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

##### 1 Answer
Dec 11, 2015

$y = - \frac{2 \sqrt{2}}{3} x + \frac{7 \sqrt{2}}{3}$

#### Explanation:

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

We can determine what point on $f \left(x\right)$ the normal line will intersect by finding that $f \left(2\right) = \sqrt{2}$, so the point is $\left(2 , \sqrt{2}\right)$.

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 $f ' \left(2\right)$. Since the normal line is perpendicular to the tangent line, its slope will be the opposite reciprocal of the tangent line's.

$f \left(x\right) = {\left({x}^{2} - x\right)}^{\frac{1}{2}}$

Finding $f ' \left(x\right)$ will require use of the chain rule.

$f ' \left(x\right) = \frac{1}{2} {\left({x}^{2} - x\right)}^{- \frac{1}{2}} \frac{d}{\mathrm{dx}} \left[{x}^{2} - x\right]$

$f ' \left(x\right) = \frac{1}{2} {\left({x}^{2} - x\right)}^{- \frac{1}{2}} \left(2 x - 1\right)$

$f ' \left(x\right) = \frac{2 x - 1}{2 \sqrt{{x}^{2} - x}}$

Find the slope of the tangent line.

$f ' \left(2\right) = \frac{2 \left(2\right) - 1}{2 \sqrt{{2}^{2} - 2}} = \frac{3}{2 \sqrt{2}}$

Take the opposite reciprocal to find that the slope of the normal line is $- \frac{2 \sqrt{2}}{3}$. Remember that it passes through the point $\left(2 , \sqrt{2}\right)$.

Write the equation in point-slope form:

$y - \sqrt{2} = - \frac{2 \sqrt{2}}{3} \left(x - 2\right)$

In slope-intercept form:

$y = - \frac{2 \sqrt{2}}{3} x + \frac{7 \sqrt{2}}{3}$