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

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Answer 1 · 2015-11-27T18:08:41

$y = - \frac{1}{3} x + \frac{2}{3}$ $y = -1/3x+2/3$

$f (x) = x^{2} - x - 2$ $f(x)=x^2-x-2$ at $x = 2$ $x=2$ we get $y = f (2) = 0$ $y=f(2) = 0$

$f'(x) = 2x-1$ , so the slope of the tangent at $(2,0)$ is $f'(2)=3$ .

Therefore the slope of the normal line is $-1/3$ . And the equation of the line through $(2,0)$ with slope $-1/3$ is $y = -1/3x+2/3$ .

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