Question

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

Answer 1

Equation of normal is `x-17y-35=0`

Explanation:

As `f(x)=-x^4-2x^3-5x^2+3x+3` and at `x=1` its value is `-1-2-5+3+3=-2` and hence curve passes through `(1,-2)` and we need to find equation of normal at this point.

Slope of tangent is given by function's derivative. It is `f'(x)=-4x^3-6x^2-10x+3` and at `x=1` slope is `-4-6-10+3=-17`. Hence the slope of tangent is `-17` and as normal is perpendicular to it, slope of normal would be `1/17`.

Now using point slope form, equation of normal is

`(y+2)=1/17(x-1)` or `17y+34=x-1` or

`x-17y-35=0`