Tangent Lines
Textbook Reference
| Primary source | OpenStax Calculus Volume 1, Section 3.1: “Defining the Derivative” |
| Book URL | https://openstax.org/details/books/calculus-volume-1 |
Freely available and openly licensed.
Key idea
The tangent line at a point on a curve is the straight line that best approximates the curve near that point. Its slope is the derivative at that point, and its equation follows from point-slope form: you know the slope $f'(a)$ and the point $(a, f(a))$.
This is not just geometry. Every linearization, every linear approximation, every Newton’s-method step, and every “use the tangent to estimate” instruction in calculus uses this single idea: the tangent line is the best linear model for $f$ near $a$.
Prerequisite Check
Before this lesson, make sure you can do all of the following:
Quick Reference
Tangent line at $(a, f(a))$: \[ y = f(a) + f'(a)(x - a). \]
Normal line at $(a, f(a))$ (perpendicular to tangent, provided $f'(a) \neq 0$): \[ y = f(a) - \frac{1}{f'(a)}(x - a). \]
Steps:
- Find $f(a)$: the $y$-coordinate of the point of tangency.
- Find $f'(a)$: the slope of the tangent.
- Write point-slope form: $y - f(a) = f'(a)(x - a)$.
Key Concepts
1. The Equation of the Tangent Line
Example 1. Find the tangent line to $f(x) = x^2$ at $x = 3$.
See It: Slide the Secant Into the Tangent
Move the slider to bring the second point toward the first on the graph of $x^2$. The secant line rotates and its slope settles on one value. When the two points meet, the secant has become the tangent. Name the slope the secant slopes approached.
Step 1: $f(3) = 9$. The point is $(3, 9)$.
Step 2: $f'(x) = 2x$, so $f'(3) = 6$. (Using the power rule; or from the definition: $\lim_{h\to 0}\frac{(3+h)^2-9}{h} = \lim_{h\to 0}(6+h) = 6$.)
Step 3: $y - 9 = 6(x - 3)$, which simplifies to $y = 6x - 9$.
Example 2. Find the tangent line to $g(x) = \sqrt{x}$ at $x = 4$.
$g(4) = 2$. $g'(x) = \dfrac{1}{2\sqrt{x}}$, so $g'(4) = \dfrac{1}{4}$.
Tangent line: $y - 2 = \dfrac{1}{4}(x - 4)$, i.e., $y = \dfrac{x}{4} + 1$.
Example 3. Find the tangent line to $h(x) = x^3 - 3x$ at the point where $x = -1$.
$h(-1) = -1 + 3 = 2$. Point: $(-1, 2)$.
$h'(x) = 3x^2 - 3$, so $h'(-1) = 3 - 3 = 0$.
When the slope is zero, the tangent line is horizontal: $y = 2$.
2. The Normal Line
The normal line at a point is perpendicular to the tangent. Perpendicular lines have slopes whose product is $-1$, so if the tangent slope is $m \neq 0$, the normal slope is $-1/m$.
Example 4. Find the normal line to $f(x) = x^2$ at $x = 3$.
Tangent slope: $f'(3) = 6$. Normal slope: $-\dfrac{1}{6}$.
Normal line: $y - 9 = -\dfrac{1}{6}(x - 3)$, i.e., $y = -\dfrac{x}{6} + \dfrac{1}{2} + 9 = -\dfrac{x}{6} + \dfrac{19}{2}$.
3. When the Tangent Slope Has Geometric Meaning
The slope of the tangent encodes the local behavior of the function:
- Positive slope: $f$ is increasing at $a$. The curve rises from left to right near $a$.
- Negative slope: $f$ is decreasing at $a$. The curve falls from left to right near $a$.
- Zero slope: The tangent is horizontal. This is a candidate for a local maximum or minimum.
Example 5. For $p(x) = -x^2 + 4x$, find all points where the tangent is horizontal.
$p'(x) = -2x + 4$. Set $p'(x) = 0$: $-2x + 4 = 0$, $x = 2$.
At $x = 2$: $p(2) = -4 + 8 = 4$. The tangent is horizontal at $(2, 4)$, which is the vertex of the parabola.
4. Tangent Lines From the Definition
When differentiation rules are not yet available, use the limit definition to find $f'(a)$.
Example 6. Find the tangent line to $f(x) = \dfrac{1}{x}$ at $x = 2$.
From the definition (computed in the Derivative at a Point lesson): $f'(a) = -1/a^2$.
$f'(2) = -\dfrac{1}{4}$. $f(2) = \dfrac{1}{2}$.
Tangent line: $y - \dfrac{1}{2} = -\dfrac{1}{4}(x - 2)$, i.e., $y = -\dfrac{x}{4} + 1$.
the tangent slope and the function value are the same at a given point.
This is the height-vs-slope error. At $x = a$, the function value $f(a)$ is the HEIGHT of the graph; the derivative $f'(a)$ is the SLOPE of the tangent there. For $f(x) = x^2 + 1$ at $x = 3$: $f(3) = 10$ (height) and $f'(3) = 6$ (slope). These numbers are unrelated in general. The tangent line at $x = 3$ passes through the point $(3, 10)$ with slope $6$: the equation is $y - 10 = 6(x - 3)$, which requires BOTH numbers, not either one alone.
the tangent line equation is just $y = f'(a)$.
This is the input-output-confusion error. $f'(a)$ is a slope (a number), not a line. A line needs both a slope and a point. The tangent line at $(a, f(a))$ with slope $f'(a)$ is written in point-slope form as $y - f(a) = f'(a)(x - a)$. Writing $y = f'(a)$ describes a horizontal line at height $f'(a)$ -- correct only if the tangent happens to be horizontal AND at the right height, which is almost never the case.
Common Errors
| Error | Example | Correction |
|---|---|---|
| Confusing slope with the equation | Writing $y = f'(a)$ for the tangent | The tangent line is $y = f(a) + f'(a)(x-a)$; the slope is $f'(a)$, not the full equation |
| Using the wrong point | Computing $f'(a)$ but writing the tangent through the origin | The line must pass through $(a, f(a))$, not $(0, 0)$ |
| Wrong sign for perpendicular | Normal slope $= 1/m$ instead of $-1/m$ | Perpendicular slopes multiply to $-1$; if tangent slope is $m$, normal slope is $-1/m$ |
| Forgetting to evaluate $f(a)$ | Giving the slope but not the full tangent equation | Both $f(a)$ (the point) and $f'(a)$ (the slope) are needed |
Leveled Practice
Level 1 -- Writing the Tangent Equation
Problem 1. Find the tangent line to $f(x) = 2x^2 - 3$ at $x = 1$.
Show answer
$f(1) = -1$. $f'(x) = 4x$, so $f'(1) = 4$.
Tangent: $y - (-1) = 4(x - 1)$, i.e., $y = 4x - 5$.
Problem 2. Find the tangent line to $g(x) = x^3 + x$ at $x = 0$.
Show answer
$g(0) = 0$. $g'(x) = 3x^2 + 1$, so $g'(0) = 1$.
Tangent: $y = x$.
Problem 3. Find the normal line to $f(x) = x^2$ at $x = -2$.
Show answer
$f(-2) = 4$. $f'(-2) = 2(-2) = -4$. Normal slope $= 1/4$.
Normal line: $y - 4 = \dfrac{1}{4}(x + 2)$, i.e., $y = \dfrac{x}{4} + \dfrac{9}{2}$.
Level 2 -- Geometric Questions
Problem 4. Find all points on $y = x^3 - 3x$ where the tangent line is horizontal.
Show answer
$y' = 3x^2 - 3 = 0$: $x^2 = 1$, $x = \pm 1$.
At $x = 1$: $y = 1 - 3 = -2$. Point $(1, -2)$.
At $x = -1$: $y = -1 + 3 = 2$. Point $(-1, 2)$.
Problem 5. Find the equation of the tangent to $f(x) = \sqrt{x}$ at $x = 9$. Where does this tangent cross the $x$-axis?
Show answer
$f(9) = 3$. $f'(9) = \dfrac{1}{6}$.
Tangent: $y - 3 = \dfrac{1}{6}(x - 9)$, i.e., $y = \dfrac{x}{6} + \dfrac{3}{2}$.
Set $y = 0$: $\dfrac{x}{6} = -\dfrac{3}{2}$, so $x = -9$. The tangent crosses the $x$-axis at $(-9, 0)$.
Level 3 -- Construction and Reasoning
Problem 6. Find a point on $f(x) = x^2$ such that the tangent line at that point passes through $(0, -4)$.
Show answer
Let the point of tangency be $(a, a^2)$. The tangent line is $y = 2a(x - a) + a^2 = 2ax - a^2$.
For this line to pass through $(0, -4)$: substitute $x = 0$, $y = -4$:
$-4 = -a^2$, so $a^2 = 4$, giving $a = 2$ or $a = -2$.
At $a = 2$: tangent $y = 4x - 4$. Check: at $x=0$, $y = -4$. Correct.
At $a = -2$: tangent $y = -4x - 4$. Check: at $x = 0$, $y = -4$. Correct.
Two tangent lines from $(0, -4)$ to $y = x^2$: $y = 4x - 4$ (touching at $(2,4)$) and $y = -4x - 4$ (touching at $(-2,4)$).
Mastery Checklist
You have mastered this skill when you can do all of the following without referring to notes:
Mental Model
The tangent line is the curve’s shadow at a moment. Stand at the point $(a, f(a))$ and look along the curve. The tangent line is the direction you are heading: it captures the instantaneous direction, not where the curve bends next.
All of calculus’s approximation tools are built on this: the tangent line is the best linear approximation of $f$ near $a$. Linear algebra uses it too (Jacobian matrix). Physics uses it for linearizing equations. Economics uses it for marginal analysis. The equation $y = f(a) + f'(a)(x-a)$ is the foundation of all local approximation in mathematics.
Connections
Within Calculus I (MATH161)
- Linear approximation / differentials: The tangent line at $a$ is the linear approximation $L(x) = f(a) + f'(a)(x-a)$, used to estimate $f(x)$ for $x$ near $a$.
- Newton’s method (Ch. 3 or 4): Iteratively follow tangent lines to find roots of equations.
- Implicit differentiation: Finding tangent lines to curves defined implicitly uses the same formula; finding $dy/dx$ at a point gives the tangent slope.
Back to Calculus I Skills | Previous: Rates of Change | Next: The Derivative as a Function