Question

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

Answer 1

`y=(-x/22)+(45/22)`

Explanation:

Take the derivative of `f(x)`. The derivative of `f(x)` is

`f^'(x) = 3(x^2)-5`

Plug `3` into the derivative to get

`f^(')(3) = 3(3^2)-5 = 22`

That is the slope tangent to the curve but to find the slope of the tangent line, you need to take the reciprocal of that number and then switch signs, so the slope of the normal line is `-1/22`.

Now plug `3` into the initial equation.

`f(3) = 3^3-5(3)+2 = 14`

Now that we have the point `(3,14)` and the slope of the normal line, we can plug into the equation

`(y-y_1) = dy/dx(x-x_1)`

Plug in `14` for `y_1`, `-1/22` for `dy/dx`, and `3` for `x_1`. Solve for `y` and you should get

`y = (-x/22) + (45/22)`