Question

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

Answer 1

`y=-1/5x+2`

Explanation:

Find the slope of the tangent line at the point `(5,1)` using the derivative. Then, write the equation of the line with a perpendicular slope at the point `(5,1)`.

Remember that `d/(dx)[ax^n]=nax^(n-1)`, where `a` is a constant.
`f'(x)=2x-5`
`f'(5)=5`

Therefore, if the slope of the tangent line at `x=5` is `5`, the perpendicular slope of the normal line is `-1/5`.

Write in point-slope form: `y-1=-1/5(x-5)`

In slope-intercept form: `y=-1/5x+2`