What is the equation of the line normal to $f (x) = 3 x^{2} - 12 x - 5$ $f(x)=3x^2 - 12x-5$ at $x = 1$ $x=1$ ?

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Answer 1 · 2016-01-03T15:17:56

So the equation normal to $f (x)$ $f(x)$ at $x = 1$ $x=1$ is

$y = \frac{1}{6} x - \frac{85}{6}$ $y=1/6x-85/6$

The normal is a line at right angles to the tangent.

First we need to calculate the $d \frac{f (1)}{d x}$ $df(1)/dx$ hence we have that

$f'(x)=6x-12=>f'(1)==-6$

So the normal will have a gradient of $m*(-6)=-1=>m=1/6$

hence at point $(1,f(1))$ or $(1,-14)$ the equation is

$y=m*x+k=>y=1/6*x+k$

replacing the value of point $(1,f(1))$ we get

$-14=1/6+k=>k=-14-1/6=>k=-85/6$

What is the equation of the line normal to f(x)=3x^2 - 12x-5 f(x)=3x2−12x−5f(x)=3x^2 - 12x-5 at x=1x=1x=1?