# How do you find the equation of the line tangent to the graph of y=3x^2-6x+2, at the point (3,11)?

Apr 11, 2018

$y - 12 x + 25 = 0$

#### Explanation:

$y = 3 {x}^{2} - 6 x + 2$

Apply differentiation (power rule )

$y ' = 6 x - 6$

substitute with $x$ value to get the slope $m$

$y ' = 12$

substitute in the equation of the tangent with the point and slope

$y - {y}_{1} = m \left(x - {x}_{1}\right)$

where
$\left({x}_{1} , {y}_{1}\right) = \left(3 , 11\right)$

$y - 11 = 12 \left(x - 3\right)$

Simplify

$y - 12 x + 25 = 0$

I hope I helped.