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

Jan 25, 2018

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

Explanation:

The normal line is perpendicular to the tangent line.

$f ' \left(x\right) = - 6 {x}^{2} + 4$ so $f ' \left(- 2\right) = - 20$, so the slope of the normal line is $\frac{1}{20}$, the opposite reciprocal.

$f \left(- 2\right) = - 2 {\left(- 2\right)}^{3} + 4 \left(- 2\right) - 2 = 6$.

The equation of the normal line is $y - 6 = \frac{1}{20} \left(x + 2\right)$

Jan 25, 2018

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

Explanation:

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

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

Slope of the curve is $\frac{\mathrm{dy}}{\mathrm{dx}} = - 6 {x}^{2} + 4$. The slope at point (-2,6) would be $- 6 {\left(- 2\right)}^{2} + 4 = - 20$

The slope of the normal passing through this point would thus be$\frac{1}{20}$ (Recollect the formula for slopes of two perpendicular lines ${m}_{1} {m}_{2} = - 1$)

The point slope form of the required line would be $y - 6 = \frac{1}{20} \left(x + 2\right)$