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.](https://socratic.org/calculus/basic-differentiation-rules/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
