What is the equation of the normal line of $f (x) = x^{3} - 3 x^{2} - 2 x + 6$ $f(x)= x^3-3x^2-2x+6$ at $x = 4$ $x = 4$ ?

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Answer 1 · 2016-02-03T13:10:40

Equation of normal:

$y = - \frac{1}{22} x + \frac{156}{11}$ $y = -1/22 x + 156/11$

$f (4) = 14$ $f(4) = 14$

The line of normal will pass through the point $(4, 14)$ $(4,14)$ .

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

$f'(4) = 22$

From coordinate geometry, the gradient of the line of normal, $m$ , will have a value of

$m = frac{-1}{22}$

The $y$ -intercept, $c$ , is given by

$c = y - mx$

$= 14 - (-1/22)*(4)$

$= 156/11$

Equation of normal:

$y = -1/22 x + 156/11$

Here is the graph

enter image source here

What is the equation of the normal line of f(x)= x^3-3x^2-2x+6f(x)=x3−3x2−2x+6f(x)= x^3-3x^2-2x+6 at x = 4x=4x = 4?