Secants

Motivation

Secants can be used to find the slope of a tangent. We shall therefore discuss what a secant is.

Definition 1

Consider a curve and two different points AA and BB on the curve. The straight line passing through AA and BB is called a secant of the curve. The function equation

Equation 1
s(x)=ax+bs(x)=ax+b

The equation of the secant is a linear function.

which describes this straight line is called the equation of the secant.

Insight

If the coordinates of the points AA and BB are known, it is straightforward to find the equation of the secant, s(x)=ax+bs(x)=ax+b.

Example 1: Determining the secant equation

Consider the function f(x)=(x−0.2)2+0.1f(x)=(x-0.2)^2+0.1. We want to determine the equation of the secant passing throught the points AA and BB, where Ax=0.1A_x=0.1 and Bx=0.9B_x = 0.9.

  • The yy-coordinate of AA is

    Ay=f(0.1)=(0.1−0.2)2+0.1=0.11A_y=f(0.1)=(0.1-0.2)^2+0.1 = 0.11

    The yy-coordinate of BB is

    By=f(0.9)=(0.9−0.2)2+0.1=0.59B_y=f(0.9)=(0.9-0.2)^2+0.1 = 0.59

    Thus, the points have the coordinates

    A(0.1∣0.11) and B(0.9∣0.59)A(0.1 \vert 0.11) \text{ and } B(0.9 \vert 0.59)

  • The slope aa is therefore given by

    a=ΔyΔx=By−AyBx−Ax=f(Bx)−f(Ax)Bx−Ax=0.59−0.110.9−0.1=0.6\begin{array}{lll} a &=& \frac{\Delta y}{\Delta x} \\[0.5em] &=& \frac{B_y-A_y}{B_x-A_x} \\[0.5em] &=& \frac{f(B_x)-f(A_x)}{B_x-A_x} \\[0.5em] &=& \frac{0.59-0.11}{0.9-0.1} \\[0.5em] &=& 0.6 \end{array}

    Thus, we have s(x)=0.6x+bs(x)=0.6 x+b.

  • To find the yy-intercept bb, we use the fact that the point AA (or BB) is on the curve and on the secant, so we have

    s(Ax)=f(Ax)0.6⋅0.1+b=(0.1−0.2)2+0.10.06+b=0.11b=0.05\begin{array}{rll} s(A_x) &=& f(A_x) \\ 0.6 \cdot 0.1+b &=& (0.1-0.2)^2+0.1 \\ 0.06+ b &=& 0.11 \\ b &=& 0.05 \end{array}

    Thus, the equation is s(x)=0.6x+0.05s(x)=0.6 x+0.05.

We shall highlight the most important point from the calculations above:

Theorem 1: Difference quotient

Consider the function ff, and two points AA and BB on the graph of ff whose xx-coordinates are known. The slope of the secant passing through the points AA and BB is

Equation 2
a=f(Bx)−f(Ax)Bx−Axa = \frac{f(B_x)-f(A_x)}{B_x-A_x}

The difference quotient is the slope of the secant.

The right-hand side of the equation is a fraction (or quotient) and is called the difference quotient.

Exercise 1

Consider the function f(x)=xf(x)=\sqrt{x}. The points AA and BB are on the graph of ff, where AA has xx-coordinate 11 and BB has xx-coordinate 22. Determine the equation of the secant of ff through AA and BB using the difference quotient.

Solution

The equation of the secant is s(x)=ax+bs(x)=ax+b with

a=f(2)−f(1)2−1=2−11=2−1≈0.414\begin{array}{lll} a &=& \frac{f(2)-f(1)}{2-1} \\ &=& \frac{\sqrt{2}-1}{1} \\ &=& \sqrt{2}-1 \\ &\approx& 0.414 \end{array}

Thus, s(x)=0.414x+bs(x)=0.414x+b. To find bb, solve the equation

s(1)=f(1)0.414⋅1+b=1b=0.586\begin{array}{rll} s(1) &=& f(1) \\ 0.414 \cdot 1+b &=& 1 \\ b &=& 0.586 \end{array}

Thus, s(x)=0.414x+0.586s(x)=0.414x+0.586.