What is the equation of the normal line of $f (x) = - 2 x^{3} + 18 x^{2} - 25 x$ $f(x)=-2x^3+18x^2-25x$ at $x = - \frac{1}{3}$ $x=-1/3$ ?

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Answer 1 · 2017-10-30T15:15:52

$y = - \frac{3}{113} \cdot x + \frac{31726}{3051}$ $y=-3/113*x+31726/3051$

I took diferential both sides,

$f'(x)=-6x^2+36x-25$

$f'(-1/3)=-2/3-12-25=-113/3$

Hence I found slope of normal line: $-3/113$

I found a point, which passes from it:

$f(-1/3)=2/27+18/9+25/3=281/27$

Thus,

$y-281/27=-3/113*(x+1/3)$

$y-281/27=-3/113*x-1/113$

$y=-3/113*x+31726/3051$

What is the equation of the normal line of f(x)=-2x^3+18x^2-25xf(x)=−2x3+18x2−25xf(x)=-2x^3+18x^2-25x at x=-1/3x=−13x=-1/3?