Sally Fitzgibbons Foundation

Beginning the Academic Essay

This chapter starts with the discussion of application of rational functions in
real life situation. The denition of rational functions and it’s form of equation.
Also, how to solve a rational function and how to represent a rational function
using tables of values, graph and equation. The chapter ends with graphing the
rational functions.
At the end of this chapter the students should be able to:
1. apply the concept of rational functions in real life situation.
2. identify the forms of a rational functions.
3. dierentiate rational equations, inequalities ,and functions.
4. solves problem involving rational equations, inequalities, and functions.
5. determine the domain and range.
6. graph a rational functions.
1

Introduction
Rational functions is an algebraic expression forming a fraction whose nu-
merator and denominator are both polynomials. The term rational comes from
the word ratio and also the result of a rational number is in ratio.
1.1 Representing real life situation using ratio
Ratio are used in our everyday routine. Ratio are usually used in comparing
two ob jects also it is used in a basic tasks such as cooking, buying as well as in
businesses like selling fruits, rice, etc. because it determines specic amount of
the product.
Steps in solving ratio problems.
1. Assign the variables.
2. Solve the equation.
Example 1.1.1
There are 1000 students in a school. The ratio of girls and boys in this school
is 5:3. How many girls and how many boys are in this school?
Solution.
Step 1: Assign the variables
Let 5 x= the girl students.
3 x = the boy students
The total students are 1000. So, the equation will be
5x + 3 x= 1000
Step 2: Solve the equation.
8x = 1000
8 x 8
=
1000 8
x
= 125 Substitute the value of
x.
Number of girls = 5(125) = 625
2

Number of boys = 3(125) = 375
Therefore, there are 625 girls and 375 boys in this school.
Example 1.1.2
The box contains white and yellow marbles, the ratio of the white marbles to
yellow marbles is 3:4. if the box contains 120 yellow marbles. how many white
marbles are there?
Solution.
Step 1: Assign the variables
Let x= white marbles
write this ratio in terms of fraction. white marbles yellow marbles
=
3 4
=
x 120
Step 2: Solve the equation
cross multiply 3 4
;
x 120
= 360 and
3 4
.
x 120
= 4
x
360 = 4 x
nd x 360 4
=
4
x 4
90 =
x Therefore, there are 90 white marbles in the box.
Example 1.1.3
Mel has 30 apples, 18 are red apples and 12 are green apples. Maria has 20 ap-
ples, either all of them are red or green. if the ratio of red and green apples are
both the same for Mel and Maria, then how many green apples does Maria have?
Solution.
Step 1: Assign the variables
3

Let
x= green apples for Maria
20 – x= red apples for Maria
Write the ratio of Mel
Mel has 30 apples, 18 are red apples and 12 are green apples. red green
=
18 12
=
3 2
We will use it for Maria’s ratio. red green
=
3 2
=
20
x x
Step 2: Solve the equation
cross multiply 3 2
&
20
x x
= 3
xand 3 2
.
20
x x
= 40
2x
3 x = 40 2x
Solve for x: 3x = 40 2x
3 x +2 x= 40
5 x = 40
5 x 5
=
40 5
x
= 8 Therefore, Maria have 8 green apples.
Example 1.1.4
An ob ject travels a distance of 10 meters. Express velocity vas a function v(t)
of travel time t, in seconds.
Solution.
The function v(t) = 10 t
:
4

t (seconds) 1 2 4 5 10
v (meters per second) 10 5 2.5 2 1
Table 1.1: shows the values of
vfor various values of t.
Example 1.1.5
Suppose that d(t) = 2
t t
2
+3 in (mg/ml) is the concentration of drug in a patient’s
bloodstream thours after the drug was administered. Construct a table for d(t)
and t= 0, 1, 2, 5, 10. then use the table to sketch the graph.
Solution. t 0 1 2 5 10
d(t) 0 0.5 0.57 0.36 0.19
Figure 1.1: the graph of the table 1.1
The graph shows the concentration of drug in a patient’s bloodstream oc-
curs 1 hour after the drug administered. After 1 hour the drug concentration
decreases until it is almost zero.
1.2 Rational functions, equations, and inequali- ties
This lesson discusses the denition of rational functions, equations, and inequal-
ities. To determine their forms and how they are dierent from each other.
Denition 1.2.1 A rational function is a function that is a polynomial, which
is in the form of f(x ) = p
(x ) q
(x ), where
q(x ) 6
= 0
Example 1.2.1 f(x ) = 2(3
x 5) x
1 or can be written as
y= 2(3
x 5) x
1
Example 1.2.2 f(x ) = x
2
+2 x+1 x
1 or
y= x
2
+2 x+1 x
1
5

Denition 1.2.2
A rational equation is an equation which is also a rational
expression containing at least one fraction which either on one or both sides of
the equation.
Example 1.2.3 2
x 1 x
+3 =1 2
Example 1.2.4 1 2
x +5 = 1 x
3
Denition 1.2.3 A rational inequality is a rational expression having at least
one fraction which involves inequality signs.
Example 1.2.5 1 2
x 3 x
+1
Therefore, a rational inequality and equation can be solve for all values of
x. we do no solve a rational function but it can be represented by the table of
values and a graph.
1.3 Solving rational equations and inequalities
This lesson contains of solving rational equations and inequalities step by step.
To solve a rational equation
1. nd the Least Common Denominator(LCD),
2. multiply the LCD both sides of the equation add or subtract the given equation,
3. solve for the value of x.
Example 1.3.1 1 x
2 3
x = 1 5
Solution
Step 1: Find the LCD of the given equation.
1 x

2 3
x =
1 5
Take a look at the denominator, the values are x, 3 x, and 5. since xand 3 x
have a common value which is x, we will disregard x. Also since 3 xand 5 does
not have a common value, we will going to multiply both of them. 3 x 5 =
15 x. So, the LCD is 15x.
Step 2: multiply the LCD both sides of the equation, then add or subtract
the given equation.
6

(15
x)1 x

2 3
x =
1 5
(15
x)
15 x x

30
x 3
x =
15
x 5
15 10 = 3 x
5 = 3 x
Step 3: Solve for x.
5 = 3x(Divide both sides by 3)
5 3
=
3
x 3
5
3
=
x Therefore, the value of
xis 5 3
or approximately 1.66666667.
Example 1.3.2 2+
x 3
x = 1 3
Solution.
Step 1: Find the LCD
2 +x 3
x=
1 3
The LCD of this equation is 3 (3 – x) which is equal to 9-3x.
Step 2: Simplify the equation.
Multiply both sides by 9 – 3 x
(9 3x )( 2 +
x 3
x) =
1 3
(9
3x )
(9 3x )(2 + x) 3
x =
(9
3x ) 3
By foil method,
(93x )(2 + x) = (18 + 9 x 6x 3x 2
)
= 3x 2
+ 3 x+ 18
Factor 3x 2
+ 3 x+ 18
3x 2
+ 3 x+ 18
3( x2
x 6)
3( x + 3)( x 2)
7

Going back to the equation,
3x 2
+ 3 x+ 18 3
x =
(9
3x ) 3
3( x + 3)( x 2) 3
x =
3(3
x) 3
3( x 2) = 3 x
3x 6 = 3 x
3x + x= 3 + 6
2x
2 =
9
2 x
= 9 2
Therefore, the value of
xis 9 2
or approximately 4.5.
To solve a rational inequalities, we must nd the critical value the interval
of the given problem.
1. make the given inequality in general form.
2. to nd the critical value, we will equate the numerator to zero, and also the denominator to zero.
3. partition the zeros in number line,
4. pick a test point for each interval. and write the answer in interval form.
Example 1.3.3 x
+3 x
1
0
Solution.
Step 1: make the given inequality in general form.
Since the given is in the general form we will now proceed to step 2.
Step 2: nd the critical value
Numerator Denominator
x + 3 = 0 x 1 = 0
x = 3 x= 1
8

Step 3:
partition the zeros in number line. Step 4:
pick a test point for each interval.
If x= 4 x= 1 x= 2
( 4) + 3 (
4) 1
0 (1) + 3 (1)
1
0 (2) + 3 (2)
1
0
1
5
0 0 0 5 0
if we test x=-3 this is a true statement because it will become 0 0. and if
we text x=1, the value will be undened. Therefore the interval will be -3,1)
1.4 Representations of rational functions
1.5 Graphing rational functions
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