AP Calculus AB and BC

Tangent and Normal Line Finder (Teaching Mode)

Q: How do I find the normal line once I have the tangent?

Keep the same point (a, f(a)) and use the negative reciprocal of the tangent slope. If the tangent slope is f'(a), the normal slope is -\frac{1}{f'(a)}, and the normal line is y - f(a) = -\frac{1}{f'(a)}(x - a). Perpendicular slopes always multiply to -1, which is what negative reciprocal means.

Q: What if the derivative is zero at the point?

Then the tangent is horizontal: f'(a) = 0 makes the tangent line y = f(a), a flat line through the point. The normal is perpendicular to horizontal, so it is the vertical line x = a. You cannot write it as y = mx + b because -\frac{1}{0} has no value, so state the normal as x = a.

Q: Do I plug in the point before or after differentiating?

After. Differentiate first to get the whole function f'(x), then substitute x = a to get the slope as one number. Substituting into f(x) too early loses the variable you need to differentiate, and it is another common mistake that produces a wrong slope on this problem type.

The tangent line at x = a has slope f'(a) and passes through the point (a, f(a)), so its equation is y = f'(a)(x - a) + f(a). The normal line is perpendicular there, so its slope is the negative reciprocal -1/f'(a). Method: find f'(x), evaluate it at a to get a number, then use point-slope form.

Problem

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Tangent and normal line

f(x) = x^2 \quad \text{at} \quad x = 1

f(1) = 1^2 = 1 \quad \text{so the point is } (1, 1)

A tangent line touches a curve at a single point and matches its steepness there; a normal line passes through that same point at a right angle to the tangent. To build either one you need two things: a slope and a point. The point is $(a, f(a))$ , and the slope is $f'(a)$ , the derivative evaluated at $x = a$ . That single fact, that the derivative at a point equals the slope of the tangent line there, is AP Topic 2.2 (CHA-2.C.1), and it is what this tool makes visible: it computes $f'(x)$ , evaluates the slope at the point you pick, and assembles both the tangent and the normal equation from point-slope form.

Here is why the tangent slope is the derivative and not a separate formula you have to memorize. The average rate of change of $f$ between $x = a$ and a nearby point $x$ is the difference quotient:

\frac{f(x) - f(a)}{x - a}

This is exactly the slope of the secant line through those two points (Topic 2.1). As the second point slides toward $x = a$ , the secant pivots into the tangent, and that limit of secant slopes is the definition of $f'(a)$ . So the tangent slope and the derivative are the same number by construction. That same tangent line doubles as the local linear approximation of $f$ near $x = a$ (Topic 4.6), so the equation you build here gets reused when you estimate function values.

y - f(a) = f'(a)(x - a) \quad\Longrightarrow\quad y = f'(a)(x - a) + f(a)

The normal line shares the point $(a, f(a))$ but runs perpendicular to the tangent. Perpendicular lines have slopes that are negative reciprocals, so if the tangent slope is $f'(a)$ , the normal slope is $-\frac{1}{f'(a)}$ . Everything else is identical: same point, point-slope form, solve for $y$ . The one case to watch is a horizontal tangent, where $f'(a) = 0$ : the normal is then the vertical line $x = a$ , because $-\frac{1}{0}$ is not a number.

y - f(a) = -\frac{1}{f'(a)}(x - a)

Evaluate the slope to a number first

The most common lost point on this problem type is writing the slope as $f'(x)$ with $x$ still in it. The slope of one specific tangent line is a single number, so differentiate to get $f'(x)$ , then substitute $x = a$ before you write the equation. On calculator-active AP questions you may evaluate $f'(a)$ numerically, but you still report the line, not the derivative.

Behavior at $x = a$	Tangent slope $f'(a)$	What the two lines look like
$f'(a)$ is a nonzero number	that number	slanted tangent; normal slope $-\frac{1}{f'(a)}$
$f'(a) = 0$ : horizontal tangent	$0$	tangent is $y = f(a)$ ; normal is vertical $x = a$
$f'(a)$ does not exist: corner or cusp	undefined	no tangent line by the standard method (Topic 2.4)
$f'(a)$ does not exist: vertical tangent	undefined	the tangent line exists but is vertical, $x = a$ , with no slope, so the slope-based point-slope method cannot build it (Topic 2.4)

Worked example

Tangent and normal line to a parabola

Find the tangent line and the normal line to $f(x) = x^2 + 1$ at $x = 2$ .

Find the derivative function with the power rule (Topic 2.5). For $f(x) = x^2 + 1$ , the derivative is $f'(x) = 2x$ . This is a function of $x$ , not yet a slope.
Evaluate the derivative at the point to turn it into a slope. At $x = 2$ , $f'(2) = 2 \cdot 2 = 4$ . This $4$ is the slope of the tangent line, because the derivative at a point is the slope of the tangent there (CHA-2.C.1). Do not leave $x$ in the slope.
Find the point of tangency on the curve by evaluating $f$ , not $f'$ : $f(2) = 2^2 + 1 = 5$ . So both lines pass through $(2, 5)$ .
Write the tangent line in point-slope form with slope $4$ and point $(2, 5)$ : $y - 5 = 4(x - 2)$ . Distribute and solve for $y$ : $y = 4x - 8 + 5 = 4x - 3$ .
Get the normal slope as the negative reciprocal of the tangent slope: $-\frac{1}{4}$ . The normal line is perpendicular to the tangent at $(2, 5)$ .
Write the normal line in point-slope form with slope $-\frac{1}{4}$ and the same point $(2, 5)$ : $y - 5 = -\frac{1}{4}(x - 2)$ . Solve for $y$ : $y = -\frac{1}{4}x + \frac{1}{2} + 5 = -\frac{1}{4}x + \frac{11}{2}$ .

Tangent line: $y = 4x - 3$ . Normal line: $y = -\frac{1}{4}x + \frac{11}{2}$ . Both pass through the point of tangency $(2, 5)$ .

Frequently asked questions

Why is the slope of the tangent line the derivative?

Because the derivative is defined as the limit of average rates of change. The slope between $x = a$ and a nearby point is the secant slope (the difference quotient from Topic 2.1); as the nearby point approaches $a$ , the secant lines approach the tangent line, and their slopes approach $f'(a)$ . The tangent slope is that limit, so it equals $f'(a)$ (Topic 2.2, CHA-2.C.1).

How do I find the normal line once I have the tangent?

Keep the same point $(a, f(a))$ and use the negative reciprocal of the tangent slope. If the tangent slope is $f'(a)$ , the normal slope is $-\frac{1}{f'(a)}$ , and the normal line is $y - f(a) = -\frac{1}{f'(a)}(x - a)$ . Perpendicular slopes always multiply to $-1$ , which is what negative reciprocal means.

What if the derivative is zero at the point?

Then the tangent is horizontal: $f'(a) = 0$ makes the tangent line $y = f(a)$ , a flat line through the point. The normal is perpendicular to horizontal, so it is the vertical line $x = a$ . You cannot write it as $y = mx + b$ because $-\frac{1}{0}$ has no value, so state the normal as $x = a$ .

Do I plug in the point before or after differentiating?

After. Differentiate first to get the whole function $f'(x)$ , then substitute $x = a$ to get the slope as one number. Substituting into $f(x)$ too early loses the variable you need to differentiate, and it is another common mistake that produces a wrong slope on this problem type.