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

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Answer 1 · 2017-02-02T15:28:38

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

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 $= -1/(y')=-1/16$ .

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

$y-1=-1/16(x-1)$ , giving

$x +16y-17=0$ .

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

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