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Activity 1.2 Secant Line to a Curve

One of the building blocks in differential calculus is the secant line to a curve. The only requirement for a line to be considered a secant line to a curve is that the line must intersect the curve in at least two points.

In Figure 1.2.1, we see a secant line to the curve \(y=\fe{f}{x}\) through the points \(\point{0}{3}\) and \(\point{4}{-5}\text{.}\) Verify that the slope of this line is \(-2\text{.}\)

The formula for \(f\) is \(\fe{f}{x}=3+2x-x^2\text{.}\) We can use this formula to come up with a generalized formula for the slope of secant lines to this curve. Specifically, the slope of the line connecting the point \(\point{x_0}{\fe{f}{x_0}}\) to the point \(\point{x_1}{\fe{f}{x_1}}\) is derived in the following example.

an upside down parabola with vertex at (1,4) and x-intercepts at -1 and 3; a downward slanting line intersects the parabola at (0,3) and (4,-5)
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Figure 1.2.1. A secant line to \(y=\fe{f}{x}=3+2x-x^2\)
Example 1.2.2. Calculating Secant Slope.
\begin{align*} m_{\text{sec}}&=\frac{\fe{f}{x_1}-\fe{f}{x_0}}{x_1-x_0}\\ &=\frac{\left(3+2x_1-x_1^2\right)-\left(3+2x_0-x_0^2\right)}{x_1-x_0}\\ &=\frac{3+2x_1-x_1^2-3-2x_0+x_0^2}{x_1-x_0}\\ &=\frac{\left(2x_1-2x_0\right)-\left(x_1^2-x_0^2\right)}{x_1-x_0}\\ &=\frac{2\left(x_1-x_0\right)-\left(x_1+x_0\right)\left(x_1-x_0\right)}{x_1-x_0}\\ &=\frac{\left[2-\left(x_1+x_0\right)\right]\left(x_1-x_0\right)}{x_1-x_0}\\ &=2-x_1-x_0\text{ for }x_1\neq x_0 \end{align*}

We can check our formula using the line in Figure 1.2.1. If we let \(x_0=0\) and \(x_1=4\) then our simplified slope formula gives us: \(2-x_1-x_0=2-4-0\text{,}\) which simplifies to \(-2\) as we expected.

Subsection Exercises

Let \(\fe{g}{x}=x^2-5\text{.}\)

1.

Following Example 1.2.2, find a formula for the slope of the secant line connecting the points \(\point{x_0}{\fe{g}{x_0}}\) and \(\point{x_1}{\fe{g}{x_1}}\text{.}\)

2.

Check your slope formula using the two points indicated in Figure 1.2.3.

That is, use the graph to find the slope between the two points and then use your formula to find the slope; make sure that the two values agree!

an upward opening parabola with vertex at (0,-5) and x-intercepts at a little less than -2 and a little greater than 2; an upward slanting line intersects the parabola at (-2,-1) and (3,4)
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Figure 1.2.3. \(y=\fe{g}{x}=x^2-5\) and a secant line
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