RATIONAL FUNCTIONS TEST REVIEW Name: ______________________

Question ANSWER A ANSWER B

1 When do you cross multiply? When a fraction equals another

fraction.

When the denominators are

the same

2 Find the LCM: 3x 5 and 2x 6x 5 6x 6

3 Find the LCM: x −3 and x + 2 (x −3)(x + 2) x −6

4 Find the LCM: (x +1)2 and (x +1)(x −1) (x +1)(x −1) (x +1)2(x −1)

5 Find the LCM: 2 and x and 4x3 4x3 4x4

6 Factor: x 3 +8 (x + 2)3 (x + 2)(x 2 − 2x + 4)

7 Factor: x 2 − 4 (x +2)(x −2) (x − 2)2

8 Factor: 3x 2 +6x 3x(x + 2) Prime

9 Which expression is equivalent to x – 8 x + 8 – (8 – x)

10 What is the first step:

4x

+ 52

= − 11x

Make all the

denominators common

Multiply all numerators by the

LCD

11 What factors can cancel?

x − 6x

x Nothing

12 What factors can cancel?

3x

+ x + 4

2 =

54x

x x + 2

13 Do you need an LCD?

6x(x + 4)

i 1

x(x + 4) Yes No

14 Do you need an LCD?

8(x + 3)x

+ x 2

(x + 3) Yes No

15 Do you need an LCD?

x + 2x(x + 2)

Yes No

16 When can you clear the denominators? When solving When adding

17 Formula for joint variation? z = kxy y = k

x

18 Formula for inverse variation? y = k

x y = kx

19 Formula for direct variation? z = kxy y = kx

20 Write an equivalent expression: 5 – x2 (x2 + 5) – (x2 – 5)

21 What cannot be the value of x?

x 2 − 2x +5x −8

x ≠ 8 x ≠ −8

22 What is the constant of variation?

X 1 2 4 10 Y 20 10 5 2

2 20

23 What type of variation?

X 2 3 4 5 Y 24 36 48 60

12 48

24 You need a common denominator when you ________ to rational expressions

Subtract Divide

25 You need a common denominator when you ________ to rational expressions

Simplify Solve

26 Find the LCM: 4x 3 2x +1( )2 and 10x 5

20x 3 2x +1( )2

20x8 2x +1( )2

27 Find the LCD:

1x

and

1x + 2

x + 2 x (x + 2)

28 Simplify:

7x2x

− x +32x

6x +32x

6x −32x

29 Simplify:

x +3x

3

x +3x

30 First step in simplifying complex expressions:

Change to multiplication and

reciprocate the second factor

Write the numerator and denominator as

single fractions

End Behavior: Find the end behavior of each (without a calculator!):

31. f (x )= −3x 2 + 2

x −1 32.

As x → +∞ then f (x)→ _____As x → −∞ then f (x)→ _____

As x → +∞ then f (x)→ _____As x → −∞ then f (x)→ _____

33. 34.

As x → +∞ then f (x)→ _____As x → −∞ then f (x)→ _____

As x → +∞ then f (x)→ _____As x → −∞ then f (x)→ _____

WHAT IS THE NEXT STEP?

1. 6 − 8

x − 5 = 3

x Cross multiply. Find the LCD. Put a 1 under the 6.

2.

3x

− 12

= 12x

Cancel out the x’s. Get a common denominator. Multiply by the LCD.

6. 0 = x2 – 2x – 35 Combine x2 and x. Use the zero product property. Factor. Square root both sides.

4.

12x 2 −1

+ 3x −1

Multiply by the LCD. Cross multiply. Factor and make the denominators common.

5.

3x(x + 2)

i 10xx 3

Cross multiply. Cancel common factors. Multiply by the LCD. Get a common denominator.

3.

x −2x +2 =

3x

Cancel out the x’s. Multiply by the LCD. Cross multiply.

7. x2 – 5x – 8 = 3x – 15 Combine like terms. Move all terms to one side and leave “0” on the other side. Divide both sides by x and then combine like terms.

8. x = 4 Check for extraneous solutions. Don't forget the ± sign! Circle your answer, you are done – yay!

FILL IN THE BLANKS

1. Direct variation formula: ____________________________ 2. Inverse variation formula: ____________________________ 3. Joint variation formula: ____________________________ 4. Where does the function have a removable discontinuity: _________________ 5. How do you find a vertical asymptote when graphing: ______________________________ _____________________________________________________________________________ 6. How do you find a horizontal asymptote when graphing if the degree is larger on the denominator: ________________________________________________________________ 7. How do you find a horizontal asymptote when graphing if the degree of the numerator is the same as the degree of the denominator: _______________________________________ _____________________________________________________________________________ Factor: 7. x2 – 100

8. x3 – 125

9. 6x2 – 7x – 5 10. x3 – 64

11. 3x2 – 24 12. 27x3 – 64

Complete the operation or simplify.

13.

4x+12x2 − 2x−15

14.

15x3y5

2xy • 6xy3

9x4 y

15.

x2 +6xx2 +8x+12

• x2 + 2x−8x3−8

16.

x 2 − 9x − 22x 2 + 5x − 24

÷ x + 2x − 3

17.

2xx +3 − 5

x2+3x 18.

xx 2 − x −12

+ 512x − 48

19.

x2− 5

6+ 3x

20.

1− xx 4

x −2 − 2x 3 + x 2

Solve the equations.

21. 22.

23.

3x2 + 4x

+ 2 = 1x + 4

24.

2xx + 2

− 4 = 6x

Identify the vertical asymptote, horizontal asymptote, domain and range of each equation.

Vertical Asymptote

Horizontal Asymptote

Domain Range

25.

26.

27.

28.

Identify the vertical asymptote, horizontal asymptote, domain and range of the graph. 29.

Vertical Asymptote: _________________ Horiztonal Asymptote: _________________ Domain: _________________ Range: _________________

30. Vertical Asymptote: _________________ Horiztonal Asymptote: _________________ Domain: _________________

Range: _________________

31.

VA:

Domain:

HA:

Range:

Table

Graph

32. f (x)= 5

x−2+3

VA:

Domain:

HA:

Range:

Table

Graph

33. f (x)= 3x2 −3

x2 −9 VA:

Domain:

HA:

Range:

Table

Graph

34.

VA:

Domain:

HA:

Range:

Table

Graph

VARIATION Determine if the equation or situation represents direct, inverse, or joint variation.

35. d = kst 36. m = k

r 37. s = kpr

Translate each situation into an equation (do not solve!) 38. An equation shows m is directly proportional to n and inversely proportional to s cubed. When m = 5, then n = 160 and s = 2. 39. The weight, w, that a column of a bridge can support varies directly as the fourth power of its diameter, d, and inversely as the square of its length, l. 40. The number, n, of grapefruit that can fit into a box is inversely proportional to the cube of the diameter, d, of each grapefruit. Solve. 41. The amount of money earned at your job ( m ) varies directly with the number of hours (h) you work. The first day of work you earned $57 after working 6 hours. You are trying to save money to go to buy a new car to take to college next year. How many hours will you need to work in order to save $ 4750? 42. The force needed to keep a car from skidding on a curve varies directly as the weight of the car and the square of the speed and inversely as the radius of the curve. Suppose a 3,960 lb. force is required to keep a 2,200 lb. car traveling at 30 mph from skidding on a curve of radius 500 ft. How much force is required to keep a 3,000 lb. car traveling at 45 mph from skidding on a curve of radius 400 ft.?