What is the equation of the normal line of $f(x)=x^3-8x^2+9x-3$ at $x=-1$ ?

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Answer 1 · 2016-11-22T08:07:45

$f(-1)=-21$

$f'(x) = 3x^2-16x+9$

$f'(-1) = 3+16+9=28 = m_tan$

The slope of the normal line at $x=-1$ is the negative reciprocal of the slope of the tangent line $m=28$ which would be $-1/28$ .

So, using a slope formula:

$y-y_1=m(x-x_1)$

We get:

$y+21=-1/28(x+1)$

$y=-1/28x-589/28$

What is the equation of the normal line of f(x)=x^3-8x^2+9x-3f(x)=x^3-8x^2+9x-3 at x=-1x=-1?