# What is the equation of the line normal to f(x)=6x^2 + 4x - 9  at x=1?

Feb 2, 2017

$x + 16 y - 17 = 0$. See the normal-inclusive Socratic graph.

#### Explanation:

y at x = 1 is 1. So, the foot of the normal is P(1, 1).

y'=12x+4=16, at P.

The slope of the normal $= - \frac{1}{y '} = - \frac{1}{16}$.

So, the equation to the normal at P(1, 1) is

$y - 1 = - \frac{1}{16} \left(x - 1\right)$, giving

$x + 16 y - 17 = 0$.

graph{(6x^2+4x-9-y)(x+16y-17)((x-1)^2+(y-1)^2-.1)=0 [-22, 22, -11, 11]}