4.7 Review

​Unit 4

Rational Expressions and Equations

​When multilying rational expressions, factor everything you can, cancel out common factors in numerator and denominator and then multiply across

To find restricted values:

make any denominator with a variable in it equal to zero and solve. This is easier when the denominators are factored

Multiple Choice

Complete Step 1:  FACTOR $$\frac{x^2\ -5x\ +\ 6}{x^2-4}\cdot\frac{\left(x\ +6\right)\left(x\ +\ 5\right)}{x^2+2x\ -15}$$  

• $$\frac{\left(x\ -\ 3\right)\left(x\ -\ 2\right)}{\left(x\ +\ 2\right)\left(x\ -\ 2\right)}\cdot\frac{\left(x\ +\ 6\right)\left(x\ +\ 5\right)}{\left(x\ +\ 5\right)\left(x\ -\ 3\right)}$$  

• $$\frac{\left(x\ +\ 6\right)\left(x\ -\ 1\right)}{\left(x\ +\ 2\right)\left(x\ -\ 2\right)}\cdot\frac{\left(x\ +\ 6\right)\left(x\ +\ 5\right)}{\left(x\ -5\right)\left(x\ -3\right)}$$  

Multiple Choice

Complete Step 2: State the Restrictions $$\frac{\left(x-3\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\cdot\frac{\left(x+6\right)\left(x+5\right)}{\left(x+5\right)\left(x-3\right)}$$  

• $$x\ne-2,\ 2,\ -5,\ 3$$  

• $$x\ne2,\ -2,\ 5,\ -3$$  

• $$x\ne3,\ 2,\ -6,\ -5$$  

• $$x\ne-3,\ 2,\ -6,\ -5$$  

Multiple Choice

Complete Step 3:  CANCEL $$\frac{\left(x-3\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\cdot\frac{\left(x+6\right)\left(x+5\right)}{\left(x+5\right)\left(x-3\right)}$$  

• $$\frac{x+6}{x+2}$$  

• $$\frac{x+2}{x+6}$$  

• $$\frac{\left(x-3\right)}{\left(x+2\right)}\cdot\frac{\left(x+6\right)}{\left(x-3\right)}$$  

​When dividng rational expressions, switch it to multiplication times the reciprocal of the second fraction.

Then treat it like a multiplication problem

restricted values for divsion or rational expressions

Remember that since you are going to take the reciprocal of the fraction you divide by, both its numerator and denominator will need to be checked for restricted values.

Multiple Choice

Complete Step 1: we do not divide by frations, instead we chooswe $$\frac{3\left(x-9\right)}{x^2-36}\div\frac{x-9}{x^2-x-30}$$  

• $$\frac{x^2-36}{3\left(x-9\right)}\cdot\frac{x^2-x-30}{x-9}$$  

• $$\frac{3\left(x-9\right)}{x^2-36}\cdot\frac{x^2-x-30}{x-9}$$  

Multiple Choice

Complete Step 2:  FACTOR $$\frac{3\left(x-9\right)}{x^2-36}\cdot\frac{x^2-x-30}{x-9}$$  

• $$\frac{3\left(x-9\right)}{\left(x+6\right)\left(x+6\right)}\cdot\frac{\left(x-5\right)\left(x-6\right)}{x-9}$$  

• $$\frac{3\left(x-9\right)}{\left(x+6\right)\left(x-6\right)}\cdot\frac{\left(x-6\right)\left(x\ +5\right)}{x-9}$$  

Multiple Choice

Complete Step 3:  State Restrictions (recall this was originally a division question)

$$\frac{3\left(x-9\right)}{\left(x+6\right)\left(x-6\right)}\cdot\frac{\left(x-6\right)\left(x+5\right)}{x-9}$$  

• $$x\ne-6,\ 6,\ 9,\ -5$$  

• $$\frac{3\left(x-9\right)}{\left(x+6\right)\left(x-6\right)}\cdot\frac{\left(x-6\right)\left(x\ +5\right)}{x-9}$$  

• $$x\ne9,\ -6,\ -5$$  

• $$x\ne-9,\ -6,\ 5$$  

Multiple Choice

Complete Step 4:  CANCEL $$\frac{3\left(x-9\right)}{\left(x+6\right)\left(x-6\right)}\cdot\frac{\left(x-6\right)\left(x+5\right)}{x-9}$$  

• $$\frac{3\left(x+5\right)}{x+6}$$  

• $$\frac{x+5}{x+6}$$  

• $$\frac{3}{\left(x+6\right)\left(x+5\right)}$$  

Multiple Choice

Subtract $$\frac{6}{x-1}-\frac{5x}{4}$$  

• $$\frac{11x}{4\left(x-1\right)}$$  

• $$\frac{1x}{4\left(x-1\right)}$$  

• $$\frac{6-5x}{4\left(x-1\right)}$$  

• $$\frac{-5x^2+5x+24}{4\left(x-1\right)}$$  

Multiple Choice

Add $$\frac{2}{x-3}+\frac{3}{x+2}$$  

• $$\frac{5x-5}{\left(x-3\right)\left(x+2\right)}$$  

• $$\frac{2x-1}{\left(x-3\right)\left(x+2\right)}$$  

• $$\frac{5}{x-1}$$  

• $$\frac{1-2x}{\left(x-3\right)\left(x+2\right)}$$  

Multiple Select

Solve      $$\frac{3c}{c^2-4}+\frac{1}{c-2}=\frac{2}{c+2}$$  

• -3

• 3

• 2

• no solution

Multiple Select

Solve for x.

$$\frac{4}{x^2-8x+12}=\frac{x}{x-2}+\frac{1}{x-6}$$  

• -1

• 2

• 6

• No solution

Multiple Choice

Solve for x:
hint: when it is a

fraction = fraction

cross multiply

• x = 0 , 4
• x = 1 , 3
• x = -4 , 0
• x = 5 , 2

Multiple Choice

Solve

• 20 min
• 40 min
• 1 hour
• 2 hour

Multiple Choice

After a heavy snowfall, Brian can shovel all of the driveway in 30 minutes. If his younger brother Allen helps, the job takes only 20 minutes. How long would it take Allen to do the job by himself?

• 60 minutes
• 45 minutes
• 50 minutes
• 555 minutes

4.7 Review

Finding Horizontal Asymptotes

There are 3 situations you need to worry about.

1. degree is bigger on the bottom,

HA is y = 0

​

2. degrees are the same,

divide leading coefficients

​

3. degree is bigger on top,

there is no HA

Multiple Choice

Find the horizontal asymptote.

• None

• y=-2

• y=2

• y=0

Multiple Choice

Find the horizontal asymptote.

• $$y=-2$$

• none

• $$y=0$$

• $$y=\frac{1}{2}$$

Multiple Choice

• y = 0

• y = 1/2

• y = 2

• None

Multiple Choice

Find the horizontal asymptote.

• None

• $$y=0$$  

• $$y=\frac{-1}{3}$$  

• $$y=-3$$  

Points of Discontinuity: Holes and VA

1. factor both numerator and denominator

2. any factors that cancel out that could cause the denominator to equal zero will create a hole

3. any remaining factors in the denominator that can cause the denominator to = 0 will create vertical asymptotes.

Multiple Choice

What is an asymptote?

• an imaginary line that your function never touches

• a part of your function

• a point on your graph

• a type of fruit

Multiple Choice

What causes a hole in a graph

• any squared factors

• factor in denominator that cancels out

• factor in numerator = 0

• a factor that cancels out = 0

Multiple Choice

What causes a vertical asymptote in a rational graph

• any squared factors

• factor = 0 in denominator after holes have been identified

• factor in numerator = 0

• a factor that cancels out = 0

Multiple Choice

What is the Vertical Asymptotes?

• x= -5

• x= 5

• x= 6

• x= -6

Multiple Choice

What is the hole?

• x= 4

• x= -4

• x= 5

• x= -5

Multiple Choice

What are the Vertical Asymptotes of...

• x = 0, -5

• x=0, 5

• x = 2, -7

• x = -2, 7

Multiple Choice

What are the domain?

• x > 5

• $$x\ne5$$

• x < 5

Multiple Choice

What is the domain?

• $$x\ne1$$

• $$x\ne1,\ 3$$  

• $$x\ne-1,\ \ -3$$  

• $$x\ne3$$

​Points of discontinuity are holes and vertical aysmptotes.

Factor the denominator and try to answer the question.

​Points of discontinuity are holes and vertical aysmptotes.

Factor the denominator and try to answer the question.


the denominator will be

(x + 6)(x - 1)

hole will be at x = 1
VA will be at x = -6

pts of discontinuity are
x = 1, x = -6

Multiple Choice

Identify the transformations & Asymptotes for $$f\left(x\right)=-\frac{1}{x+3}$$  

• Move Right 3 and Up 0

  VA: x = 3

  HA: y = 0

• Move Left 3 and Up 0

  VA: x = -3

  HA: y = 0

• Move Right 3 and Down 1

  VA: x = 3

  HA: y = -1

• Move Left 3 and Down 1

  VA: x = -3

  HA: y = -1

Multiple Choice

What are the points of discontinuity?

(holes or vertical asymptotes)

• x = {5, -5}

• x = { -5, 4}

• x= {1, 5, 4}

• x = {5, -4}

Multiple Choice

What are the points of discontinuity?

(Holes or vertical asymptotes)

• x = { 5, 2}

• x = {5, 0}

• x = {-5, 5}

• x = {-5, -2}

Multiple Choice

Referring to f(x)=  $$\frac{1}{x}$$  how does f(x) =  $$\frac{1}{x+3}$$  translate?. 

• Shifts right 3

• Shifts up three

• Shifts left three

• Shifts down three

Multiple Choice

Referring to f(x) =  $$\frac{1}{x}$$  how does f(x) =   $$\frac{1}{x-4}+1$$ Translate. 

• Shifts left 4 units and up 1 unit.

• Shifts left 4 units and down 1 unit.
  

• Shifts right 4 units and up 1 unit.
  

• Shifts right 4 units and down 1 unit.

Multiple Choice

Identify the transformations & Asymptotes for $$f\left(x\right)=\frac{3}{x-2}+2$$  

• Move Right 2 and Up 2

  VA: x = 2

  HA: y = 2

• Move Left 2 and Up 3

  VA: x = -2

  HA: y = 3

• Move Right 2 and Down 2

  VA: x = 2

  HA: y = -2

• Move Left 2 and Down 3

  VA: x = -2

  HA: y = -3

Multiple Choice

What is the vertical asymptote? 

• x = 2
• x = 7
• y = 2
• y = 7

Multiple Choice

What is the horizontal asymptote? 

• x = 2
• x = 7
• y = 2
• y = 7