Constructing a Tangent to a Curve

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 P Q m P Q = f ( x 0 + Δ x ) f ( x 0 ) Δ x
Slope of tangent at P m tangent = lim Δ x 0 f ( x 0 + Δ x ) f ( x 0 ) Δ x
Tangent line equation y y 0 = m tangent ( x x 0 )

Why Tangent Lines Matter

Read more 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 l is a tangent line to the curve at P because it "touches" the curve, even though it meets the curve at more than one point. On the other hand, the line d is not a tangent line, despite intersecting the curve at a single point.

Line l is tangent to the curve at point P but line d is not, illustrating that one-point intersection does not define tangency

In the seventeenth century, mathematicians devised a method to construct the tangent line. To draw the tangent line to a curve at a point P on it, we choose a second nearby point Q on the curve and draw the secant line P Q . (The line connecting P and Q 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 P is the limiting position of the secant P Q as Q moves along the curve toward P , that is, as Q approaches P (see the following figure).

The limit of the secant slopes as Q approaches P from either side is the slope of the tangent line to the curve at P

In the above discussion, we have assumed that there is a tangent to the curve at P , or that there is a definite limiting position of the straight line connecting P and Q . 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 P lies on the line.

Let P ( x 0 , y 0 ) be a fixed point on the curve y = f ( x ) (see the following figure). Since P is on the graph of f , we know y 0 = f ( x 0 ) . Here, we use the subscript 0 to emphasize that x 0 and y 0 are held constant through the discussion. Let Q ( x 1 , y 1 ) be another point on the curve; because Q is on the curve y = f ( x ) , we have y 1 = f ( x 1 ) .

The slope of the straight line connecting P and Q is

m P Q = y 1 y 0 x 1 x 0 = f ( x 1 ) f ( x 0 ) x 1 x 0 .

If we denote the difference of the independent variable x by Δ x and the difference of the function y ( = f ( x ) ) by Δ y , 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 Δ is an abbreviation for difference and not a factor. Using these notations, the slope of the secant line P Q is

m P Q = Δ y Δ x = f ( x 0 + Δ x ) f ( x 0 ) Δ x .
Read more
  • Notice that m P Q = tan ( R P Q ) = tan α .
  • Notice that Δ x can be positive or negative (but not zero since Q does not coincide with P ). If Δ x > 0 , the point R lies to the right of P ; if Δ x < 0 , R lies to the left of P . Depending on the curve, x 0 , and the sign of Δ x , Δ y can be positive, negative, or even zero.
The slope of PQ is tan alpha equals the difference quotient of f at x_0

If we let Δ x 0 , the point Q will move along the curve and gradually approach P , the secant P Q will turn about P and approach the tangent as a limiting position. In this case, the limit of the slope of P Q is called the slope of the tangent line at P :

\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

m tangent = lim Δ x 0 f ( x 0 + Δ x ) f ( x 0 ) Δ x .
As Q approaches P, the secant line approaches the tangent line

The slope of the tangent line to the curve at P is called the slope of the curve at P .

slope at  ( x 0 , f ( x 0 ) ) = m tangent = lim Δ x 0 f ( x 0 + Δ x ) f ( x 0 ) Δ x .

Examples

Find the equation of the tangent line to the curve y = x at $(4, 2)$.

Solution Let f ( x ) = x . Here x 0 = 4 , and thus the slope of the tangent line is given by m tangent = lim Δ x 0 f ( 4 + Δ x ) f ( 4 ) Δ x = lim Δ x 0 4 + Δ x 2 Δ x . Direct substitution of Δ x = 0 produces the indeterminate form 0 / 0 . To get rid of the radical in the numerator, multiply numerator and denominator by the conjugate 4 + Δ x + 2 : \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 m tangent = 1 4 , and we know the tangent line passes through $(4, 2)$, its equation is y 2 = 1 4 ( x 4 ) or y = 1 4 x + 1. The following figure shows the graph of y = x and its tangent line at $(4, 2)$.
Graph of y equals square root of x with tangent line at (4,2) having slope 1/4

Consider the curve y = 2 x 2 x + 1 . Find the slope of this curve at an arbitrary point.

Solution To find the slope at an arbitrary point ( x 0 , f ( x 0 ) ) , we use the definition Slope = lim Δ x 0 f ( x 0 + Δ x ) f ( x 0 ) Δ x , where f ( x ) = 2 x 2 x + 1 . We calculate f ( x 0 + Δ x ) : \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 f ( x 0 ) = 2 x 0 2 x 0 + 1 : f ( x 0 + Δ x ) f ( x 0 ) = 4 x 0 Δ x Δ x + 2 ( Δ x ) 2 . 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 y = 2 x 2 x + 1 at any point ( x 0 , y 0 ) is 4 x 0 1 . Since ( x 0 , y 0 ) can be any point on the curve, we can remove the subscript and state that m = 4 x 1 gives the slope at any point P ( x , y ) on the curve.

Consider the curve y = 2 x 3 6 x + 1 .

  1. Find the slope of this curve at an arbitrary point.
  2. Find the equation of the tangent line to this curve at the point ( 2 , 3 ) .
  3. Does this curve have a horizontal tangent? If so, at what points is the tangent line horizontal?
Solution (a) Using f ( x ) = 2 x 3 6 x + 1 , we compute \begin{aligned} f(x_0 + \Delta x) &= 2(x_0 + \Delta x)^3 - 6(x_0 + \Delta x) + 1 \\ &= 2[x_0^3 + 3x_0^2 \Delta x + 3x_0 (\Delta x)^2 + (\Delta x)^3] - 6x_0 - 6\Delta x + 1. \end{aligned} Subtracting f ( x 0 ) = 2 x 0 3 6 x 0 + 1 : f ( x 0 + Δ x ) f ( x 0 ) = 6 x 0 2 Δ x + 6 x 0 ( Δ x ) 2 + 2 ( Δ x ) 3 6 Δ x . Therefore, \begin{aligned} \text{Slope} &= \lim_{\Delta x \to 0} \frac{6x_0^2 \Delta x + 6x_0 (\Delta x)^2 + 2(\Delta x)^3 - 6\Delta x}{\Delta x} \\ &= \lim_{\Delta x \to 0} (6x_0^2 + 6x_0 \Delta x + 2(\Delta x)^2 - 6) \\ &= 6x_0^2 - 6. \end{aligned} Removing the subscript, the slope at any point P ( x , y ) is m tangent = 6 x 2 6. (b) First, we verify that ( 2 , 3 ) is on the curve: y = 2 ( 2 ) 3 6 ( 2 ) + 1 = 16 + 12 + 1 = 3. The slope at x = 2 is: m tangent = 6 ( 2 ) 2 6 = 24 6 = 18. Using point-slope form: y ( 3 ) = 18 ( x ( 2 ) ) y = 18 x + 33. (c) The tangent is horizontal when m tangent = 0 : 6 x 2 6 = 0 x 2 = 1 x = ± 1. At x = 1 : y = 2 ( 1 ) 6 ( 1 ) + 1 = 3 , giving the point ( 1 , 3 ) . At x = 1 : y = 2 ( 1 ) 6 ( 1 ) + 1 = 5 , giving the point ( 1 , 5 ) . The curve has horizontal tangent lines at ( 1 , 3 ) and ( 1 , 5 ) .

Frequently Asked Questions

What is the difference between a secant line and a tangent line? A secant line passes through two distinct points on a curve. A tangent line touches the curve at exactly one point (locally) and represents the limiting position of the secant as the second point approaches the first. The slope of the tangent is the limit of the slope of the secant.

Why can't we just substitute Δ x = 0 in the difference quotient? Substituting Δ x = 0 produces 0 0 , which is an indeterminate form with no defined value. Instead, we simplify the difference quotient algebraically to cancel the Δ x in the denominator, and then take the limit as Δ x 0 .

Why does the slope of the tangent equal the slope of the curve? The "slope of the curve" at a point is defined to be the slope of the tangent line at that point. A curve does not have a single direction, so we use the tangent line to capture its instantaneous direction at each point. The slope of the tangent is therefore the natural way to measure how steeply the curve rises or falls at that location.