Question

What is the equation of the normal line of `f(x)=-x^4+4x^3-x^2+5x-6` at `x=2`?

Answer 1

Slope of normal is `x+17y-818=0`

Explanation:

As the function is `f(x)=-x^4+4x^3-x^2+5x-6`, the slope of tangent at any point will be the value of `f'(x)` at that point.

As `f'(x)=-4x^3+12x^2-2x+5`, the slope of the tangent at `x=2` will be

`-4*2^3+12*2^2-2*2+5=-32+48-4+5=17`.

And slope of normal would be `-1-:17=-1/17`

Note that value of function at `x=2` is `f(x)=2^4+4*2^3-2^2+5*2-6`

= `16+32-4+10-6=48`

Hence, slope of normal is `-1/17` and it passes through `(2,48)`

its equation of normal is `y-48=-1/17(x-2)` or `17(y-48)=-x+2` or

`x+17y-818=0`