# How do you find the equation of the line tangent to f(x) = 6x^2 - 1 at x = 3?

May 28, 2018

$y = 36 x - 55$

#### Explanation:

$f \left(x\right) = 6 {x}^{2} - 1$ , $\textcolor{w h i t e}{a a}$ $x$$\in$$\mathbb{R}$

$f ' \left(x\right) = 12 x$

$f \left(3\right) = 53$

$f ' \left(3\right) = 36$

The equation of the tangent line at $A \left(3 , f \left(3\right)\right)$ will be

$y - f \left(3\right) = f ' \left(3\right) \left(x - 3\right)$ $\iff$

$y - 53 = 36 \left(x - 3\right)$ $\iff$

$y = 36 x - 55$

graph{(y-6x^2+1)(y-36x+55)=0 [-41.1, 41.1, -20.55, 20.55]}