Transcript College Algebra with Modeling and Visualizaitions

Chapter 3
Section 3.5
Rational Functions and
Models
Rational Functions

Definition
p(x)
f(x) is a rational function if and only if f(x) = q(x)
where p(x) and q(x) are polynomial functions with q(x)  0

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Examples

3x2 + 4x + 1
1. f(x) =
x3 – 1

3(x – 3)(x + 3)
3x2 – 27
= 3x + 9 , for x ≠ 3
=
2. f(x) =
x–3
x–3

x3/2 – 8
3. f(x) =
x2 + 1
Question: Is this a rational function ?
Section 3.4
2
Rational Functions
 Domains
 Where are rational functions defined?
 Examples

3x2 + 4x + 1
3x2 + 4x + 1
1. f(x) =
=
x3 – 1
(x – 1)(x2 + x + 1)
Dom f(x) = { x | x ≠ 1 }

3x2 – 27
2. f(x) =
x–3

6x2 – x – 2
6x2 – x – 2
3. f(x) = 2
=
x +x–6
(x + 3)(x – 2)
Dom f(x) = { x | x ≠ 3 }
Dom f(x) = { x | x ≠ – 3, x ≠ 2 }
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Section 3.4
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Rational Functions

Asymptotes

Asymptotes are lines that the graph of a function f(x)
approaches closely as x approaches some k or ±

Vertical Asymptote: line x = k such that either f(x)
or f(x)
– ∞ as x approaches k

Horizontal Asymptote: line y = k such that f(x)
as x
–∞
∞ or x
∞
∞
k
Question: Can the graph of f(x) cross its asymptotes ?
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Section 3.4
4
Rational Functions

Asymptote Examples

1
1. f(x) = x
y
y=0
x
x=0
y

2. f(x) =
1
x+1
x
x = –1
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Section 3.4
5
Rational Functions

Asymptote Examples

x+1
3. f(x) =
x–5
y
y=1
x
x=5
y
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x2 – 4
4. f(x) =
x–2
●
2
Section 3.4
x
6
Rational Functions

Asymptote Examples

5. f(x) =

y
4x – 10
2x – 5
(0, 2)
●
Function is linear, but
undefined at x = 5/2
●
Sketch asymptotes,
intercepts and the graph
2(2x – 5)
4x – 10
=
2x – 5
2x – 5
=2
… provided x ≠ 5/2
x
5/2
f(x) =
No asymptotes !
One intercept: (0, 2)
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Domain ?
Domain = { x │ x ≠ 5/2 }
= ( – ∞ , 5/2 )  ( 5/2 , ∞ )
Range ?
Range = { 2 }
Section 3.4
7
Rational Functions

Asymptote Examples

6. f(x) =
y
11x – 2
x3 – 1
f(1) does not exist since
x3 – 1 = 0 when x = 1
f(x)
as x
0
±∞
Line y = 0
Horizontal asymptote

x
Line x = 1
Vertical asymptote
Question: Does the graph cross an asymptote ? YES !
2
f(x) = 0 at x =
so the graph cuts the asymptote y = 0 at
11
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Section 3.4
(
2
,0
11
)
8
Rational Functions

Asymptote Review

Vertical Asymptote: line x = k such that either f(x)
or f(x)
– ∞ as x approaches k

Horizontal Asymptote: line y = k such that f(x)
as x
–∞
∞ or x
∞
k
Question: Can the graph of f(x) cross its asymptote ?
Consider f(x) =
5x2 + 8x – 3
3x2 + 2
and g(x) =
11x + 2
2x3 – 1
... and their horizontal asymptotes
What about vertical asymptotes ?
Question: Are asymptotes always vertical or horizontal ?
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Section 3.4
9
Rational Functions

Asymptote Examples

y
5x2 + 8x – 3
7. f(x) =
3x2 + 2
3x2
Since
+ 2 is never zero
for any real x there is no
vertical asymptote
Rewriting
f(x) =
8
3
5 + x – x2
2
3 + x2
Horizontal
asymptote
y = f(x)
●
y=
x
5
3
( 2419 , 53)
5
3
as x
±∞
5
Thus f(x) has a horizontal asymptote of y =
3
Question: Where does the graph cross its asymptote ?
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Section 3.4
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Rational Functions

Asymptote Examples

y
x2 + 2x + 1
8. f(x) =
x–1
By synthetic division ... f(x) = x + 3 +
1 1
2
1
3
1
3
4
4
x–1
1
Thus
f has a vertical asymptote at x = 1
and
f(x)
y = x + 3 as x
±
Oblique Asymptote: line y = ax + b such that f(x)
as x
±∞
y=x+3
Oblique (slant)
Asymptote
x
x=1
Vertical
Asymptote
y
Occurs when deg (numerator) = deg (denominator) + 1
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Section 3.4
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Rational Functions

Finding Vertical and Horizontal Asymptotes

p(x)
Vertical Asymptotes for f(x) =
q(x)

Find values of x , say x = k, where q(x) = 0



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f(x) fails to exist at x = k
… BUT might not have an asymptote there
if (x – k) is also a factor of p(x)
Ensure f(x) is reduced to lowest terms
Check x = 0 for vertical intercept
Horizontal and Oblique Asymptotes for f(x) =
p(x)
q(x)

Determine what f(x) approaches as x approaches ± ∞

Check f(x) = 0 for horizontal intercept
Section 3.4
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Rational Functions

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Find Horizontal / Oblique Asymptotes

4x3 – 2
9. f(x) =
x+2

6x2 – x – 2
10. f(x) =
2x2 – 3x + 4

x2 + 2x + 1
11. f(x) =
x–1
Section 3.4
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Rational Functions

Rational Functions for “Large” x

For the function R(x) =
x
10
100
1,000
10,000
100,000 1,000,000
3x + 1 10 31
301
3,001
30,001
300,001 3,000,001
98
998
9,998
99,998
3.007
3.0007
3.00007 3.000007
x–2
R(x)

1


8
10 3.9 3.07
Thus as x
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3
3x + 1
fill in the table
x–2
3x + 1
x–2
R(x)
999,998
1,000,000 we see that
3,000,001 ≈ 3,000,000 = 3x
999,998 ≈ 1,000,000 = x
3.000007 ≈ 3 = (3x)/x
Section 3.4
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Solving Rational Equations

f(x)
How do we solve equations of form: h(x) = g(x)

Method 1: Clear Fractions
Examples:

1. Solve:
23
= 15
x+2
23
· (x + 2) = 15 · (x + 2)
x+2
23 = 15x + 30
–7 = 15x
–7
x=
15
Solution Set:
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Section 3.4
{
7
–
15
}
15
Solving Rational Equations

Method 1: Clear Fractions

2. Solve:
x+5
x+1
=
x+7
x+2
x+5
x+1
· (x + 2)(x + 7) =
· (x + 2)(x + 7)
x+7
x+2
(x + 1)(x + 7) = (x + 5)(x + 2)
x2 + 8x + 7 = x2 + 7x + 10
8x + 7 = 7x + 10
x= 3
Solution Set: { 3 }
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Section 3.4
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Solving Rational Equations

Method 2: Cross Multiplication
a
c
 Basic Principle:
=
if and only if ad = bc
b
d

Examples:

1. Solve:
x+5
x+1
=
x+7
x+2
(x + 1)(x + 7) = (x + 2)(x + 5)
x2 + 8x + 7 = x2 + 7x + 10
8x + 7 = 7x + 10
x= 3
Solution Set: { 3 }
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Section 3.4
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Solving Rational Equations

Method 2: Cross Multiplication


1
x–3
=
x+3
7
Cross multiplying
(x – 3)(x + 3) = 7
2. Solve:
x2 – 9 = 7
Factoring and Zero Product Property
Square Root Property
x2 – 16 = 0
(x + 4)(x – 4) = 0
x + 4 = 0 OR x – 4 = 0
x =–4
x= 4
x2 = 16
√ x2 = ± √ 16
x = ±4
Solution Set: { – 4, 4 }
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Section 3.4
18
Solving Rational Equations

Method 2: Cross Multiplication

3. Solve:
2
x+1
=
3
5x – 3
Cross multiplying
3(x + 1) = 2(5x – 3)
3x + 3 = 10x – 6
9 = 7x
9
x=
7
Solution Set:
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{
Section 3.4
9
7
}
19
Solving Rational Equations

Method 3: Graphical Approach

1. Solve:
Let y1 =
x+1
=2
x–5
x+1
x–5
y
y1
and y2 = 2
3
Horizontal
Asymptote
y=1
y2

(11, 2)
So y1 = y2
For what x is
this true ?
Intercepts for y1 :
Horizontal : ( –1, 0 )
Vertical : ( 0, –1/5 )
–2

3
6
9
x
11
Vertical Asymptote
x=5
–3
Intersection at (11, 2)
Hence: x = 11
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Section 3.4
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Direct and Inverse Variation

Direct Variation



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Output varies directly with input
y
OR
k
Example: y = kx
x =
Inverse Variation

Output varies inversely with input

Example: y = kx–1
OR
yx = k
k is the constant
of variation
Inverse Variation Functions

Output varies inversely with xn

Example: y = kx–n
OR
yxn = k
Section 3.4
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Variation Examples

Direct Variation

The resultant force acting on an object of mass m is
directly proportional to the acceleration of the object
F
=m
F = ma
OR
a

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F varies directly with a -Constant of variation is m
Section 3.4
Newton’s Second Law
22
Variation Examples

Direct Variation Example

Weight and Mass

Excluding other external forces, the only force acting
on an object of mass m is the force of gravity mg, where
g is the acceleration due to gravity ; that is
F = mg

To lift the object, the force of gravity must be overcome
 This force is called weight and given by F = W = mg
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Clearly weight varies directly with mass
Section 3.4
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Variation Examples

Inverse Variation

At constant temperature the volume of n moles of gas
is inversely proportional to the pressure of the gas
PV = nRT OR P = (nRT)V–1

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P varies inversely with V -Constant of variation is nRT
Section 3.4
Ideal Gas Law
24
Variation Examples

Function of Variation

The earth’s gravitational force acting on an object of
mass m is inversely proportional to the square of the
distance between the mass and the center of the earth
GMm
=
OR Fr2 = GMm
F
2
r
2
 F varies inversely with r
-- Law of Gravity

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Constant of variation is GMm, where G is the earth’s
gravitational constant, M is the mass of the earth, and
m is the mass of the object
This is just one of many inverse square laws
Section 3.4
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Think about it !
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Section 3.4
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