Question

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

Answer 1

`x-6y=-52`

Explanation:

Given `f(x)=(5-x)^2`
`color(white)("XXX")rarr f(x)=25-10x+x^2`
Its general slope is given by the derivative:
`color(white)("XXX")f'(x) = 2x-10`

At `x=2` this slope will be
`color(white)("XXX")2(2)-10 = -6`
and a line normal to it (i.e. perpendicular to it) will have a slope
`color(white)("XXX")=-(1/(-6))=1/6`

At `x=2, f(2)= (5-2)^2=9`
giving a point `(2,9)`

Using the slope-point form for the normal line
`color(white)("XXX")y-9=1/6(x-2)`

We can simplify this as
`color(white)("XXX")6y-54=x-2`
or (in standard form)
`color(white)("XXX")x-6y=-52`
graph{(y-(5-x)^2)(x-6y+52)=0 [-7.83, 14.67, -0.585, 10.665]}