Question

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

Answer 1

Slope of normal is `x=1`

Explanation:

The slope of tangent at a point `x=x_0` on `f(x)` is `f'(x_0)=((df)/(dx))_(x=x_0)`

and as normal is perpendicular to slope, its slope is `-1/(f'(x_0))`

Here we are seeking normal at `x=1` on curve `f(x)=(x-3)^2+4x` i.e. at `(1,(1-3)^2+4*1)` or `(1,8)`.

As `(df)/(dx)=2(x-3)+4=2x-2`, slope of tangent at `x=1` is `0` i.e. tangent at `(1,8)` is `y=8`

and normal at `(1,8)` would be `x=1`