Question

What is the equation of the line normal to `f(x)=4x^2-7x+6` at `x=1`?

Answer 1

Equation of normal to `f(x)` at `x=1` is `x+y=4`.

Explanation:

At `x=1`, `f(x)=4-7+6=3`, hence normal will be at point `(1,3)` on the curve `f(x)=4x^2-7x+6`.

The slope of tangent will be given by value of `f'(x)` at `x=1`/

As `f'(x)=8x-7`, slope of tangent is `1`.

As normal is perpendicular to tangent, slope of normal will be `-1/1=-1`;

Hence equation of normal given by point slope form will be

`(y-3)=-1*(x-1)` or `y-3=-x+1` or `x+y=4`.