Question

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

Answer 1

`y=-1/5x+6/5`

Explanation:

The derivative is `f'(x)=8x-3` so that `f'(1)=5`. The slope of the normal line is the opposite reciprocal of this, which is `-1/5`.

Since the point `(1,f(1))=(1,1)` is on the line, the equation of the line is `y-1=-1/5(x-1)`, which is equivalent to `y=-1/5(x-1)+1` and `y=-1/5x+6/5`.