What is the equation of the normal line of $f (x) = - x^{4} + 2 x^{3} - x^{2} - 9 x - 3$ $f(x)=-x^4+2x^3-x^2-9x-3$ at $x = - 1$ $x=-1$ ?

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Answer 1 · 2015-11-24T02:12:17

$y = - \frac{1}{3} x + \frac{5}{3}$ $y = -1/3x+5/3$

$f (- 1) = - 1 - 2 - 1 + 9 - 3 = 2$ $f(-1) = -1-2-1+9-3 =2$

$f'(x) = -4x^3+6x^2-2x-9$

so $f'(-1) = 4+6+2-9 = 3$ .

The slope of the tangent line at $(-1,2)$ is $3$ , so the normal line has slope $-1/3$ .

The line with slope $m=-1/3$ through the point $(-1,2)$ is

$y-2 = -1/3(x-(-1))$

$y-2 = -1/3x-1/3$

$y = -1/3x+5/3$

What is the equation of the normal line of f(x)=-x^4+2x^3-x^2-9x-3f(x)=−x4+2x3−x2−9x−3f(x)=-x^4+2x^3-x^2-9x-3 at x=-1x=−1x=-1?