5.2: Generic Tangent line Equation and Properties

Learning Objectives

1. Explain the generic form of the tangent line equation (5.1) and be able to connect it to the geometry of the tangent line.
2. Find the coordinate of the point at which the tangent line intersects the \(x\)-axis (important for Newton’s Method later on in Section 5.4).

Generic tangent line equation

We can find the general equation of a tangent line to an arbitrary function \(f(x)\) at a point of tangency \(x_{0}\). (The result is Equation (5.1).)

Figure 5.4: The graph of an arbitrary function \(y = f(x)\) and a tangent line at \(x = x_0\). The equation of this generic tangent line is Eqn. (5.2).

Shown in Figure 5.4 is a continuous function \(y=f(x)\), assumed to be differentiable at some point \(x_{0}\) where a tangent line is attached. We see:

1. The line goes through the point \(\left(x_{0}, f\left(x_{0}\right)\right)\).
2. The line has slope given by the derivative evaluated at \(x_{0}\), that is, \(m=f^{\prime}\left(x_{0}\right)\).

Then from the slope-point form of the equation of a straight line,

\[\frac{y-f\left(x_{0}\right)}{x-x_{0}}=m=f^{\prime}\left(x_{0}\right) . \nonumber \]

Rearranging and eliminating the notation \(m\), we have the desired result.

Summary, Tangent Line equation: The equation of a tangent line at \(x=x_{0}\) to the graph of the differentiable function \(y=f(x)\) is

\[y=f\left(x_{0}\right)+f^{\prime}\left(x_{0}\right)\left(x-x_{0}\right) . \nonumber \]

Mastered Material Check

6. Circle the point (x0, f(x0)) on Figure 5.4.
7. Circle where the tangent line depicted in Figure 5.4 would cross the x-axis

Quick video with a derivation of the generic equation of a tangent line. Summary, Tangent Line

Where a generic tangent line intersects the \(x\)-axis

From the generic tangent line equation (5.2) we can determine the (generic) coordinate at which it intersects the \(x\)-axis. The result is key to Newton’s method for approximating the zeros of a function, explored in Section 5.4.

Example 5.5

Let \(y=f(x)\) be a smooth function, differentiable at \(x_{0}\), and suppose that Equation (5.2) is the equation of the tangent line to the curve at \(x_{0}\). Where does this tangent line intersect the \(x\)-axis?

Solution

At the intersection with the \(x\)-axis, we have \(y=0\). Plugging this into Equation (5.1) leads to

\[\begin{aligned} 0=f\left(x_{0}\right)+f^{\prime}\left(x_{0}\right)\left(x-x_{0}\right) & \Rightarrow\left(x-x_{0}\right)=-\frac{f\left(x_{0}\right)}{f^{\prime}\left(x_{0}\right)} \\ & \Rightarrow x=x_{0}-\frac{f\left(x_{0}\right)}{f^{\prime}\left(x_{0}\right)} \end{aligned} \nonumber \]

Thus the desired \(x\) coordinate, which we refer to as \(x_{1}\) is

\[x_{1}=x_{0}-\frac{f\left(x_{0}\right)}{f^{\prime}\left(x_{0}\right)} \nonumber \]

The calculations for Example 5.5. We show how to find the coordinate \(x_{1}\) where a tangent line intersects the \(x\) axis.