Lesson Objectives

Dear learner,

By the end of this lesson, you will be able to:

• Define rational functions
• Determine the domain of a given rational function
• Determine the range of a given rational function
• Draw the graph of rational functions
• Identify the asymptote types that the graph of a given rational function may have
• Use graph of rational functions to solve rational inequalities

Key Terms

• Rational functions
• Graph of rational functions
• Vertical asymptote
• Horizontal asymptote
• Oblique assymptote
• Hole

Brainstorming Activity

Dear learner, give your answer for the following Questions

a. How do you describe rational expressions?

b. Identify rational expressions i),. $\frac{x+0.5}{x-\sqrt 2}$ ii). $\frac{\sqrt(x+1)}{x+2}$

Solution:

your discussion may be wider but it may keep basic points.

a). a rational expression is an expression which is written of the form $\frac{P(x)}{Q(x)}$, where $P(x), Q(x)$ are all polynomials.

b). i. $\frac{x+0.5}{x-\sqrt 2}$ is a rational expression since $x+0.5$ and $x- \sqrt 2$ are polynomials

ii. $\frac{\sqrt(x+1)}{x+2}$ is not a rational expression since $\sqrt(x+1)$ is not a polynomial expression.

3.1. Rational Functions

Definition:   A function  f  which is written of the form $\frac{P(x)}{Q(x)}$where $P(x)$ and $Q(x)$ are polynomials, $Q(x) \ne 0$  is a rational function.

$\text{Domain of}~f(x)=\frac{P(x)}{Q(x)}\; is\;{x\in\mathbb{R}:Q(x)\neq0}$.

Example 1:

$\text{1. Find}~f(0),~f(1)~\text{and}~f(2)~\text{for the function}~f(x)=\frac{5x-8}{2-3x}.$

$\text{2. Determine the domain of}~f(x)=\frac{5-3x}{1-x+2x^{2}-2x^{3}}$

$\text{3. If the ordered pair}~(1,3)~\text{and}~(0,6)~\text{belongs to the function}$

$f(x)=\frac{ax^{3}-2bx+3}{ax^{2}+b},~\text{then what is the value of}~a~\text{and}~b?$

$\text{4. Find the number or the numbers that makes the denominator zero?}$

$\text{a.}~f(x)=\frac{x^{2}+1}{x^{2}-1} ~ \; \text{b.}~f(x)=\frac{x^{2}-x-12}{x^{2}+x-6}$.

Solution:

$\text{1. Find}~f(0),~f(1)~\text{and}~f(2)~\text{for the function}~f(x)=\frac{5x-8}{2-3x}?$

$\Rightarrow f(0)=\frac{5(0)-8}{2-3(0)}=-4$

$\Rightarrow f(1)=\frac{5(1)-8}{2-3(1)}=3$

$\Rightarrow f(2)=\frac{5(2)-8}{2-3(2)}=\frac{-1}{2}.$

$\text{2. Domain of}~f(x)=\frac{5-3x}{1-x+2x^{2}-2x^{3}}$.

$\text{Steps to obtain: i. Factorize the denominator. \text{Find the zeros of denominator}}$

$\Rightarrow Q(x)=0 \Rightarrow1-x+2x^{2}-2x^{3}=0 \Leftrightarrow1-x+2x^{2}(1-x)=0$

$\Leftrightarrow(1-x)(1+2x^{2})=0. \Rightarrow x=1~\text{and}~2x^{2}+1\ne0$.

$\text{Domain}=\mathbb{R}/\{1\}$.

$\text{3. Find}~f(x)=\frac{ax^{2}-2bx+3}{ax^{2}+b}~\text{and}~f(1)=3,~f(0)=6.$

$\Rightarrow f(0)=\frac{a(0)-2b(0)+3}{a(0)+b}=6\Rightarrow\frac{3}{b}=6\Rightarrow b=\frac{1}{2}$

$\Rightarrow f(1)=\frac{a(1)-2b(1)+3}{a(1)+b}=\frac{a-2b+3}{a+b}=3$

$\Rightarrow a-2(\frac{1}{2})+3=3(a+\frac{1}{2})$

$\Rightarrow a+2=3a+\frac{3}{2}\Rightarrow2a=2-\frac{3}{2}\Rightarrow a=\frac{1}{4}.$

$\Rightarrow(a,b)=(\frac{1}{4},\frac{1}{2}).$

$\text{4. a.}~f(x)=\frac{x^{4}+1}{x^{4}-1}\Rightarrow x^{4}-1=0\Rightarrow x=\pm1$.

$\text{b.}~f(x)=\frac{x^{2}-x-12}{x^{2}+x-6}\Rightarrow x^{2}+x-6=0$

$\Rightarrow(x+3)(x-2)=0\Rightarrow x=-3~\text{or}~x=2.$

Graph of Rational Functions

Procedures following to draw graph of $f(x) = \frac{p(x)}{q(x)}$, where $p(x)$= numerator, $q(x)$ =  denominator are polynomials.

• Determine domain
• Find x and y intercepts if any
• Identify asymptotes
• Check holes if any
• Study behavior of  the graph near asymptotes 
• Sketch the graph

$\text{Let}~f(x)=\frac{p(x)}{q(x)}~\text{be a rational function:}$

$\bullet~\text{domain}=\mathbb{R}/{q(x)\neq0}.$

$\bullet~\text{x-intercept, at}~y=0~\text{and}~y\text{-intercept at}~x=0.$

Asymptotes:

Definition:

Given a rational function $f(x) = \frac{p(x)}{q(x)}$:

• The line $x = a$ is a vertical asymptote of the graph of $f$ if $f(x) \to \infty$ or $f(x) \rightarrow -\infty as x \to a$ either from the right or from the left.
• The line $y = b$ is a horizontal asymptote of the graph of $f$ if $f(x) \to b as x \to \infty$ or $x \rightarrow -\infty$

Consider a rational function$~f(x)=\frac{p(x)}{q(x)},~\text{where}~p(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0},$

$~\text{with}~a_{n}\neq0~\text{and}~q(x)=b_{m}x^{m}+b_{m-1}x^{m-1}+\cdots+b_{0},~\text{with}~b_{m}\neq0.$

$\text{1. If}~p(a)\neq0~\text{and}~q(a)=0,~\text{then the line}~x=a~\text{is a vertical asymptote to the graph of}~f(x).$

$\text{2. If both}~p(a)=0~\text{and}~q(a)=0,~\text{then the graph of}~f(x)~\text{has a “hole” at}$

$x=a$ or we have to make further simplification to decide.

$\text{3. If}~n<m,~\text{then the line}~y=0~\text{is a horizontal asymptote to the graph of}~f(x).$

$\text{4. If}~n=m,~\text{then the line}~y=\frac{a_{n}}{b_{n}}~\text{is a horizontal asymptote to the graph of}~f(x).$

$\text{5. If}~n=m+1,~\text{then the graph of}~f(x)~$has an asymptote $y=ax+b,~a\neq0,$ and because the line is an oblique line,$~y=ax+b$~ is called an oblique asymptote to the graph of$~f(x).$

$\text{6. If}~n>m+1,~\text{the graph of}~f(x)~\text{has no horizontal asymptote and it has no oblique asymptote.}$

Example 2:

$\text{1. Find the vertical asymptotes for the following functions}.$

$\text{a.}~f(x)=\frac{2x+1}{x^{2}-x-2}$

$\Rightarrow q(x)=0 \Rightarrow x^{2}-x-2=0.$

$\Rightarrow(x-2)(x+1)=0 \Rightarrow x=2~\text{or}~x=-1$

$\text{Therefore vertical asymptotes is}~x=2~\text{and}~x=-1.$

$\text{b.}~f(x)=\frac{x^{2}+5x+6}{x^{2}-9}$

$f(x)=\frac{x^{2}+5x+6}{x^{2}-9}=\frac{(x+2)(x+3)}{(x-3)(x+3)}=\frac{x+2}{x-3}$

$\Rightarrow q(x)=0 \Rightarrow x-3=0$

$\Rightarrow x=3$

$\text{Thus, the vertical asymptotes is}~x=3.$

$\text{2. Find the horizontal asymptotes for the following functions}$

$\text{a.}~f(x)=\frac{5x}{x^2+1}$.

$\text{Here, degree of}~p(x)=5x<\text{degree of}~q(x)=x^2+1$

$\Rightarrow y=0~\text{is the horizontal asymptote}.$

$\text{b.}~f(x)=\frac{5x^{2}+2x-4}{2x^{2}-6x+1}$.

$\text{Here, degree of}~p(x)=5x^{2}+2x-4=\text{degree of}~q(x)=2x^{2}-6x+1$

$\Rightarrow\text{horizontal asymptote is}~y=\frac{5x^{2}}{2x^{2}}=\frac{5}{2}.$

$\text{Hole of the graph:}$

$\bullet~\text{It is detected when factors of}~p(x)~\text{and}~q(x)~\text{are cancelled to each other.}$

Example 3

$\text{Identify whether the function has a hole or not:}$

$\text{1.}~f(x)=\frac{x^{2}-5x-6}{x^{2}-1}$

$\Rightarrow f(x)=\frac{x^{2}-5x-6}{x^{2}-1}=\frac{(x+1)(x-6)}{(x-1)(x+1)}=\frac{x-6}{x-1}.$

$\Rightarrow f~\text{has a hole at}~(x,y)=(-1,\frac{-1-6}{-1-1})=(-1,\frac{7}{2})~\text{because}~x+1~$ is cancelled from both the numerator and denominator.

$\text{2.}~f(x)=\frac{3x+1}{3x^{2}+4x+1}=\frac{3x+1}{(3x+1)(x+1)}=\frac{1}{x+1}.$

$\Rightarrow f~\text{has a hole at}~(x,y)=(-\frac{1}{3},\frac{1}{-\frac{1}{3}+1})=(-\frac{1}{3},\frac{3}{2})~\text{because}~3x+1~$ is the cancelled factor from both the numerator and denominator.

Example 4:

$\text{1. For each of the following functions find:}$

$\text{i. Domain}$

$\text{ii. x- and y-intercepts}$

$\text{iii. asymptotes}$

$\text{iv. holes if any}$

$\text{v. and draw the graph}$

$\text{a.}~f(x)=\frac{x+1}{x^2-1} ~ ~ ~\text{b.}~f(x)=\frac{x^{2}-x-6}{x-2} ~ ~ ~\text{c.}~f(x)=\frac{x^{3}-1}{x^{2}-1}$

Solution:

$\text{a.}~f(x)=\frac{x+1}{x^{2}-1}$

$\text{i. domain:}~x^{2}-1=0$

$\Rightarrow(x-1)(x+1)=0\Rightarrow x=\pm1 \Rightarrow\text{Dom}~f=\mathbb{R}\backslash\{-1,1\}.$

$\text{ii. intercepts:}$

$\text{x-intercept:}~\text{when}~y=0.$

$\Rightarrow0=\frac{x+1}{x^{2}-1}=\frac{x+1}{(x+1)(x-1)}=\frac{1}{x-1}\Rightarrow1=0 \text {~ is a false statement}.$

$\text{Hence, it has no x-intercept.}$

$\text{y-intercept:}~\text{when}~x=0.$

$\Rightarrow y=\frac{0+1}{0-1}=-1.$

$\Rightarrow\text{y-intercept at}~(0,-1).$

$\text{iii. asymptotes:}$
$\text{vertical asymptote}, ~\text{horizontal asymptote or oblique}$
$f(x)=\frac{x+1}{x^{2}-1}=\frac{x+1}{(x-1)(x+1)}=\frac{1}{x-1}$ & $\text{degree of}~p(x)=1<\text{degree of}~q(x)=x-1 .$
$q(x)=0\Rightarrow x-1=0\Rightarrow x=1$ & $\text{horizontal asymptote is}~y=0. $
$\text{Hence, Vertical asymptote is}~x=1$ & $\text{No oblique asymptote}. $

$\text{iv. Hole:}$

$f(x)=\frac{x+1}{x^{2}-1}=\frac{x+1}{(x-1)(x+1)}.$

$\Rightarrow x+1~\text{is the cancelled factor} \Rightarrow\text{Hole at}~(-1,-\frac{1}{2}).$

$\text{v. Behavior of the graph:}$
$ \text{As x is getting closer and closer to vertical asymptote from the right, y is increasing indefinitely} $
$ \text{As x is getting closer and closer to vertical asymptote from the left, y is decreasing indefinitely}$
$ \text{As x approaches infinity, y approaches horizontal asymptote} $
$ \text{As x approaches negative infinity, y approaches horizontal asymptote}. $

Fig.3.1.Graph of $f(x) =\frac{x+1}{x^{2}-1}$

$\text{b.}~f(x)=\frac{x^{2}-x-6}{x-2}$.

$\text{i. domain:}~x-2=0\Rightarrow x=2 \Rightarrow\text{dom}~f=\mathbb{R}\backslash\{2\}.$

$\text{ii. intercepts:}$

$\text{x-intercept:}~\text{when}~y=0.$

$\Rightarrow0=\frac{x^{2}-x-6}{x-2}\Leftrightarrow x^{2}-x-6=0 \Rightarrow(x-3)(x+2)=0\Rightarrow x=3,-2.$

$\Rightarrow\text{x-intercept at}~(-2,0)~\text{and}~(3,0).$

$\text{y-intercept:}~\text{when}~x=0.$

$\Rightarrow y=\frac{0-0-6}{0-2}=3$. Hence y-intercept is at $(0,3)$.

$\text{iii. Asymptotes:}$
$\text{Vertical} ~and ~\text{Oblique Asymptote}$.
$ Q(x)=x-2=0 ~ \text{using long division} .$
$\text{vertical asymptote at}~x=2 $.

$f(x)=\frac{x^{2}-x-6}{x-2}=x+1-\frac{4}{x-2}.$
$\text{No horizontal asymptote} ~\text{since degree of}~p(x)>\text{degree of}~q(x).$

$ \text{Oblique asymptote is}~y=x+1.$

$\text{iv. Behavior of the graph:}$

As $ x$ approaches 2 from the left, y approaches to positive infinity and as $x$ approaches 2 from the right, $y$ approaches to negative infinity.

Fig.3.2. Graph of $f(x) = \frac{x^{2}-x-6}{x-2}$

$\text{c.}~f(x)=\frac{x^{3}-1}{x^{2}-1}$

$f(x)=\frac{x^{3}-1}{x^{2}-1}=\frac{(x-1)(x^{2}+x+1)}{(x-1)(x+1)}=\frac{x^{2}+x+1}{x+1}$

$\text{i. domain:}~(x-1)(x+1)=0\Rightarrow x=-1,1$

$\Rightarrow\text{Dom}~f=\mathbb{R}\backslash\{-1,1\}$

$\text{ii. intercepts:}$

$\text{x-intercept:}~\text{when}~y=0.$

$\Rightarrow0=\frac{x^{2}+x+1}{x+1}\Rightarrow x^{2}+x+1=0, x=\empty set$

$\Rightarrow\text{Hence, it has no x-intercept}.$

$\text{y-intercept:}~\text{when}~x=0 \Rightarrow ~y=\frac{0-1}{0-1}=1$.

$\text{Hence, it has y-intercept at (0,1)}.$

$\text{iii. Asymptotes:}$

$\text{Vertical asymptote}$

$q(x)=0\Leftrightarrow x+1=0 \Rightarrow x=-1 \text{ is a vertical asymptote.}$

$ \text{Oblique Asymptote} $
$ \text{using long division we get}$ $y=x$ $\text{as an oblique asymptote and no horizontal asymptote.} $

Fig.3.3. Graph of $f(x) = \frac{x^{3}-1}{x^{2}-1}$

$\text{2. Show that the graph of}~f(x)=\frac{x^{3}+x-1}{x^{2}}~\text{crosses
its oblique asymptote.}$

$\text{Solution:}$

$\text{Determine its oblique asymptote}.$

$\bullet~\text{Deg}~p(x)>\text{Deg}~q(x)~\text{by one}.$

$\bullet~\text{Using the long division, we get the oblique asymptote}~y=x.$

$\bullet~\text{if it crosses O.A.}~f(x)=0.A$
$\Rightarrow~x=\frac{x^{3}+x-1}{x^{2}}\Leftrightarrow x^{3}=x^{3}+x-1 \Rightarrow~0=x-1\Rightarrow x=1.$

$\text{Hence, the graph crosses its oblique asymptote at}~(1,1)$

$\text{See the graph below}.$

Fig.3.4 Graph of $f(x) = \frac{x^{3} +x -1}{x^{2}}$

https://www.geogebra.org/m/ucjx2rjy

Note:

• The graph of any rational function never crosses its vertical asymptote.

2. The graph of any rational function may or may not cross its horizontal and oblique asymptotes.

https://www.geogebra.org/m/jqmvkeux

3.2. Applications of rational Expressions and Functions

Work rate and shared problems

General illustration to solve such kind of problems

Let A be the time taken to complete a certain job by worker one and B be the time taken by the second worker to complete the job.

Rate of accomplishing the job of the first worker is equal to $\frac{1}{A}$   and rate of second worker is equal to $\frac{1}{B}$ Rate of working together is $\frac{1}{A}+\frac{1}{B}=\frac{1}{x}$  , if $x$is the time it takes to complete the job together.

Example 5:

• Roman can do her home-work assignments in 100 minutes. It takes Dergu about two hours to complete a given assignment. How long will it take two of them working together to complete the assignment?

• Ojullu and Tolla working together can do ajob in 6 hours. Ojullu working alone could have completed the job in 5 hours earlier than Tolla could have done alone. How much time could each worker need to complete the whole job alone?.

Solution: Given Roman’s rate of doing her assignment = $\frac{1}{100}$ in a minute. Dergu’s rate of doing the same assignment = $\frac{1}{120}$  in a minuteAnd let $x$ be the time taken in minute to complete the assignment when they do togetherThen $\frac{1}{100} +\frac{1}{120} = \frac{1}{x}$   is the general equationImplies $\frac{1}{100} +\frac{1}{120} = \frac{1}{x}$ $\iff$ $\frac{120+100}{12000}= \frac{1}{x}$ $220x = 12000$.

Therefore$x = \frac{12000}{220} = 54.54 minutes$

Ans.:- Roman and Dergu take 54.54 minutes to complete their assignment together.

2. let $x$ be the time taken by  Tolla to do the job. Ojullu and Tolla take 6 hours to do the job together. Ojullu can do the job 5 hours earlier than Tolla  can do alone.  Implies Ojullu’s time to do the job$x-5$ hours.

$\text{General equation:}$

$\frac{1}{x}+\frac{1}{x-5}=\frac{1}{6},~\text{where}~x>5.$

$\Rightarrow\frac{(x-5)+x}{x(x-5)}=\frac{1}{6} \Rightarrow\frac{2x-5}{x^{2}-5x}=\frac{1}{6}$

$\Rightarrow6(2x-5)=x^{2}-5x \Rightarrow12x-30=x^{2}-5x$

$\Rightarrow x^{2}-17x+30=0 \Rightarrow(x-2)(x-15)=0.$

$\Rightarrow x=2~\text{or}~x=15.$

Implies Tolla’s time could be  15 hours,  2 cannot be the solution  since $ x > 5$. Then Ojulu can do the job in  15-5 = 10 hours

Variation problems

Let $x$, $y$ be variables and $K$ be the constant:

Direct variation $y = Kx$  means x and y are directly across from each other.
Inverse variation$y = \frac{K}{x}$ means that x is inverted on the other side of y.

Joint variation$y = Kwx$  means w and x jointly are directly varied with y.

Example 6:

• A plane flies 910 km with the wind in the same time it can go 660 km against the wind. The speed of the plane in still air is 305 km per hour. What is the speed of the wind?

2. The area of a trapezoid (𝐴) is jointly proportional to its height and the sum of the lengths of the parallel sides. Express the previous statement as a formula, using the trapezoid below on the left and a constant of variation 𝑘. Find the constant of variation 𝑘, if the area of the trapezoid on the right is 49 

Solution:

• Let $x$= the speed attained by the two moving objects(according to their character, it varies) and $y$ = the distance covered by the two moving objects. implies the two variables distance and speed are directely proportional:

$y = K x$, where $ k$ is the constant. assume that the two objects are flying at the same time, that is $K = time$, then we formulate

$K = \frac{y}{x}$ $\implies$ $\frac{910}{305}= \frac{660}{x}$ where $x > 0$ $\implies$ $x = \frac{660(305)}{910}$,

then $x = 221.21 km/h$, therefore the speed of wind is 221.21 km/h

2. let $A$ = Area of trapezium, $b,a$ = the two bases while $h$ stands for height of the trapezium. Area is jointely proportional top height and the sum of the two bases:

$A$ =K (h)(a+b), $\implies$ $A =K \frac{(h)(a+b)}{2}$.

from the picture on the right a = 5, b = 9, h = 7, and Area(A) = 49

$49 = K\frac{7(5+9)}{2}$ $\implies$ $49 = K(49)$, therefore $K = 1$