Question

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

Answer 1

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

Explanation:

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

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

Explanation:

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)`