Alg unit 7 notes part 2

Algebra 1
Unit 7 (Part 2)
11.2 rational functions
11.8 rational equations
Notes Packet
Name __________________________
Period ____
Teacher ____________
1
Lesson 11-2 Rational Function Notes
π‘π‘œπ‘™π‘¦π‘›π‘œπ‘šπ‘–π‘Žπ‘™
A rational function can be written in the form f(x) = π‘π‘œπ‘™π‘¦π‘›π‘œπ‘šπ‘–π‘Žπ‘™ where the value of the denominator
cannot be zero.
β€’ Dividing by zero is undefined so any value of a variable that makes the denominator equal to zero
in a rational function is excluded from the domain of the function.
β€’ These values are called excluded values.
To find Excluded values:
1. Set the denominator equal to zero.
2. Solve for the variable as necessary.
3. Write as: β€œx = ____ is an excluded value” because x CANNOT be this value.
Examples
Find the excluded values for each function.
3
y=π‘₯
Set the denominator equal to zero and solve for the variable
x=0
0 is excluded from the domain
1
y = π‘₯βˆ’2
Set the denominator equal to zero and solve for the variable
x–2=0
x=2
2 is excluded from the domain
3
g(x) = 2π‘₯βˆ’4
Set the denominator equal to zero and solve for the variable
2x – 4 = 0
2x = 4
x=2
2 is excluded from the domain
2
Examples: Find the excluded values of the domain for each function.
1. y =
3
2. y =
π‘₯
1
3. g(x) =
π‘₯βˆ’2
Practice:
Find the excluded value(s) of the function
1. g(x) =
2. y =
5
3π‘₯βˆ’9
9
βˆ’2βˆ’3π‘₯
3. f(x) =
1
5βˆ’π‘₯
3
3
2π‘₯βˆ’4
Graphing Rational Functions –
The parent function of all rational functions is
𝟏
f(x) = 𝒙
x
-10
-1
-.5
-.1
0
.1
.5
1
10
y
-.1
1
2
-10
UND
10
2
1
.1
Notice the outputs (y-values) in the table approach but never touch the excluded value of x = 0. This is
called an asymptote.
Asymptote: a line that the graph of a function approaches, but NEVER touches.
There are 2 asymptotes to Rational Functions:
1. Vertical asymptote is the excluded value of the domain.
a. Domain is written as β€œxβ‰ 
β€œ (write in the x-value of the vertical asymptote.)
π‘Ž
2. Horizontal asymptote can be found where y = c in the form y = π‘₯βˆ’π‘ + c.
a. In the parent rational function, this occurs at y = 0.
b. In all other functions, this occurs at y = c
c. RANGE is written as β€œyβ‰ 
β€œ (write in the y-value of the horizontal asymptote.)
4
Examples: Identify the asymptotes of each function.
1. y =
𝟐
π’™βˆ’πŸ
Vertical Asymptote
x–1=0
x=1
x = 1 is the vertical asymptote
Domain: x≠
2. y =
πŸ“
𝒙+𝟏
Horizontal Asymptote
The value of c = 0 so,
y = 0 is the horizontal asymptote
Range: y≠
+πŸ‘
Vertical Asymptote
Horizontal Asymptote
x+1=0
The value of c = 3 so,
x = -1
y = 3 is the horizontal asymptote
x = -1 is the vertical asymptote
Domain: x≠
Range: y≠
5
11-2
Graphing Rational Functions
To Graph Rational Functions:
1. Find the asymptotes of the function.
2. State the Domain and Range.
3. Complete the input output table using 3 x-values on BOTH sides of the
VERTICAL asymptote.
a. Your table will have 7 x-values in it. (vertical asymptote and 3 x-values
to each side of it)
4. Sketch the asymptotes on the graph as dashed lines.
5. Plot the points to create the hyperbola
a. Make sure the graph does not cross the ASYMPTOTES!
Examples: Identify the asymptotes of each function.
1. y =
𝟐
π’™βˆ’πŸ
Vertical Asymptote
Horizontal Asymptote
x–1=0
x=1
x = 1 is the vertical asymptote
The value of c = 0 so,
y = 0 is the horizontal asymptote
X
-9
0
.5
1
2
1.5
11
Y
-.5
-2
-4
UND
2
4
.5
6
2. y =
πŸ“
𝒙+𝟏
+πŸ‘
Vertical Asymptote
Horizontal Asymptote
X
-11
-2
-1.5
-1
-.5
0
1.5
Y
3.5
-2
13
UND
13
8
5
Practice
3. y =
𝟏
π’™βˆ’πŸ
Vertical Asymptote
Horizontal Asymptote
X
Y
UND
7
4. f(x) =
𝟐
π’™βˆ’πŸ
βˆ’πŸ
Vertical Asymptote
Horizontal Asymptote
X
Y
3.
UND
y=
𝟏
4.
π’™βˆ’πŸ
Vertical Asymptote Horizontal Asymptote
f(x) =
𝟐
π’™βˆ’πŸ
βˆ’πŸ
Vertical Asymptote Horizontal Asymptote
8
11-2b Skills Practice
State the excluded value for each function.
1. y =
4. y =
7. y =
6
π‘₯
π‘₯βˆ’3
π‘₯+4
π‘₯
3π‘₯ + 21
2. y =
5. y =
8. y =
2
π‘₯βˆ’2
3. y =
3π‘₯ βˆ’ 5
π‘₯+8
6. y =
π‘₯βˆ’1
9π‘₯ βˆ’ 36
9. y =
π‘₯
π‘₯+6
βˆ’5
2π‘₯ βˆ’ 14
9
5π‘₯ + 40
Identify the asymptotes of each function. Then graph the function.
10. y =
13. y =
1
π‘₯
3
π‘₯βˆ’2
11. y =
14. y =
3
π‘₯
2
π‘₯+1
12. y =
βˆ’1
15. y =
9
2
π‘₯+1
1
π‘₯βˆ’2
+3
11-2c Practice
Rational Functions
State the excluded value for each function.
1. y =
4. y =
βˆ’1
π‘₯
π‘₯βˆ’1
12π‘₯ + 36
2. y =
5. y =
3
π‘₯+5
3. y =
π‘₯+1
2π‘₯ + 3
6. y =
2π‘₯
π‘₯βˆ’5
1
5π‘₯ βˆ’ 2
Identify the asymptotes of each function. Then graph the function.
7. y =
10. y =
1
π‘₯
2
π‘₯+2
8. y =
11. y =
3
π‘₯
1
π‘₯βˆ’3
9. y =
+2
12. y =
13. AIR TRAVEL Denver, Colorado, is located approximately 1000 miles from
Indianapolis, Indiana. The average speed of a plane traveling between the
1000
two cities is given by y =
, where x is the total flight time. Graph the
π‘₯
function.
10
2
π‘₯βˆ’1
2
π‘₯+1
–1
11-8
Rational Equations
Solve each proportion.
1
π‘₯
=
3
5
3
𝑦
=
5
π‘₯
2
3
=
27
6
A Rational Equation is one that contains one or more rational expressions.
One type of rational equation is a rational proportion. One way to solve equations of this
type is by cross multiplying.
Steps
1. Put parentheses around any binomials
2. Write the cross products.
a. This shows you what you have to multiply.
3. Multiply – this may require any or all of the following:
a. Distributive Property
b. FOIL method to multiply binomials.
4. Solve equation using inverse operations.
a. May need to incorporate any or all of the following along the way:
i. Set equation equal to 0
ii. Factoring using the Zero Product Property
iii. Quadratic Formula
5. ALWAYS check your answers to look for any EXTRANEOUS SOLUTIONS.
11
4
5
=
π‘₯+3 π‘₯
5(x) = (x+3)4
Write cross products
Distribute, then use inverse operations to solve for the variable.
5x = 4x + 12
x = 12
1
4
=
π‘₯2 π‘₯ + 8
4(x+8) = 1(x2)
Write cross products
4x + 32 = x2
Distribute
0 = x2 – 4x – 32
Factor
0 = (x – 8)(x + 4) Zero Product Property
x = -4, 8
Practice:
Solve
πŸπŸ“
π’™πŸ βˆ’ 𝟏
=
πŸ“
𝟐(𝒙 βˆ’ 𝟏)
.
12
11.8a
Solve each equation. State any extraneous solutions.
1.
4.
7.
9.
π‘₯βˆ’5
5
π‘₯
4
8
π‘›βˆ’1
π‘ž+4
π‘žβˆ’1
3
π‘₯
+ =8
=
+
π‘š+1
π‘šβˆ’1
–
2. =
10
𝑛+1
π‘ž
π‘ž+1
5. t –
=2
π‘š
1βˆ’π‘š
=1
6
π‘₯+1
4
𝑑+3
3.
=t+3
8.
6.
5 βˆ’ 2π‘₯
2
10.
13
–
π‘₯2 βˆ’ 9
π‘₯βˆ’3
4π‘₯ + 3
6
=
+ π‘₯2 = 9
7π‘₯ + 2
6
π‘₯βˆ’1
5
π‘š+4
π‘š
=
+
2π‘₯ βˆ’ 2
15
π‘š
3
=
π‘š
3
11.
13.
2
π‘₯ 2 βˆ’ 36
4
4βˆ’π‘
–
–
1
π‘₯βˆ’6
𝑝2
π‘βˆ’4
=0
12.
=4
14.
4𝑧
𝑧 2 + 4𝑧 + 3
π‘₯ 2 βˆ’ 16
π‘₯βˆ’4
11-8b Skills Practice
Rational Equations
Solve each equation. State any extraneous solutions.
5
𝑐
1. =
3.
5.
2
𝑐+3
7
π‘š+1
𝑦
π‘¦βˆ’2
=
=
12
π‘š+2
𝑦+1
π‘¦βˆ’5
3
π‘ž
2. =
4.
6.
5
π‘ž+4
3
π‘₯+2
π‘βˆ’2
𝑏
=
=
5
π‘₯+8
𝑏+4
𝑏+2
14
=
6
𝑧+3
+ π‘₯ 2 = 16
+
4
𝑧+1
7.
9.
11.
13.
15.
17.
3π‘š
2
1
4
– =
2π‘Ž + 5
6
𝑐+2
𝑐
βˆ’2
π‘₯+1
–
–
2π‘Ž
3
=–
𝑐+3
𝑐
+
π‘šβˆ’4
π‘š
π‘Ÿ+3
π‘Ÿβˆ’1
–
10π‘š
8
2
π‘₯
1
2
10.
=7
π‘š βˆ’ 11
π‘š+4
π‘Ÿ
π‘Ÿβˆ’3
8.
=
=0
+ =1
12.
1
π‘š
14.
16.
18.
7𝑔
9
1
3
+ =
π‘›βˆ’3
10
𝑓+2
𝑓
π‘›βˆ’5
5
+
3𝑏 βˆ’ 4
𝑏
𝑓+1
𝑓+5
5
π‘šβˆ’4
–
–
=
π‘βˆ’7
𝑏
–
–
𝑒+1
π‘’βˆ’2
5𝑔
6
𝑒
𝑒+1
=1
=
1
𝑓
=0
π‘š
2π‘š βˆ’ 8
15
1
2
=1
11-8
Rational Equations
Find the Least Common Denominator (LCD) of each group of expressions.
3
1
2
1
, ,𝑛
4π‘₯ 2
2
4
1
, , 3π‘₯
3π‘₯ 5
,
1
8𝑦 𝑦 2
5
,6
The denominators of a rational expression may contain more than one term.
The LCD of these denominators in a rational equation may also contain more than one term.
To solve rational equations that are not set up like proportions:
1. Put parentheses around any binomials you have in the problem
2. Find the LCD of all terms in the problem
3. Multiply ALL TERMS in the problem by the LCD to simplify the equation.
a. This could look like the Distributive Property.
4. Simplify each new term/fraction.
a. You should not have fractions left in the equation after this step.
5. Use inverse operations to solve for the variable.
a. This could include Factoring, Completing the Square, or Quadratic Formula
6. Check for extraneous solutions!
Examples
A.
1
3
1
+ 10 = 5π‘₯
2π‘₯
1
3
The LCD of 2x, 10, and 5x is 10x
1
10π‘₯ οΏ½ 2π‘₯ + 10 οΏ½ = 10π‘₯( 5π‘₯ )
10π‘₯
2π‘₯
+
30π‘₯
10
=
10π‘₯
5π‘₯
5 + 3x = 2
3x = -3
x = -1
Multiply both sides by 10x
Distribute
Simplify fractions as needed.
Don’t forget to plug -1 back into the original equation to check!
16
1
B.
3
1
1
1
6π‘₯
6π‘₯
The LCD of 3, 3x, and 6 is 6x
+ 3π‘₯ = 6
1
1
Multiply both sides by 6x
6π‘₯ οΏ½ 3 + 3π‘₯ οΏ½ = 6π‘₯( 6 )
6π‘₯
3
C.
+ 3π‘₯ =
Distribute
6
2x + 2 = x
2 = -x
x = -2
Simplify fractions as needed.
Don’t forget to plug -2 back into the original equation to check!
8
1
= π‘₯+1
(π‘₯+3)
8
The LCD of x+3, x, and 1 is x(x+3)
1
π‘₯(π‘₯ + 3) (π‘₯+3) = π‘₯(π‘₯ + 3)( π‘₯ + 1)
8 π‘₯(π‘₯+3)
(π‘₯+3)
=
π‘₯(π‘₯+3)1
π‘₯
+ 1π‘₯(π‘₯ + 3)
8x = (x+3) + x(x+3)
8x = x + 3 + x2 + 3x
8x = x2 + 4x + 3
0 = x2 – 4x + 3
0 = (x – 1)(x – 3)
x = 1, x = 3
Multiply both sides by x(x+3)
SHOW new fractions after you distribute
Simplify fractions
Distribute and simplify
Simplify by combining like terms
Rearrange so quadratic equals 0
Factor, Quadratic Formula, or Complete the Square
Zero Product Property
Don’t forget to plug 1 and -3 back into the original equation to check!
17
Something to watch out for: The denominator of a Rational Expression can NEVER be 0 because you
can’t divide by zero.
10
π‘₯(π‘₯βˆ’2)
+
4
π‘₯
5
= π‘₯βˆ’2
The common denominator will be x(x – 2), and x cannot be zero or 2.
Multiply both sides of the equation by the LCD.
10 + 4(x – 2) = 5(x)
10 + 4x – 8 = 5x
4x + 2 = 5x
2=x
The solution is 2 however, one of the denominators in the original equation was x – 2. In this case x = 2 is
an excluded value. Since the equation yielded only 2 as a solution, there is not a valid solution to this
problem. The correct answer is that there is no solution.
18
Practice
1.
3.
2
=
1
+ =
π‘₯
2
5. 1 +
1
2.
π‘₯+4
2
π‘₯
1
π‘₯+2
1
4.
π‘₯
=
28
3
π‘₯
5
=
2π‘₯
π‘₯βˆ’2
3
+ =
4
6.
π‘₯ 2 +2π‘₯
19
5
9
4π‘₯
2π‘₯βˆ’ 5
π‘₯βˆ’2
βˆ’2=
3
π‘₯+2
11-8c Practice
Rational Equations
Solve each equation. State any extraneous solutions.
1.
4.
7.
5
=
2β„Ž
=
𝑛+2
β„Žβˆ’1
2π‘ž βˆ’ 1
6
7
2.
2β„Ž + 1
5.
𝑛+6
β„Ž+2
βˆ’
π‘ž
3
=
π‘ž+4
18
8.
π‘₯
π‘₯βˆ’5
4𝑦
3
5
=
π‘₯+4
3.
π‘˜+5
=
π‘˜βˆ’1
1
5𝑦
6.
π‘¦βˆ’2
–
𝑦+2
π‘₯βˆ’6
+ =
π‘βˆ’1
2
–
6
3
𝑝+2
=0
20
9.
π‘˜
4
3𝑑
3𝑑 βˆ’ 3
π‘˜+9
–
5
1
= –1
9𝑑 + 3
=1
10.
13.
16.
4π‘₯
–
2
–
π‘š+2
=
2𝑝
+
𝑝+2
2π‘₯ + 1
π‘š+2
π‘βˆ’2
2π‘₯
=1
11.
π‘‘βˆ’3
–
π‘‘βˆ’4
7
14.
𝑛+2
+
𝑛+5
=1
17.
π‘₯+7
2π‘₯ + 3
π‘šβˆ’2
𝑝2
βˆ’4
3
𝑑
𝑛
π‘₯2
βˆ’9
21
π‘‘βˆ’2
=
𝑛+3
–
π‘₯
π‘₯+3
1
12.
𝑑
=βˆ’
=1
1
𝑛
15.
18.
3𝑦 βˆ’ 2
π‘¦βˆ’2
+
1
–
2𝑛
–
𝑧+1
π‘›βˆ’4
𝑦2
2βˆ’π‘¦
6βˆ’π‘§
6𝑧
= –3
=0
𝑛+6
𝑛2 βˆ’ 16
=1
11.8d PRACTICE Rational Equations
Solve each equation. Check your solution. If there is no solution, write no solution.
1.
+
4.
7. x +
10.
=
2.
=
=β€”7
=
+
=
3.
5.
=
6.
8.
=
9.
11.
β€”
=
22
=
β€”
=
+
12. 4 β€”
=
=
4π‘₯
–
15.
3𝑦 βˆ’ 2
+
17.
𝑛+2
+
𝑛+5
𝑛+3
=βˆ’
19.
2𝑝
+
𝑝+2
=1
13.
2π‘₯ + 1
π‘¦βˆ’2
𝑛
π‘βˆ’2
2π‘₯
=1
14.
π‘‘βˆ’3
–
π‘‘βˆ’4
𝑦2
= –3
16.
2
–
π‘š+2
1
18.
2π‘₯ + 3
2βˆ’π‘¦
𝑝2
βˆ’4
𝑛
20.
23
𝑑
π‘š+2
1
𝑧+1
π‘₯+7
π‘₯2
–
βˆ’9
π‘‘βˆ’2
π‘šβˆ’2
6βˆ’π‘§
–
=
6𝑧
π‘₯
1
𝑑
=
7
3
=0
π‘₯+3
=1
Algebra 1
Unit 7 TEST Review
Name: _____________________ Per ______
1. How does the graph of y = √π‘₯ + 3 compare to the parent graph?
A) translated up 3
C) translated right 3
B) translated down 3
D) translated left 3
2. Which expression has a domain of {x β‰₯ 2} ?
A) y = √π‘₯ + 2
B) y = √π‘₯ – 2
C) y = √π‘₯ + 2
D) y = √π‘₯ βˆ’ 2
3. What is the equation of the graph?
A) y = √π‘₯ + 1 - 3
B) y = √π‘₯ βˆ’ 1 - 3
C) y = √π‘₯ βˆ’ 1 + 3
D) √π‘₯ + 1 + 3
For questions 4 – 8, simplify each expression.
4. 8 2 βˆ’ 2
A) 8
5. √18 – √54 + 2√50
A) 13√2 – 3√6
6.
C) 7 2
B) 16
D) βˆ’8 2
B) –4√3 + 4√5
C) –4√3 – 4√5
D) 8√2 – 3√6
B) 15 βˆ’ 3 2
C) 8 βˆ’ 3 2
D) 15 βˆ’ 2 3
B) 256
C) 1768 βˆ’ 24 7
D) 256 + 24 7
3(5 3 βˆ’ 6)
A) 5 9 βˆ’ 18
7. (2 βˆ’ 6 7) 2
A) 256 βˆ’ 24 7
24
8. ( 5 + 11) 2
A) 16
B) 4
9. Solve √3π‘₯ βˆ’ 2 + 4 = 8.
A) 12
B) 6
C)
10. Solve √7π‘Ž + 32 = a + 2.
A) –4
B) 7
C) –4, 7
12. Solve
3
5
5π‘₯
3π‘₯ + 1
A) 58
B)
–
10
3π‘₯ +1
2
3
3
5
3
2
D) –7, 4
8
5
D) βˆ’
C) -3
1
5
= .
B) βˆ’
18
19
C)
13. Which value is an extraneous solution of
A) 5
D)
x + 1 2x
?
=
15
5
11. Which is the solution of
A) βˆ’
D) 16 + 2 55
C) 146
B) 0
π‘₯
π‘₯+1
C) –1
–
18
5
D) -
6
π‘₯ 2 βˆ’ 4π‘₯ βˆ’ 5
25
= 1?
D) 6
18
5
14. Given the rational equation
3
1
5
which polynomial would you multiply by to eliminate the
βˆ’
=
2x + 8 x + 4 2 ,
denominators?
A) 2
B) x + 4
C) 2x + 4
D) 2x + 8
15. Which function is graphed?
A)
y
=
1
+3
x+2
C=
) y
1
βˆ’3
xβˆ’2
B)
y
=
1
βˆ’3
x+2
D)
y
=
1
+3
xβˆ’2
16. The velocity V of an object that has fallen d meters can be found using the equation V =
2
2gd , where g
represents the gravitational constant and is equal to 9.81 m/s . Determine how fast a penny would be traveling
when it hits the ground below after being dropped off the Willis Tower, which has a height of 520 meters.
A) 101 m/s
B) 7214.2 m/s
C) 10202.4 m/s
26
D) 71.4 m/s
Algebra 1
Unit 7 Free Response Review
18. Graph 𝑦 = √π‘₯ βˆ’ 5 + 3
Domain ________________
Range _________________
1
19. Graph 𝑦 = π‘₯+1 βˆ’ 3
Vertical Asymptote: _________________
Horizontal Asymptote: ___________________
27
Name __________________________