangent line has slope `m=f'(a)` and goes through the point `(a,f(a))`. So the tangent line has point slope form:

`y-f(a) = f'(a)(x-a)`.

The linearization at `x=a`, is:

`y = f'(a)x + f(a)-af'(a)`

`= (6a^2+8a)x - (4a^3+4a^2-6)`