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

Nov 9, 2015

$y = \left(6 {a}^{2} + 8 a\right) x - \left(4 {a}^{3} + 4 {a}^{2} - 6\right)$

#### Explanation:

$f ' \left(a\right) = 6 {a}^{2} + 8 a$

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

$y - f \left(a\right) = f ' \left(a\right) \left(x - a\right)$.

The linearization at $x = a$, is:

$y = f ' \left(a\right) x + f \left(a\right) - a f ' \left(a\right)$

$= \left(6 {a}^{2} + 8 a\right) x - \left(4 {a}^{3} + 4 {a}^{2} - 6\right)$