What is the equation of the normal line of f(x)=x2−x at x=2?

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titemomo8gjz · Accepted Answer

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
$f (2) = \sqrt{2}$ , so the point is $(2, \sqrt{2})$ .
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.
$f (x) = {(x^{2} - x)}^{\frac{1}{2}}$
Finding f'(x) will require use of the chain rule.
$f^{'} (x) = \frac{1}{2} {(x^{2} - x)}^{- \frac{1}{2}} \frac{d}{d x} [x^{2} - x]$
$f^{'} (x) = \frac{1}{2} {(x^{2} - x)}^{- \frac{1}{2}} (2 x - 1)$
$f^{'} (x) = \frac{2 x - 1}{2 \sqrt{x^{2} - x}}$
Find the slope of the tangent line.
$f^{'} (2) = \frac{2 (2) - 1}{2 \sqrt{2^{2} - 2}} = \frac{3}{2 \sqrt{2}}$
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 $(2, \sqrt{2})$ .
Write the equation in point-slope form:
$y - \sqrt{2} = - \frac{2 \sqrt{2}}{3} (x - 2)$
In slope-intercept form:
$y = - \frac{2 \sqrt{2}}{3} x + \frac{7 \sqrt{2}}{3}$

What is the equation of the normal line

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