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Inverse Variation Is the relationship between the values in each table an inverse variation or not? If it is inverse variation, write an equation to model the inverse variation. Then find y when x = 10. 1. 2. 3. Each function below is one where x varies inversely with y. Complete each table of values, then find the constant of variation for each table. 4. If x and y vary inversely, as y increases, what happens to the value of x? 5. If x and y vary inversely, as x decreases, what happens to the value of y? x 1 2 5 7 y 6 12 30 42 x 3 1.5 0.5 0.3 y 5 10 30 50 x y 2 4 8 6 4 x y 20 15 5 10 20 x y 2 4 8 6 120

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Page 1: Inverse Variationodomqchs.weebly.com/uploads/1/0/9/5/10959249/... · Inverse Variation Is the relationship between the values in each table an inverse variation or not? If it is inverse

Inverse Variation

Is the relationship between the values in each table an inverse variation or not? If it is inverse variation,

write an equation to model the inverse variation. Then find y when x = 10.

1.

2.

3. Each function below is one where x varies inversely with y. Complete each table of values, then find the constant of variation for each table.

4. If x and y vary inversely, as y increases, what happens to the value of x? 5. If x and y vary inversely, as x decreases, what happens to the value of y?

x 1 2 5 7

y 6 12 30 42 x 3 1.5 0.5 0.3

y 5 10 30 50

x y

2

4

8

6 4

x y

20

15

5

10

20

x y

2

4

8

6

120

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6. y varies inversely with x. If y = 40 when x = 16, find x when y = -5.

7. y varies inversely with x. If y = 7 when x = -4, find y when x = 5. 8. y varies inversely with x. If x = - 4 when y = 3/2, find y when x = 5.

Graph the following functions.

9. 𝑦 =2

𝑥

10. 𝑦 = −3

𝑥

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Graphing Rational Functions

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Rewriting Rational Functions

Use synthetic division to rewrite each equation in the form 𝑓(𝑥) =𝑎

𝑥−ℎ+ 𝑘. Then state the

domain, range and its asymptotes

1. 𝑦 =𝑥+2

𝑥−1

2. 𝑦 =𝑥

𝑥−6

3. 𝑦 =4𝑥−3

𝑥+8

4. 𝑦 =3𝑥−2

𝑥+2

Factor each of the following completely

5. 𝑓(𝑥) = 𝑥2 + 7𝑥 + 12

6. 𝑦 = 2𝑥2 − 𝑥 − 1

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Simplifying Rational Functions

1. Write each of the following ratios in simplest form.

(a) 8

2

5

20

x

x (b)

3

12

12

8

y

y

(c)

10 2

4 5

6

15

x y

x y (d)

3 7

6 10

24

12

x y

x y

2. Which of the following is equivalent to the expression 6 4

2 6

4

12

x y

x y? (Multiple choice)

(1) 4

23

x

y (2)

3

2

3x

y (3)

2

3

3y

x (4)

3

23

x

y

3. Simplify each of the following rational expressions.

(a) 2 25

4 20

x

x

(b)

2

2

11 24

9

x x

x

(c)

2

2

4 1

5 10

x

x x

(d) 2

2

9 4

3 4 4

x

x x

(e)

2

2

7 42

2 48

x x

x x

(f)

2

2

2 3 5

25 4

x x

x

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4. Which of the following is equivalent to the fraction 2

2

9 18

15 5

x x

x x

?

(1) 3

5

x

x

(2)

6

5

x

x

(3)

6

5

x

x

(4)

6

5

x

x

5. The rational expression 2

2

2 7 6

4

x x

x

can be equivalently rewritten as

(1) 2 3

2

x

x

(2)

2 3

2

x

x

(3)

2 1

6

x

x

(4)

3 2

2

x

x

6. Written in simplest form, the fraction 2 2

5 5

y x

x y

is equal to

(1) 5 5y x (2)

5

x y (3)

5

y x (4)

5

x y

7. When we simplify an algebraic fraction, we are producing equivalent expressions for most

values of x. Consider the expressions 2 4 2

and 2 4 2

x x

x

.

(c) Clearly these two expressions are not

equivalent for an input value of 2x

. Explain why.

(a) Show by simplifying the first expression that these two are equivalent.

(b) Use your calculator to fill out the value for both of these expressions to show their equivalence.

x

2 4

2 4

x

x

2

2

x

0

1

2

3

4

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Multiplying and Dividing Rational Functions

1. Express each of the following products in simplest form.

(a) 4

8 2

12 15

5 30

x y

y x (b)

2 3

9 6

14 10

15 21

a b

b a (c)

3 7 2

5 2 3

4 3 30

9 10 8

x y z

z x y

2. Write each of the following products in simplest form.

(a) 2

2

9 16 8 8

12 16 3 4

x x

x x x

(b)

2 2

2 2

12 2 15

8 15 16

x x x x

x x x

(c) 2 2

3 2 2

2 7 4 12 24

8 4 6 8

x x x x

x x x x

(d)

2 2

2 2

7 8 3 4 1

1 9 1

x x x x

x x

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3. When 1024

2

x

y is divided by

2

8

36

6

x

y the result is

(1) 8 72x y (3) 8

73

x

y

(2) 5

7

3

2

x

y (4)

4

72

x

y

4. Express the result of each division problem below in simplest form.

(a) 3 2 2

2 2

5 10 5 6

10 40 12

x x x x

x x x x

(b)

2

2 2

24 18 2 2

9 16 3 7 4

x x x

x x x

(c) 2 2

4 3 3 2

6 8 4 1

3 6 2

x x x

x x x x

(d)

2 2

2

49 2 35

9 14 6 3

x x x

x x x

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Solving Rational Functions I 1. Determine if

1

2 is a solution to the following rational equation. Show or explain how you arrived at your

answer.

2. A student solved the equation 5

𝑥−4=

𝑥

𝑥−4 and got the solutions 4 and 5. Which, if either of these is an

extraneous solution? Explain your reasoning.

Solve the following equations, don’t for get to check for extraneous sol’ns

3. 2 =𝑥+2

𝑥−3 4.

1

𝑥+5=

2

7𝑥

5.2𝑥+4

5𝑥=

2

𝑥 6.

2𝑥−6

𝑥−6=

𝑥

𝑥+2

7. 𝑥−5

−3=

4

𝑥+2 8.

𝑥+1

𝑥−2=

𝑥−3

𝑥

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In numbers 11-13, a is a nonzero real number. Tell whether the statement is always true, sometimes true or never true. Explain your reasoning.

11.

12. 13. From the beginning (𝑡 = 0) of a chemical reaction, P, of one of the compounds in the mix is represented by the equation below. How many times during the reaction did the compound comprise 20% of the mix (𝑃 = 20)?

a. One time

b. Two times

c. Three times

d. The percentage cannot equal 20

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Solving Rational Functions II

Monique solves the equation −15

𝑥+5=

𝑥2+8𝑥

𝑥+5 as shown. She identified -5 and -3 as solutions

−15

𝑥 + 5=

𝑥2 + 8𝑥

𝑥 + 5

Step 1: −(𝑥 + 5)15

𝑥+5= (𝑥 + 5)

𝑥2+8𝑥

𝑥+5

Step 2: −15 = 𝑥2 + 8𝑥

Step 3: 0 = 𝑥2 + 8𝑥 + 15

Step 4: 0 = (𝑥 + 5)(𝑥 + 3)

Step 5: 𝑥 + 5 = 0; 𝑥 + 3 = 0

Step 6: 𝑥 = −3 𝑎𝑛𝑑 𝑥 = −5

Monique’s teacher tells her that she made an error somewhere

1. In which step did Monique first make her error

o Step 1

o Step 2

o Step 3

o Step 4

o Step 5

o Step 6

2. Why is the step selected in part A incorrect?

o Monique did not factor correctly

o Monique did not check for extraneous solutions. There is no solution to the equation

o The denominator on the left side of the equation cannot be divided out because of the negative

sign.

o Monique should have distributed the binomial to the terms in the numerator on the right side of

the equation.

o Monique should not have multiplied both sides by the denominator because the denominator

has a variable.

3. State any possible extraneous solutions to the following: 𝑥 + 2

𝑥 − 5=

12

𝑥2 − 7𝑥 + 10

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4. From the beginning (𝑡 = 0) of a chemical reaction, P, of one of the compounds in the mix is

represented by the equation below. How many times during the reaction did the compound

comprise 5% of the mix (𝑃 = 5)?

a. One time

b. Two times

c. Three times

d. The percentage cannot equal 5

5. Select the two solutions of the equation below.

o 4−√32

2

o -4

o -2

o -1

o 0

o 1

o 2

o 4

o 4+√32

2

6. Solve the following rational equations

a) 15 + 𝑥

22 + 𝑥= 0.5

b) 12 + 𝑥

16 + 𝑥= 0.75

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Inverse of Rational Functions

.

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