How do you find the equation of a normal line to a curve at a given point?

Question

lildeutsch11xq2j · Accepted Answer

The equation of a normal line will have the form
y=mx+b
and its slope will be the negative reciprocal of the curve's derivative at the point. That is to say, take the value of the derivative at the point, divide 1 by it, and then multiply that value by -1.
After incorporating the point's x, y, and m coordinates, you must then solve for b.
This is much better illustrated with an example:
Let's say that we are expected to find the equation of a line normal to the curve $f (x) = x^{2}$ at the point (2,4). Since the normal line is perpendicular to the tangent line, we must first determine the slope of the tangent line using the derivative of f(x), and then we must get the slope of the normal line using the negative reciprocal of the slope of the tangent line.
$\frac{d}{d x} f (x) = 2 x$
$\frac{d}{d x} f (2) = 2 \cdot 2 = 4$
The negative reciprocal of 4 is $- \frac{1}{4}$ . We now have a value for m:
$y = - \frac{1}{4} x + b$
The last step is to plug in our point's coordinates and find b.
$4 = - \frac{1}{4} \cdot 2 + b$
$4 = - \frac{2}{4} + b$
$4 + \frac{1}{2} = b$
$b = \frac{9}{2}$
We currently have everything required to complete our entire equation:
$y = - \frac{1}{4} x + \frac{9}{2}$

Drantumcem0 · Accepted Answer

In the point-slope formula for the equation of the line, enter n and the specified point,&amp;nbsp;(y−y1)=n(x−x1)(y−y1)=n(x−x1)(y−y1)=n(x−x1). Simplify the equation by solving for y.

How do you find the equation of a

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New Questions in Calculus 1