Finding Tangent Lines Using the Derivative

MATH161

Reference: Stewart 2.7 • Chapter: 2 • Section: 7

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Finding Tangent Lines Using the Derivative

The Line That "Just Touches"

A tangent line to a curve at a point has a special property: it touches the curve at that point and has the same "steepness" as the curve. If you zoom in far enough on the point, the curve and the tangent line become indistinguishable.

But how do we actually find the equation of this tangent line? We need two things:

A point on the line (the point of tangency)
The slope at that point (this is the derivative!)

Once we have these, the point-slope form gives us the tangent line equation.

Prerequisite Map

Quick Reference

Property	Value
Section	Stewart 2.1
Course	MATH161
Difficulty	Intermediate
Time	~15 minutes

Key Concepts

The Tangent Line Formula

The tangent line to $y = f(x)$ at the point $(a, f(a))$ has equation:

$$\boxed{y - f(a) = f'(a)(x - a)}$$

or equivalently:

$$y = f(a) + f'(a)(x - a)$$

The ingredients:

$(a, f(a))$ = the point of tangency
$f'(a)$ = the slope (computed from the derivative definition)

Visual Understanding

        Tangent line: y - f(a) = f'(a)(x - a)
              ↗
             /
            /  slope = f'(a)
        ───●───────────── curve y = f(x)
          /
         / Point of tangency: (a, f(a))
        /

As you zoom in on the point $(a, f(a))$, the curve looks more and more like the tangent line.

The Complete Process

Step 1: Identify the point of tangency $(a, f(a))$

Step 2: Compute $f'(a)$ using the limit definition

Step 3: Write the tangent line equation: $y - f(a) = f'(a)(x - a)$

Step 4: Simplify to slope-intercept form if requested

Worked Example: Tangent to $y = x^2 + 2x$ at $x = 1$

Step 1: Find the point of tangency.

At $x = 1$: $f(1) = 1 + 2 = 3$

Point: $(1, 3)$

Step 2: Compute $f'(1)$.

$$f'(1) = \lim_{h \to 0} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0} \frac{[(1+h)^2 + 2(1+h)] - 3}{h}$$

Expand: $$= \lim_{h \to 0} \frac{1 + 2h + h^2 + 2 + 2h - 3}{h} = \lim_{h \to 0} \frac{4h + h^2}{h} = \lim_{h \to 0}(4 + h) = 4$$

Step 3: Write the tangent line.

$$y - 3 = 4(x - 1)$$

Step 4: Simplify.

$$y = 4x - 4 + 3 = 4x - 1$$

The tangent line is $y = 4x - 1$.

Practice Problems

Level 1 Writing the Tangent Line Equation

If $f(3) = 7$ and $f'(3) = -2$, write an equation for the tangent line to $y = f(x)$ at $x = 3$.

Thought Process

You already have both pieces of information needed:

Point: $(3, f(3)) = (3, 7)$
Slope: $f'(3) = -2$

Just plug into point-slope form: $y - y_1 = m(x - x_1)$

Show Answer

Using point-slope form with point $(3, 7)$ and slope $-2$:

$$y - 7 = -2(x - 3)$$

Simplifying: $$y = -2x + 6 + 7 = \boxed{-2x + 13}$$

Level 2 Finding a Tangent Line (Polynomial)

Find an equation of the tangent line to the curve $y = x^2 - 4x + 3$ at the point $(2, -1)$.

Thought Process

The point $(2, -1)$ is given (verify: $4 - 8 + 3 = -1$ ✓)
Compute $f'(2)$ using the limit definition
Write the tangent line using point-slope form

Show Answer

Step 1: Point is $(2, -1)$. ✓

Step 2: Find $f'(2)$.

$$f'(2) = \lim_{h \to 0} \frac{[(2+h)^2 - 4(2+h) + 3] - (-1)}{h}$$

$$= \lim_{h \to 0} \frac{4 + 4h + h^2 - 8 - 4h + 3 + 1}{h} = \lim_{h \to 0} \frac{h^2}{h} = \lim_{h \to 0} h = 0$$

Step 3: Tangent line: $$y - (-1) = 0(x - 2)$$ $$y = \boxed{-1}$$

The tangent line is horizontal! This makes sense because $(2, -1)$ is the vertex of the parabola.

Level 3 Finding a Tangent Line (Rational Function)

Find an equation of the tangent line to the curve $y = \dfrac{2}{x+1}$ at the point where $x = 1$.

Thought Process

First find the y-coordinate: $f(1) = \frac{2}{2} = 1$, so the point is $(1, 1)$
Compute $f'(1)$ using the definition (you'll need to combine fractions)
Write the tangent line equation

Show Answer

Step 1: Point of tangency: $(1, f(1)) = (1, 1)$

Step 2: Find $f'(1)$.

$$f'(1) = \lim_{h \to 0} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0} \frac{\frac{2}{(1+h)+1} - 1}{h}$$

$$= \lim_{h \to 0} \frac{\frac{2}{2+h} - 1}{h} = \lim_{h \to 0} \frac{\frac{2 - (2+h)}{2+h}}{h}$$

$$= \lim_{h \to 0} \frac{-h}{h(2+h)} = \lim_{h \to 0} \frac{-1}{2+h} = -\frac{1}{2}$$

Step 3: Tangent line: $$y - 1 = -\frac{1}{2}(x - 1)$$

$$y = -\frac{1}{2}x + \frac{1}{2} + 1 = \boxed{-\frac{1}{2}x + \frac{3}{2}}$$

Or equivalently: $x + 2y = 3$

Level 4 Finding a Point Given the Tangent Slope

The tangent line to $y = f(x)$ at the point $(4, 3)$ passes through the point $(0, 1)$. Find $f(4)$ and $f'(4)$.

Thought Process

This problem works backwards from the usual process!

We know the tangent line passes through $(4, 3)$ and $(0, 1)$
Since $(4, 3)$ is on the curve, $f(4) = 3$
The slope of the tangent line through these two points is $f'(4)$

Use the slope formula with the two given points.

Show Answer

Since $(4, 3)$ is the point of tangency (on the curve): $$f(4) = \boxed{3}$$

The tangent line passes through $(4, 3)$ and $(0, 1)$, so its slope is: $$f'(4) = \frac{3 - 1}{4 - 0} = \frac{2}{4} = \boxed{\frac{1}{2}}$$

Level 5 Finding Where the Tangent is Horizontal

For the function $f(x) = x^3 - 6x^2 + 9x + 2$:

(a) Use the limit definition to find a formula for $f'(a)$.

(b) Find all points on the curve where the tangent line is horizontal.

Thought Process

(a) Use the h-form definition to find $f'(a)$ as a function of $a$.

(b) A horizontal tangent has slope 0, so solve $f'(a) = 0$.

Show Answer

(a) Finding $f'(a)$:

$$f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$$

Expand $f(a+h) = (a+h)^3 - 6(a+h)^2 + 9(a+h) + 2$:

$= a^3 + 3a^2h + 3ah^2 + h^3 - 6(a^2 + 2ah + h^2) + 9a + 9h + 2$

$= a^3 + 3a^2h + 3ah^2 + h^3 - 6a^2 - 12ah - 6h^2 + 9a + 9h + 2$

And $f(a) = a^3 - 6a^2 + 9a + 2$.

Difference: $$f(a+h) - f(a) = 3a^2h + 3ah^2 + h^3 - 12ah - 6h^2 + 9h$$

$$= h(3a^2 + 3ah + h^2 - 12a - 6h + 9)$$

Divide by $h$: $$\frac{f(a+h) - f(a)}{h} = 3a^2 + 3ah + h^2 - 12a - 6h + 9$$

Take limit as $h \to 0$: $$f'(a) = 3a^2 - 12a + 9 = 3(a^2 - 4a + 3) = \boxed{3(a-1)(a-3)}$$

(b) Solve $f'(a) = 0$: $$3(a-1)(a-3) = 0$$ $$a = 1 \text{ or } a = 3$$

Points of horizontal tangency:

At $a = 1$: $f(1) = 1 - 6 + 9 + 2 = 6$, so $(1, 6)$
At $a = 3$: $f(3) = 27 - 54 + 27 + 2 = 2$, so $(3, 2)$

(c) Horizontal tangent lines: $$\boxed{y = 6} \text{ (at } x = 1\text{)}$$ $$\boxed{y = 2} \text{ (at } x = 3\text{)}$$

Conceptual Questions (CCI-Style)

Level 2 Understanding Tangent Line Properties

Which statement is TRUE about the tangent line to a curve at a point?

(A) The tangent line crosses the curve exactly once

(B) The tangent line never crosses the curve

(D) The tangent line is always horizontal

Thought Process

Think about what "tangent" means geometrically. Consider counterexamples:

Can a tangent line cross the curve elsewhere? (Yes - think of $y = x^3$ at the origin)
Is a tangent line always horizontal? (No - only when $f'(a) = 0$)

The defining property is about matching the curve's slope at the point of tangency.

Show Answer

(C) is correct.

(A) is false: The tangent line to $y = x^3$ at $(0,0)$ is $y = 0$, which crosses the curve at the origin.

(B) is false: The tangent line to $y = \sin x$ at $(0,0)$ is $y = x$, which crosses the curve infinitely many times.

(D) is false: Only tangent lines at local max/min have slope 0.

The key property of a tangent line is that it matches the curve's instantaneous slope at the point of tangency.

Mastery Checklist

[ ] I can find the point of tangency $(a, f(a))$ given the $x$-coordinate
[ ] I can compute $f'(a)$ using the limit definition to find the tangent slope
[ ] I can write the tangent line equation using point-slope form
[ ] I can simplify to slope-intercept form
[ ] I can work backwards from tangent line information to find $f(a)$ and $f'(a)$
[ ] I can find where tangent lines are horizontal by solving $f'(a) = 0$

Mental Model

The Best Approximation: The tangent line is the "best straight-line approximation" to a curve near a point. If you can only use a straight line to estimate the curve, the tangent line gives the smallest error near the point of tangency.

This is why tangent lines matter for applications: near $x = a$, we have $f(x) \approx f(a) + f'(a)(x - a)$, which is just the tangent line!

Connections

Looking back:

The slope $f'(a)$ comes directly from the derivative definition
Point-slope form from algebra is the tool for writing the line equation

Looking ahead:

Linear approximation (Section 2.9) formalizes the idea of using tangent lines to estimate function values
Finding horizontal tangents leads to critical points for optimization (Chapter 3)

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Last updated: 2026-01-22