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]}