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

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Answer 1 · 2018-01-25T17:17:05

$y - 6 = \frac{1}{20} (x + 2)$ $y-6=1/20(x+2)$

The normal line is perpendicular to the tangent line.

$f'(x)=-6x^2+4$ so $f'(-2)=-20$ , so the slope of the normal line is $1/20$ , the opposite reciprocal.

$f(-2) = -2(-2)^3+4(-2)-2=6$ .

The equation of the normal line is $y-6=1/20(x+2)$

Answer 2 · 2018-01-25T17:28:08

$y-6= 1/20 (x+2)$

At x=-2, the y co-ordinate of the point would be $-2(-2)^3 +4(-2) -2 =6$

the given point through which the required normal would pass is (-2,6)

Slope of the curve is $dy/dx =-6x^2+4$ . The slope at point (-2,6) would be $-6(-2)^2 +4 = -20$

The slope of the normal passing through this point would thus be $1/20$ (Recollect the formula for slopes of two perpendicular lines $m_1 m_2 =-1$ )

The point slope form of the required line would be $y-6= 1/20 (x+2)$

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