# Tangent Line to a Curve

The **tangent line to a curve** at a given point is the line which intersects the curve at the point and has the same instantaneous slope as the curve at the point. Finding the tangent line to a point on a curved graph is challenging and requires the use of calculus; specifically, we will use the derivative to find the slope of the curve.

For any point on the curve we are interested in, it is easy to find *a* line through the point, but to find the tangent line, we will need to find the slope of the curve at the point we are interested in. Let's consider a function \(f(x)\) and the point on the function \(\big(c,f(c)\big).\)

## Introduction

We need to find the slope of the line that passes through \(\big(c,f(c)\big)\) that is tangent to \(f(x)\). In general, we know that we can find the slope, \(m\), through any two points by calculating the change in \(y\, (\Delta y)\) divided by the change in \(x\, (\Delta x),\) which is described by the formula

\[m = \frac{\Delta y}{\Delta x}=\frac{y_2 - y_1}{x_2 - x_1}.\]

However, in this case, we are trying to find the slope at the single point \(\big(c, f(c)\big)\), which means that, if we were to use the formula above, we would find \(m=\frac00,\) which is undefined. To find the slope of our tangent line, we will need to use a difference quotient.

## Difference Quotient

A *difference quotient* is an expression that describes the slope of a line at a single point. Let's consider our slope formula again:

\[m = \frac{\Delta y}{\Delta x}=\frac{y_2 - y_1}{x_2 - x_1}.\]

We want to find the change in \(y\) divided by the change in \(x\). Let's consider what happens if we use our point \(\big(c,f(c)\big)\) and one other point close to it, \(\big(c + \Delta x,f(c+\Delta x)\big)\). We can make a secant line that intersects the function twice (represented below by the dotted red line):

Note that this secant line does not have the same slope as the tangent line, but it has a slope that is *close* to the slope of the tangent line. We can calculate the slope of the secant line easily:

\[m = \frac{\Delta y}{\Delta x}=\frac{y_2 - y_1}{x_2 - x_1}=\frac{f(c+\Delta x) - f(c)}{c+\Delta x - c}=\frac{f(c+\Delta x) - f(c)}{\Delta x}.\]

The big insight here of the difference quotient is that as \(\Delta x\) gets smaller, the slope of the secant line gets closer and closer to the slope of the tangent line we are trying to find. In fact, using limits, we can find the exact slope of the tangent line to the curve \(f(x)\) at \(x=c\) using the difference quotient:

The slope \(m\) of the curve \(f(x)\) at \(x=c\) is defined to be

\[m = \lim_{\Delta x \to 0} \frac{f(c+\Delta x) - f(c)}{\Delta x}.\]

Note: The above assumes that the function \(f(x)\) is continuous and differentiable at \(c\).

What is the slope of \(f(x)=x^2\) at the point \((3,9)?\)

Using the definition above, we can see that the slope is

\[ \begin{align} m &= \lim_{\Delta x \to 0} \frac{f(3+\Delta x) - f(3)}{\Delta x} \\\\ &= \lim_{\Delta x \to 0} \frac{(3+\Delta x)^2-3^2}{\Delta x} \\\\ &= \lim_{\Delta x \to 0} \frac{6\Delta x + (\Delta x)^2}{\Delta x} \\\\ &= \lim_{\Delta x \to 0} 6 + \Delta x\\\\ &= 6. \end{align}\]

Thus, the slope of the line tangent to \(f(x)=x^2\) at the point \((3,9)\) will have slope \(m=6\). \(_\square\)

This method of finding slope is known as the *derivative*, and you can read more about it here: Derivative by First Principle.

## Finding the Tangent Line

Once we have found the slope (using the derivative shown above), we can find the equation of the line with that slope through the point quite easily, using linear equations and point-slope form.

What is the equation of the line tangent to \(y=x^3\) at the point \((1,1)?\)

First, we want to find the slope. Here \(f(x) = x^3 \) and \(c=1\). Using the difference quotient, we can write\[\begin{align} m &= \lim_{h \to 0} \frac{ f(1 + h) - f(1) }{h}\\ & = \lim_{h \to 0} \frac{ (1 + h)^3 - (1)^2 }{h} \\ & = \lim_{h \to 0} \frac{ 1 + 3h + 3h^2 + h^3 - 1 }{h} \\ & = \lim_{h \to 0} \frac{ 3h + 3h^2 + h^3 }{h} \\ & = \lim_{h \to 0} (3 + 3h + h^2) \\ & = 3. \end{align}\]

Thus the slope of the tangent line will be 3. Using the point \((1,1)\) and the slope \(m-3\), we have \(y=3x-2\). \(_\square\)

What is the equation of the line that passes through the point \((3,24)\) and is perpendicular to the tangent line of the curve \( y=x^2+5x\) at that point?

## See Also

- Finding the Slope of a Line
- Linear Equations and Point-Slope Form
- Derivative by First Principle
- To learn about this material more thoroughly, consider starting the course Calculus Done Right.

**Cite as:**Tangent Line to a Curve.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/tangent-line-point/