13.2: Breaking Up is NOT Hard to Do
This activity is intended to supplement Algebra I, Chapter 12, Lesson 6.
ID: 11934
Time Required: 20 minutes
Activity Overview
In this activity, students will split rational functions into sums of partial fractions. Graphing is utilized to verify accuracy of results and to support the understanding of functions being represented in multiple ways.
Topic: Rational Functions & Equations
- Least common denominator
- Sum of partial fractions
- Equivalent functions
Teacher Preparation and Notes
- Problems 1-3 should be done in class as guided practice or small group work. Several problems are provided on the student worksheet for additional practice.
- As an extension, the teacher could include a discussion of the placement of vertical asymptotes.
- Before beginning the activity, make sure that all plots have been turned off and all equations have been cleared from the \begin{align*}Y=\end{align*} screen.
Associated Materials
- Student Worksheet: Breaking Up is NOT Hard to Do, http://www.ck12.org/flexr/chapter/9622, scroll down to the second activity.
Problem 1 – Introduction
This part of the activity involves an exploration of equivalent ways to express a rational function. Students will generate function graphs from which they will learn that a rational function can be represented as the sum of individual fractions, known as partial fractions.
To make it clear to students that the graphs of \begin{align*}Y1\end{align*} and \begin{align*}Y2\end{align*} are identical, show students how to place the tracing circle in front of \begin{align*}Y2=\end{align*} by using the arrows to move to the left of \begin{align*}Y2=\end{align*} and pressing ENTER until the tracing circle appears.
Students will answer questions regarding their observations of the graphs of the given equations. They are also asked to observe the denominators of the two functions.
The denominators of the fractions in \begin{align*}Y2\end{align*} are the factors of the denominator in \begin{align*}Y1\end{align*}.
Since the graphic results show that the two functions are equivalent, they are set equal to each other and a framework is established for finding the numerators of the partial fractions of a rational function. Directions are provided to help students through the process.
\begin{align*}Y1(x)& =Y2(x)\\ \frac{7x+3}{x^2-9}& =\frac{A}{x+3}+\frac{B}{x-3}\end{align*}
Students proceed to solve for \begin{align*}A\end{align*} and \begin{align*}B\end{align*} by substituting in values for \begin{align*}x\end{align*} that will simplify the work to be done. For example, substituting \begin{align*}-3\end{align*} for \begin{align*}x\end{align*} will eliminate the \begin{align*}B\end{align*} term and simplify the process of solving for \begin{align*}A\end{align*}. Similarly, substituting \begin{align*}3\end{align*} for \begin{align*}x\end{align*} will simplify solving for \begin{align*}B\end{align*}.
Discuss with students why it might be helpful to decompose a rational expression into a sum of partial fractions. Students may note that the partial fractions, being less complex, will be easier to work with for certain mathematical applications.
\begin{align*}&(x+3)(x-3)\left(\frac{7x+3}{x^2-9} = \frac{A}{x+3}+\frac{B}{x-3}\right)\\ & \qquad \qquad \qquad \quad \ 7x-3 = A(x-3)+B(x+3)\\ & \qquad \text{Let} \ x=3; 7(3)-3 = A(3-3)+B(3+3)\\ &\qquad \qquad \qquad \qquad \ \quad 18 = 6B\\ &\qquad \qquad \qquad \qquad \quad \ \ 3 = B\\ &\text{Let} \ x=-3; 7(-3)-3 \ = A(-3-3)+B(-3+3)\\ &\qquad \qquad \qquad \qquad \quad 24 = -6A\\ &\qquad \qquad \qquad \quad \ \quad -4 = B\end{align*}
Problem 2 – Practice
Students apply what was learned in Problem 1 to find a sum of partial fractions equivalent to a given rational function.
Once the algebraic work is completed, students can verify the equivalence of their solution to the original function via graphing the two functions. Remind students to use the show/hide feature to the left in the function entry bar of the graph page to be certain that the graphs of the two functions are identical.
\begin{align*}& \qquad \qquad \qquad \quad \frac{7x-4}{x^2+x-6} = \frac{A}{(x+3)}+\frac{B}{(x-2)}\\ &(x+3)(x-2)\left(\frac{7x-4}{x^2+x-6} = \frac{A}{(x+3)}+\frac{B}{(x-2)}\right)\\ &\qquad \qquad \qquad \qquad \quad 7x-4 = A(x-2)+B(x+3)\\ & \qquad \quad \ \ \text{Let} \ x=2; 7(2)-4 = A(2-2)+B(2+3)\\ &\qquad \qquad \qquad 10=5B, \quad B = 2\\ & \qquad \ \text{Let} \ x=-3; 7(-3)-4 = A(-3-2)+B(-3+3)\\ & \qquad \qquad \quad \ 25=-5B, \quad B = -5\end{align*}
Problem 3 – The Next Level
Students again apply what has been learned, but the challenge level increases.
In this situation, the denominator has a constant factor in addition to two binomial factors. A hint is given to prompt students to use the algebraic binomial factors as denominators for the partial fractions to be determined.
When students solve for \begin{align*}A\end{align*} and \begin{align*}B\end{align*}, they will find that the value for \begin{align*}B\end{align*} is a fraction, which will result in a need for simplification of the partial fractions obtained.
\begin{align*}& \qquad \qquad \qquad \qquad \frac{5x-7}{4x^2-8x-12} = \left(\frac{A}{(x-3)}+\frac{B}{(x+1)}\right)\\ &4(4x-3)(x+1)\left(\frac{5x-7}{4x^2-8x-12}=\frac{A}{(x-3)}+\frac{B}{(x+1)}\right)\\ & \qquad \qquad \qquad \qquad \qquad \quad \ \ 5x-7 = 4A(x+1)+4B(x-3)\\ & \qquad \qquad \ \text{Let} \ x=-1; \ 5(-1)-7 = 4A(-1+1)+4B(-1-3)\\ & \qquad \qquad \qquad -12=-16B, \quad B = \frac{3}{2}\\ & \qquad \qquad \qquad \text{Let} \ x=3; \ 5(3)-7 = 4A(3+1)+4B(3-3)\\ & \qquad \qquad \qquad \qquad \ \ 8=16A; \quad A = \frac{1}{2}\end{align*}
Solutions
- The graphs are the same.
- The functions appear to be equal.
- The denominators of \begin{align*}f2\end{align*} are factors of the denominator of \begin{align*}f1\end{align*}.
- \begin{align*}x^2-9\end{align*} or \begin{align*}(x-3)(x+3)\end{align*}
- \begin{align*}7x+3 = A(x-3) + B(x+3)\end{align*}
- \begin{align*}3\end{align*}
- \begin{align*}4\end{align*}
- \begin{align*}\frac{7x+3}{x^2-9}=\frac{3}{x+3}+\frac{4}{x-3}\end{align*}
- The results verify algebraically that the two functions are equivalent.
- \begin{align*}\frac{7x-4}{x^2+x-6}=\frac{2}{x-2}+\frac{5}{x+3}\end{align*}
- Yes; Same graph result verifies equivalent algebraic result.
- \begin{align*}\frac{5x-7}{4x^2-8x-12}=\frac{1}{2x-6}+\frac{3}{4x+4}\end{align*}
- Yes; Same graph result verifies equivalent algebraic result.
- \begin{align*}\frac{-7x-11}{x^2+4x+3}=\frac{-2}{x+1}-\frac{5}{x+3}\end{align*}
- \begin{align*}\frac{2x+42}{x^2+2x-24}=\frac{5}{x-4}-\frac{3}{x+6}\end{align*}
- \begin{align*}\frac{x}{x^2+2x-8}=\frac{2}{3x+12}+\frac{1}{3x-6}\end{align*}