Single-Variable-Calculus · Unit 1 · Video 1 · Interactive Practice

The Derivative as the Slope of the Tangent Line

IKey Formulas

Formula	Name	What it gives
$y - y_0 = m(x - x_0)$	Point-slope form	A line through $P$ — one for every $m$
$y_0 = f(x_0)$	Piece 1 — the point	Plain evaluation, no calculus
$m = f'(x_0)$	Piece 2 — the derivative	The slope of the tangent line at $P$
$y = f(x_0) + f'(x_0)(x - x_0)$	Tangent line at $P$	The two pieces assembled

Key Insight: $f'(x_0)$ is defined as a geometric quantity — the slope of one particular line — not as a formula, a rule, or a procedure. Naming that number is not the same as computing it.

IIVisualization 1 — Which Line Through $P$ Is the Tangent?

Every slope $m$ produces a line through $P$ ; exactly one of them is the tangent.

IIIVisualization 2 — The Two Pieces of Information

The tangent line at $P$ needs exactly two numbers — and they come from different worlds.

Location x₀ 2.70

Piece 1 — the point

Piece 2 — the slope

Point-slope form assembled

IVVisualization 3 — What "Matches the Curve" Means

Magnify around $P$ : only the tangent line stays glued to the curve.

Magnification 1.5× Rival line through P

💡 The definition names the number $m = f'(x_0)$ but supplies no procedure for producing it — a definition is not an algorithm, and closing that gap is the work of the rest of Unit 1.

VQuiz Questions

Problem 1 · Assemble the Tangent Line

Given: $f(4) = 3$ and $f'(4) = -2$ — find the tangent line to $y = f(x)$ at $x_0 = 4$ .

✅ Correct! Point-slope

y - 3 = -2(x - 4)

expands to

y = -2x + 11

, and

-2(4) + 11 = 3

puts the line through

P = (4, 3)

❌ The two pieces are swapped.

f(4) = 3

is the height of

P

and

f'(4) = -2

is the slope; here they trade places.

❌ Close — that treats $y_0$ as the $y$ -intercept.

3

is the height at

x_0 = 4

, not at

x = 0

-2(4) + 3 = -5 \ne 3

❌ A sign slipped while distributing.

y - 3 = -2(x - 4)

gives

y = -2x + 8 + 3

; dropping the

+3

and the sign on

8

leaves

-2x - 5

, which sends the line through

(4, -13)

❌ Not quite. Substitute

x = 4

into your line — the result must be

f(4) = 3

Show solution

The tangent line needs the point and the slope, and both are handed to you:

y_0 = f(4) = 3, \qquad m = f'(4) = -2

Substitute into point-slope form:

y - 3 = -2(x - 4)

y - 3 = -2x + 8

y = -2x + 11

Verify: at $x = 4$ , $\ -2(4) + 11 = 3 = f(4)$ ✓ — the line passes through $P = (4, 3)$ with slope $-2$ .

Problem 2 · Reading the Pieces Back Off the Line

Given: the tangent line to $y = f(x)$ at $x_0 = 5$ is $y = -x + 9$ — find $f(5)$ and $f'(5)$ .

What is $f(5)$ ?

What is $f'(5)$ ?

✅ Correct! The tangent touches the curve at

P

, so its height there is

f(5) = 4

; its slope is the derivative,

f'(5) = -1

❌ Check the height. The tangent meets the curve at

P = (5, f(5))

, so evaluate the line at

x = 5

— not at

x = 0

❌ That is the slope, not the height.

-1

f'(5)

, the coefficient of

x

in the line;

f(5)

is how high the line sits above

x = 5

❌ That is $x_0$ , not $f(5)$ .

5

is the input you feed the line;

f(5)

is the output it returns.

❌ Check the slope. Written as

y = mx + b

, the coefficient of

x

is the slope, and that slope is the derivative at

x_0

Show solution

The point. A tangent line touches the graph at $P$ , so the line and the curve share that one height:

f(5) = -(5) + 9 = 4

The slope. The line is already in slope-intercept form $y = mx + b$ with $m = -1$ , and the derivative is defined as the slope of that tangent line:

f'(5) = -1

So $P = (5, 4)$ and the curve is falling there. The number $9$ is only the $y$ -intercept of the tangent line — the height at $x = 0$ , which says nothing about $f$ at $x_0 = 5$ .

Problem 3 · Following the Tangent Away from the Point

Given: $f(3) = 4$ and $f'(3) = -2$ — find the height of the tangent line at $x = 5$ .

✅ Correct! From

P = (3, 4)

you move

2

units right, and the slope

-2

drops the line

4

units:

4 - 4 = 0

❌ Close — that is only one unit of run. From

x_0 = 3

x = 5

the run is

x - x_0 = 2

, not

1

❌ That drops $x_0$ from the formula. The slope multiplies

x - x_0 = 5 - 3

, not

x = 5

;

y_0

is the height at

x_0

, not the intercept.

❌ Check the sign. A slope of

-2

makes the line fall as

x

increases, so the height at

x = 5

must be below

4

Show solution

Step 1 — assemble the line. With $y_0 = f(3) = 4$ and $m = f'(3) = -2$ :

y = 4 - 2(x - 3)

Step 2 — evaluate it at $x = 5$ .

y = 4 - 2(5 - 3) = 4 - 4 = 0

Expanding first gives the same thing: $y = -2x + 10$ , and $-2(5) + 10 = 0$ ✓.

Careful: $0$ is the height of the tangent line at $x = 5$ , not $f(5)$ . The tangent is guaranteed to agree with the curve only at $x_0 = 3$ — away from $P$ the curve is free to drift off the line.

Problem 4 · The Derivative at a Crest

Given: the curve $y = f(x)$ above rises to a crest at $x_0 = 2.20$ , where its height is $f(2.20) = 3.38$ — find $f'(2.20)$ .

✅ Correct! At the crest the tangent line is horizontal, and a horizontal line has slope

0

— so the derivative there is

0

❌ That is $f(2.20)$ , not $f'(2.20)$ . The height of the point and the slope of the tangent at that point are two different numbers.

❌ That is $x_0$ , the input.

2.20

is where the derivative is being taken;

f'(2.20)

is the slope the tangent has there.

❌ Not quite. Nothing breaks at a smooth crest — the tangent line exists there, and you can read its slope straight off the picture.

❌ Not quite. The derivative is the slope of the tangent line at that point — so ask what the tangent does at the top of a smooth hump.

Show solution

Use the definition directly: $f'(2.20)$ is the slope of the tangent line to $y = f(x)$ at $P = (2.20,\ 3.38)$ .

At the top of a smooth crest the curve stops rising and has not yet begun to fall, so the tangent line there is horizontal:

m = f'(2.20) = 0

The tangent line itself is $y = 3.38 + 0 \cdot (x - 2.20)$ , that is $y = 3.38$ .

The recurring trap: $3.38$ answers "how high?" (Piece 1, plain evaluation) while $0$ answers "how steep?" (Piece 2, the derivative). They are different numbers about the same point.

Solved: 0 / 4

The Derivative as the Slope of the Tangent Line

IKey Formulas

IIVisualization 1 — Which Line Through PPP Is the Tangent?

IIIVisualization 2 — The Two Pieces of Information

IVVisualization 3 — What "Matches the Curve" Means

VQuiz Questions

IIVisualization 1 — Which Line Through $P$ Is the Tangent?