Determine equations of normal lines

Given a differentiable function and a specified point on its graph, the normal line is the line through that point perpendicular to the tangent, whose slope is the negative reciprocal of $f'(a)$ : $y-f(a)=-\frac{1}{f'(a)}(x-a)$ when $f'(a)\ne0$ . The understanding includes recognizing that a horizontal tangent gives a vertical normal, while treatment is limited to ordinary Cartesian function graphs rather than parametric or implicit curves.

Detailed Explanation: Determine equations of normal lines

To determine a normal line, use this process:

Find the point on the graph.
Find the derivative, $f'(x)$ .
Evaluate the derivative at the given $x$ -value to get the tangent slope.
Take the negative reciprocal to get the normal slope.
Use point-slope form.

Example

Find the equation of the normal line to

f(x)=x^2+1

at the point where $x=2$ .

First, find the point on the graph:

f(2)=2^2+1=5

So the point is $(2,5)$ .

Next, find the derivative:

f'(x)=2x

The tangent slope at $x=2$ is

f'(2)=2(2)=4.

The normal line is perpendicular to the tangent, so its slope is the negative reciprocal:

m_{\text{normal}}=-\frac{1}{4}.

Now use point-slope form, $y-y_1=m(x-x_1)$ , with the point $(2,5)$ :

y-5=-\frac14(x-2).

Therefore, the equation of the normal line is

\boxed{y-5=-\frac14(x-2)}.

If the tangent slope is $0$ , the tangent is horizontal, so the normal is vertical. In that case, the normal line has the form $x=a$ .

Learn by doing: Determine equations of normal lines

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Slope - Find Perpendicular - Graph to Slope Zero Intercept Form

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