Finding the tangent line to a curve at a point is one of the central problems that led to the invention of calculus. This section shows how to define the tangent line precisely using limits of secant lines, and how to compute its slope for any function.
| Concept | Quick Reference |
|---|---|
| Slope of secant | |
| Slope of tangent at | |
| Tangent line equation |
Why Tangent Lines Matter
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In many practical problems, it is essential to determine the maximum or minimum values of a given function. The connection between finding tangent lines to the function's graph and identifying its maxima and minima becomes apparent when we consider a typical curve. At each point along a curve, we can draw a tangent to it. At certain points where the curve reaches a maximum or minimum, the tangent line is horizontal. Consequently, the problem of locating these extreme values depends on the solution of another problem: the identification of horizontal tangents. Furthermore, when the slope of the tangent line is positive, the graph ascends to the right, meaning the function is increasing. When the slope of the tangent line is negative, the graph descends to the right, meaning the function is decreasing. Therefore, the sign of the tangent line's slope provides important information about the behavior of the function.
Defining the Tangent Line
The ancient Greeks defined a tangent to a circle as a line that intersects it at only one point. This definition cannot be extended to curves in general. The word "tangent" originates from the Latin verb tangere, which means "to touch." So, we intuitively know that the tangent line to a curve at a point is a straight line that touches the curve at that point.
For example, in the following figure, we intuitively know that the line is a tangent line to the curve at because it "touches" the curve, even though it meets the curve at more than one point. On the other hand, the line is not a tangent line, despite intersecting the curve at a single point.

In the seventeenth century, mathematicians devised a method to construct the tangent line. To draw the tangent line to a curve at a point on it, we choose a second nearby point on the curve and draw the secant line . (The line connecting and is called a "secant line" because it cuts the curve. The word secant comes from the Latin secare, meaning to cut, and does not refer to the secant function in trigonometry.) The tangent line at is the limiting position of the secant as moves along the curve toward , that is, as approaches (see the following figure).

In the above discussion, we have assumed that there is a tangent to the curve at , or that there is a definite limiting position of the straight line connecting and . This assumption is not true for all curves, or even all continuous curves.
Finding the Slope of the Tangent Line
To find the equation of the tangent line, we just need its slope because we already know that lies on the line.
Let be a fixed point on the curve (see the following figure). Since is on the graph of , we know . Here, we use the subscript 0 to emphasize that and are held constant through the discussion. Let be another point on the curve; because is on the curve , we have .
The slope of the straight line connecting and is
If we denote the difference of the independent variable by and the difference of the function by , then
\begin{aligned} \Delta x &= x_1 - x_0 \\ \Delta y = \Delta f(x_0) &= y_1 - y_0 \\ &= f(x_1) - f(x_0) \\ &= f(x_0 + \Delta x) - f(x_0). \end{aligned}Note that is an abbreviation for difference and not a factor. Using these notations, the slope of the secant line is
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- Notice that .
- Notice that can be positive or negative (but not zero since does not coincide with ). If , the point lies to the right of ; if , lies to the left of . Depending on the curve, , and the sign of , can be positive, negative, or even zero.

If we let , the point will move along the curve and gradually approach , the secant will turn about and approach the tangent as a limiting position. In this case, the limit of the slope of is called the slope of the tangent line at :
\begin{aligned} \text{slope of tangent at } P = \tan\theta &= \lim_{\Delta x \to 0} m_{PQ} \\ &= \lim_{\Delta x \to 0} \frac{f(x_0 + \Delta x) - f(x_0)}{\Delta x}, \end{aligned}or simply

The slope of the tangent line to the curve at is called the slope of the curve at .
Examples
Find the equation of the tangent line to the curve at $(4, 2)$.
Solution
Let . Here , and thus the slope of the tangent line is given by Direct substitution of produces the indeterminate form . To get rid of the radical in the numerator, multiply numerator and denominator by the conjugate : \begin{aligned} m_{\text{tangent}} &= \lim_{\Delta x \to 0} \frac{\sqrt{4 + \Delta x} - 2}{\Delta x} \cdot \frac{\sqrt{4 + \Delta x} + 2}{\sqrt{4 + \Delta x} + 2} \\ &= \lim_{\Delta x \to 0} \frac{(\sqrt{4 + \Delta x})^2 - 2^2}{\Delta x(\sqrt{4 + \Delta x} + 2)} \\ &= \lim_{\Delta x \to 0} \frac{\Delta x}{\Delta x(\sqrt{4 + \Delta x} + 2)} \\ &= \lim_{\Delta x \to 0} \frac{1}{\sqrt{4 + \Delta x} + 2} = \frac{1}{\sqrt{4} + 2} = \frac{1}{4}. \end{aligned} Since the slope of the tangent line is , and we know the tangent line passes through $(4, 2)$, its equation is The following figure shows the graph of and its tangent line at $(4, 2)$.
Consider the curve . Find the slope of this curve at an arbitrary point.
Solution
To find the slope at an arbitrary point , we use the definition where . We calculate : \begin{aligned} f(x_0 + \Delta x) &= 2(x_0 + \Delta x)^2 - (x_0 + \Delta x) + 1 \\ &= 2[x_0^2 + 2x_0 \Delta x + (\Delta x)^2] - x_0 - \Delta x + 1 \\ &= 2x_0^2 + 4x_0 \Delta x + 2(\Delta x)^2 - x_0 - \Delta x + 1 \\ &= 2x_0^2 - x_0 + 1 + 4x_0 \Delta x - \Delta x + 2(\Delta x)^2. \end{aligned} Subtracting : Therefore, \begin{aligned} \text{Slope} &= \lim_{\Delta x \to 0} \frac{4x_0 \Delta x - \Delta x + 2(\Delta x)^2}{\Delta x} \\ &= \lim_{\Delta x \to 0} (4x_0 - 1 + 2\Delta x) \\ &= 4x_0 - 1. \end{aligned} So the slope of the curve at any point is . Since can be any point on the curve, we can remove the subscript and state that gives the slope at any point on the curve.Consider the curve .
- Find the slope of this curve at an arbitrary point.
- Find the equation of the tangent line to this curve at the point .
- Does this curve have a horizontal tangent? If so, at what points is the tangent line horizontal?