 Eddie Clarke

2022-04-09

What is the equation of the normal line of $f\left(x\right)=\sqrt{{x}^{2}-x}$ at x=2? titemomo8gjz

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
$f\left(2\right)=\sqrt{2}$, 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'(2). 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'(x) will require use of the chain rule.

${f}^{\text{'}}\left(x\right)=\frac{1}{2}{\left({x}^{2}-x\right)}^{-\frac{1}{2}}\left(2x-1\right)$
${f}^{\prime }\left(x\right)=\frac{2x-1}{2\sqrt{{x}^{2}-x}}$
Find the slope of the tangent line.
${f}^{\prime }\left(2\right)=\frac{2\left(2\right)-1}{2\sqrt{{2}^{2}-2}}=\frac{3}{2\sqrt{2}}$
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:
$y=-\frac{2\sqrt{2}}{3}x+\frac{7\sqrt{2}}{3}$

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