Question

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

Answer 1

`y=-(2sqrt2)/3x+(7sqrt2)/3`

Explanation:

The normal line will be perpendicular to the tangent line when `x=2`.

We can determine what point on `f(x)` the normal line will intersect by finding that `f(2)=sqrt2`, so the point is `(2,sqrt2)`.

If we already have a point on the normal line, all we need to know is its slope. We can find of the tangent line when `x=2` by finding `f'(2)`. Since the normal line is perpendicular to the tangent line, its slope will be the opposite reciprocal of the tangent line's.

`f(x)=(x^2-x)^(1/2)`

Finding `f'(x)` will require use of the chain rule.

`f'(x)=1/2(x^2-x)^(-1/2)d/dx[x^2-x]`

`f'(x)=1/2(x^2-x)^(-1/2)(2x-1)`

`f'(x)=(2x-1)/(2sqrt(x^2-x))`

Find the slope of the tangent line.

`f'(2)=(2(2)-1)/(2sqrt(2^2-2))=3/(2sqrt2)`

Take the opposite reciprocal to find that the slope of the normal line is `-(2sqrt2)/3`. Remember that it passes through the point `(2,sqrt2)`.

Write the equation in point-slope form:

`y-sqrt2=-(2sqrt2)/3(x-2)`

In slope-intercept form:

`y=-(2sqrt2)/3x+(7sqrt2)/3`