Tangent lines & slopes | Derivative basics

What you’ll learn:

Secant vs. tangent lines
The slope of the tangent line at a point represents the derivative
How to write the equation of a tangent line
Write the equation of a normal line

The previous section introduced the derivative as an instantaneous rate of change - how fast a function $f (x)$ changes at any point $x = a$ . Another key interpretation of the derivative is as the slope of the tangent line at that point.

Before that, let’s introduce another important line: the secant line.

Secant line:

A straight line that passes through two points on a curve, like taking a “shortcut” from one to the other.

Use the formula for average rate of change when calculating the slope of the secant line.

As the two points are brought closer and closer together (bringing $x \to a$ ), eventually $x$ converges to the single point at $a$ , and the line connecting the two points is indistinguishable from a tangent line that effectively touches the curve at just point $a$ .

Tangent line:

A straight line that touches the curve at a single point but does not pass through it.

The slope of this tangent line is given by:

$x \to a lim \frac{f ( x ) - f ( a )}{x - a}$

Recall from the previous section that this is the same limit definition used for the instantaneous rate of change at $x = a$ , which tells us that the derivative at $x = a$ , or the value $f^{'} (a)$ , is exactly the slope of the tangent line at $x = a$ .

AP tip:

Whenever you see the keyword “tangent line” in a problem, immediately think “derivative.” Tangent lines always involve the derivative of a function.

Understanding the derivative as the slope of the tangent line has uses that will be covered in later sections, such as:

Optimization: A flat tangent line (slope of 0) can show where the maximum or minimum occurs, like finding the best number of products to make to maximize profit.
Graphing: The derivative tells you if a function is going up or down (if it’s increasing or decreasing as you move from left to right) and how a curve is shaped.
Approximations: The tangent line allows you to estimate the value of something like $1.0 1^{50}$ more accurately than just “about 1”.
Physics and engineering: Derivatives help analyze motion, like how fast a car is accelerating, and rates of change in systems.

Sidenote

On tangent lines

When tangent line touches $f (x)$ at a single point, it doesn’t mean the line never meets $f (x)$ again - just that it doesn’t pass through the curve at that point.

Equation of the tangent line

The equation of a line is typically written in point-slope form:

$y - y_{1} = m (x - x_{1})$

For the equation of the tangent line:

$(x_{1}, y_{1})$ is the point where it meets the curve $f (x)$ (called the point of tangency)
$m$ is the line’s slope.

The alternate form of the equation, in proper calculus notation, is called the Taylor form and replaces a few parts of the basic point-slope form:

The tangent line meets $f (x)$ at $(a, f (a))$ which replaces $(x_{1}, y_{1})$ .
The slope of the tangent line is the derivative at $x = a$ , or $f^{'} (a)$ , which replaces $m$ .

So the equation of the tangent line can also be written as:

Equation of the tangent line:

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

You may use whichever form you find easiest to remember as long as you understand what each piece means.

Equation of the tangent line to f at x = a

Example

1. Find the equation of the tangent line to the curve $f (x) = x^{2}$ at $x = - 3$ .

Solution

(spoiler)

Answer: $y = - 6 x - 9$

1. Find the derivative:
As shown in the previous section,

$f^{'} (x) = 2 x$

2. Slope of the tangent line at $x = - 3$ :

$f^{'} (- 3) = 2 (- 3) = - 6$

3. Point $(a, f (a))$ :

At $x = - 3$ ,

$f (- 3) = (- 3)^{2} = 9$

So the tangent line meets the curve at $(- 3, 9)$ .

Then the equation of the tangent line is:

$y - 9 = - 6 (x + 3)$

or in slope-intercept form,

$y = - 6 x - 9$

It’s helpful practice to plot the two functions $y = x^{2}$ and $y = - 6 x - 9$ into Desmos for a visual representation of what a tangent line to a curve looks like!

Normal lines

Another important type of line is the normal line, which is perpendicular to the tangent line at the point of tangency.

Perpendicular slopes are negative reciprocals of each other. For example, if the slope of the tangent line is $f^{'} (a) = 3$ , then the slope of the normal line at $a$ would be

$- \frac{1}{f ^{'} ( a )} = - \frac{1}{3}$

Example

1. Find the equation of the normal line to $f (x) = x^{2}$ at $x = 2$ .

Solution

(spoiler)

Answer: $y - 4 = - \frac{1}{4} (x - 2)$

1. Find the derivative:
Just like in the previous example,

$f^{'} (x) = 2 x$

2. Slope of the tangent line at $x = 2$ :

$f^{'} (2) = 2 (2) = 4$

Then the slope of the normal line is the negative reciprocal, or $- \frac{1}{4}$ .

4. Point $(a, f (a))$ :

At $x = 2$ ,

$f (2) = 2^{2} = 4$

Then the equation of the normal line is:

$y - 4 = - \frac{1}{4} (x - 2)$

Confirm visually by plotting the original function $f (x) = x^{2}$ , the equation of the tangent line, and the equation of the normal line!

Challenge problems

1. The tangent line to the graph of $y = f (x)$ at $(2, 0)$ passes through the point $(3, 4)$ . Find $f (2)$ and $f^{'} (2)$ .

Solution

(spoiler)

Answer: $f (2) = 0, f^{'} (2) = 4$

The tangent line meets $f (x)$ at $(2, 0)$ which means

$f (2) = 0$

Next, $f^{'} (2)$ represents the slope of the tangent line at $x = 2$ . Given that the tangent line passes through two points $(2, 0)$ and $(3, 4)$ , calculate its slope using the basic slope formula:

$Slope = \frac{y _{2} - y _{1}}{x _{2} - x _{1}} = \frac{( 4 - 0 )}{( 3 - 2 )} = 4$

$f^{'} (2) = 4$

2. A ball is tossed into the air and follows the parabolic curve $h (t) = - t^{2} + 4 t + 5$ , which represents the height (in meters) as a function of time (in seconds). At what time is the rate of change of the ball’s height equal to $0$ ?

Solution:

(spoiler)

Answer: $t = 2$ seconds (at the vertex)

The key words “rate of change” refers to the derivative, which is also the slope of the tangent line. We’re looking for the point on the graph where the tangent line has a slope of 0, meaning it’s a horizontal line.

As you move along the downward-facing parabola $h (t)$ from left to right, visualize the tangent lines: notice the slope of the tangent line starts off very positive (steep), becomes smaller, and eventually becomes $0$ at the vertex.

The $x$ -coordinate (or $t$ in this case) of the vertex of a parabola can be found using the formula $t = \frac{- b}{2 a}$ , where $a$ and $b$ are the coefficients in the quadratic equation.

Here, $b = 4$ and $a = - 1$ and the vertex is at $t = 2$ . At this point, the ball is neither rising nor falling and the rate of change is $0$ .

Key points

A secant line connects 2 points and represents the average rate of change.
A tangent line touches effectively touches just 1 point (and does not cross it) and represents the instantaneous rate of change.
The tangent line equation is:

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

A normal line is perpendicular to the tangent line, with a slope that is the negative reciprocal of the tangent line’s slope, or $\frac{1}{f ^{'} ( a )}$ .

What you’ll learn:

Secant vs. tangent lines
The slope of the tangent line at a point represents the derivative
How to write the equation of a tangent line
Write the equation of a normal line

Before that, let’s introduce another important line: the secant line.

Secant line:

A straight line that passes through two points on a curve, like taking a “shortcut” from one to the other.

Use the formula for average rate of change when calculating the slope of the secant line.

Tangent line:

A straight line that touches the curve at a single point but does not pass through it.

The slope of this tangent line is given by:

$x \to a lim \frac{f ( x ) - f ( a )}{x - a}$

AP tip:

Whenever you see the keyword “tangent line” in a problem, immediately think “derivative.” Tangent lines always involve the derivative of a function.

Understanding the derivative as the slope of the tangent line has uses that will be covered in later sections, such as:

Optimization: A flat tangent line (slope of 0) can show where the maximum or minimum occurs, like finding the best number of products to make to maximize profit.
Graphing: The derivative tells you if a function is going up or down (if it’s increasing or decreasing as you move from left to right) and how a curve is shaped.
Approximations: The tangent line allows you to estimate the value of something like $1.0 1^{50}$ more accurately than just “about 1”.
Physics and engineering: Derivatives help analyze motion, like how fast a car is accelerating, and rates of change in systems.

Sidenote

On tangent lines

When tangent line touches $f (x)$ at a single point, it doesn’t mean the line never meets $f (x)$ again - just that it doesn’t pass through the curve at that point.

Equation of the tangent line

The equation of a line is typically written in point-slope form:

$y - y_{1} = m (x - x_{1})$

For the equation of the tangent line:

$(x_{1}, y_{1})$ is the point where it meets the curve $f (x)$ (called the point of tangency)
$m$ is the line’s slope.

The alternate form of the equation, in proper calculus notation, is called the Taylor form and replaces a few parts of the basic point-slope form:

The tangent line meets $f (x)$ at $(a, f (a))$ which replaces $(x_{1}, y_{1})$ .
The slope of the tangent line is the derivative at $x = a$ , or $f^{'} (a)$ , which replaces $m$ .

So the equation of the tangent line can also be written as:

Equation of the tangent line:

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

You may use whichever form you find easiest to remember as long as you understand what each piece means.

Example

1. Find the equation of the tangent line to the curve $f (x) = x^{2}$ at $x = - 3$ .

Solution

(spoiler)

Answer: $y = - 6 x - 9$

1. Find the derivative:
As shown in the previous section,

$f^{'} (x) = 2 x$

2. Slope of the tangent line at $x = - 3$ :

$f^{'} (- 3) = 2 (- 3) = - 6$

3. Point $(a, f (a))$ :

At $x = - 3$ ,

$f (- 3) = (- 3)^{2} = 9$

So the tangent line meets the curve at $(- 3, 9)$ .

Then the equation of the tangent line is:

$y - 9 = - 6 (x + 3)$

or in slope-intercept form,

$y = - 6 x - 9$

It’s helpful practice to plot the two functions $y = x^{2}$ and $y = - 6 x - 9$ into Desmos for a visual representation of what a tangent line to a curve looks like!

Normal lines

Another important type of line is the normal line, which is perpendicular to the tangent line at the point of tangency.

Perpendicular slopes are negative reciprocals of each other. For example, if the slope of the tangent line is $f^{'} (a) = 3$ , then the slope of the normal line at $a$ would be

$- \frac{1}{f ^{'} ( a )} = - \frac{1}{3}$

Example

1. Find the equation of the normal line to $f (x) = x^{2}$ at $x = 2$ .

Solution

(spoiler)

Answer: $y - 4 = - \frac{1}{4} (x - 2)$

1. Find the derivative:
Just like in the previous example,

$f^{'} (x) = 2 x$

2. Slope of the tangent line at $x = 2$ :

$f^{'} (2) = 2 (2) = 4$

Then the slope of the normal line is the negative reciprocal, or $- \frac{1}{4}$ .

4. Point $(a, f (a))$ :

At $x = 2$ ,

$f (2) = 2^{2} = 4$

Then the equation of the normal line is:

$y - 4 = - \frac{1}{4} (x - 2)$

Confirm visually by plotting the original function $f (x) = x^{2}$ , the equation of the tangent line, and the equation of the normal line!

Challenge problems

1. The tangent line to the graph of $y = f (x)$ at $(2, 0)$ passes through the point $(3, 4)$ . Find $f (2)$ and $f^{'} (2)$ .

Solution

(spoiler)

Answer: $f (2) = 0, f^{'} (2) = 4$

The tangent line meets $f (x)$ at $(2, 0)$ which means

$f (2) = 0$

$Slope = \frac{y _{2} - y _{1}}{x _{2} - x _{1}} = \frac{( 4 - 0 )}{( 3 - 2 )} = 4$

$f^{'} (2) = 4$

2. A ball is tossed into the air and follows the parabolic curve $h (t) = - t^{2} + 4 t + 5$ , which represents the height (in meters) as a function of time (in seconds). At what time is the rate of change of the ball’s height equal to $0$ ?

Solution:

(spoiler)

Answer: $t = 2$ seconds (at the vertex)

The $x$ -coordinate (or $t$ in this case) of the vertex of a parabola can be found using the formula $t = \frac{- b}{2 a}$ , where $a$ and $b$ are the coefficients in the quadratic equation.

Here, $b = 4$ and $a = - 1$ and the vertex is at $t = 2$ . At this point, the ball is neither rising nor falling and the rate of change is $0$ .

Key points

A secant line connects 2 points and represents the average rate of change.
A tangent line touches effectively touches just 1 point (and does not cross it) and represents the instantaneous rate of change.
The tangent line equation is:

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

A normal line is perpendicular to the tangent line, with a slope that is the negative reciprocal of the tangent line’s slope, or $\frac{1}{f ^{'} ( a )}$ .