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 a specific point $x=a$. Another key interpretation is that the derivative gives the slope of the tangent line to the graph at that point.

To see where the tangent line comes from, it helps to start with a closely related idea: the secant line.

Secant line:

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

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

As the two points on the curve move closer and closer together (so $x\to a$), the secant line approaches a limiting position. In the limit, the two points merge into the single point at $x=a$, and the secant line becomes the tangent line at that point.

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:

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

This is the same limit definition used for the instantaneous rate of change at $x=a$. That means the derivative at $x=a$ - the value $f^{\prime }(a)$ - 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 is useful in many settings, including:

• Optimization: A flat tangent line (slope $0$) can indicate where a maximum or minimum occurs, like finding the best number of products to make to maximize profit.
• Graphing: The derivative tells you whether a function is going up or down (whether it’s increasing or decreasing as you move left to right) and gives information about how the curve behaves.
• Approximations: The tangent line can estimate values like $1.01^{50}$ more accurately than a rough guess like “about 1.”
• Physics and engineering: Derivatives help analyze motion (for example, how fast a car is accelerating) and other rates of change in systems.

Sidenote

On tangent lines

When a tangent line touches $f(x)$ at a single point, it doesn’t mean the line never meets $f(x)$ again - only 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 the line meets the curve $f(x)$ (called the point of tangency)
• $m$ is the line’s slope

An alternate form, in standard calculus notation, is often called the Taylor form. It uses the same point-slope structure, but replaces the point and slope with calculus notation:

• 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 $f^{\prime }(a)$, which replaces $m$.

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

Equation of the tangent line:

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

You can use whichever form you prefer, as long as you know what each part represents.

Example

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

Solution

(spoiler)

Answer: $y = -6x - 9$1. Find the derivative:
As shown in the previous section,

$$f^{\prime }(x)=2x$$

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

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

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,

$$\boxed{y=-6x-9}$$

It can help to plot both $y=x^2$ and $y=-6x-9$ in Desmos to see what a tangent line looks like on the graph.

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 $f^{\prime }(a)=3$, then the slope of the normal line at $a$ would be

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

Example

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

Solution

(spoiler)

Answer: $y - 4 = -\dfrac{1}{4}(x - 2)$1. Find the derivative:
Just like in the previous example,

$$f^{\prime }(x)=2x$$

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

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

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:

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

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 $f^{\prime }(2)$.

Solution

(spoiler)

Answer: $\boxed{f(2)=0,f^{\prime }(2)=4}$

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

$$\boxed{f(2)=0}$$

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

$$\begin{array}{rl}\text{Slope}& =\frac{y_2-y_1}{x_2-x_1}\\ & =\frac{(4-0)}{(3-2)}\\ & =4\end{array}$$

So

$$\boxed{f^{\prime }(2)=4}$$

2. A ball is tossed into the air and follows the parabolic curve $h(t)=-t^2+4t+5$, 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 phrase “rate of change” refers to the derivative, which is also the slope of the tangent line. So we’re looking for the time when the tangent line has slope $0$ (a horizontal tangent line).

As you move along the downward-opening parabola $h(t)$ from left to right, the slope of the tangent line starts positive, decreases, and becomes $0$ at the vertex.

The $x$-coordinate of the vertex of a parabola $a{t}^2+bt+c$ is given by $t=\frac{-b}{2a}$.

Here, $a=-1$ and $b=4$, so

$$t=\frac{-4}{2(-1)}=2.$$

At $t=2$, the ball is at its maximum height, so its instantaneous rate of change in height is $0$.