Question

What is the equation of the line that is normal to `f(x)=4x^3-3x^2+5x-2` at `x=1`?

Answer 1

`11y = -x +44`

Explanation:

The slope of the normal at a given point is the negative inverse of the slope of the function at that point. i.e. if the slope of the function is `m` then the slope of the normal is `-1/m`

To find the slope, differentiate the function.

`f'(x) = 12x^2 - 6x +5`

At `x=1` this equals `12 -6 +5 = 11`

Therefore the slope of the normal is `-1/11`

The calculate the constant `c` in the standard lien equation `y=mx+c`, substitute `x=1` back into the original function `f(x)`
`=4*1^3 -3*1^2 +5*1 - 2 = 4`

Therefore the normal is `y = (-1/11)x +4`
`11y = -x +44`