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Course by: Pradnya Bhawalkar, Kim Johnston. E-mail the authors

# Discontinuities

Module by: Pradnya Bhawalkar, Kim Johnston. E-mail the authors

Summary: Finding discontinuities - vertical asymptotes and holes - of rational functions

Vertical Asymptotes occur when factors in the denominator = 0 and do not cancel with factors in the numerator

• Vertical asymptotes are vertical lines the graph approaches
• The equation of the vertical asymptote is x = (that number which makes the denominator = 0)
Holes (Removable Discontinuities) occur when the factor in the denominator = 0 and it cancels with like factors in the numerator.
• Holes are open "points" so they have an x and y coordinate
• The x-value is the number that makes the cancelled factor = 0.
• The y-value is found by substituting x into the "reduced" equation (after cancelling) like factors.

Find the vertical asymptotes and holes (if any) for the following. Don't forget that vertical asymptotes are equations and holes are points!

## Example 1

y=1x y1x

Vertical Asymptote: x=0x0

Hole: None

## Example 2

y=x(x1)x1 y x x 1 x 1

Vertical Asymptote: None

Hole: (1,1) since (x-1) was cancelled, the hole is at x=1. To find the y-coordinate, plug 1 into the reduced equation: x(x1)x1=x=1 x x 1 x 1 x 1

## Exercise 1

y=4x+3x7 y 4 x 3 x 7

### Solution

Vertical Asymptote: x=7x7 since x7=0 x70

Hole: None

## Exercise 2

y=9x32x y9x32x

### Solution

Vertical Asymptote: x=32x32 since 32x=032x0, x=32x32

Hole: None

## Exercise 3

y=7(x9)(x+1) y 7 x 9 x 1

### Solution

Vertical Asymptote: x=9 x9, x=1 x1 since x=9 x 9 and x=1 x 1

Hole: None

## Exercise 4

y=7x2x27x+3 y 7 x 2 x 2 7 x 3

### Solution

Vertical Asymptote: x=12 x 12, x=3 x3 since 2x27x+3=0 2 x 2 7 x 3 0 , (2x1)(x3)=0 2x 1 x 3 0 , 2x1=0 2x 1 0 and x3=0 x 3 0 , x=12 x 12 and x=3 x 3

Hole: None

## Exercise 5

y=2x+1x+52-1 y 2 x 1 x 5 2 -1

### Solution

Vertical Asymptote: x=5 x 5 since x+52=0 x5 2 0 , x+5=0 x5 0 , x=5 x 5

Hole: None

## Exercise 6

y=x+3x2+25-1 y x 3 x 2 25 -1

### Solution

Vertical Asymptote: None since x2+25=0 x 2 25 0 , x2=25 x2 25 , a number squared will never be negative

Hole: None

## Exercise 7

y=x7x2+2-1 y x 7 x 2 2 -1

### Solution

Vertical Asymptote: None since x2+2=0 x 2 2 0 , x2=2 x2 2 and any number squared will never be a negative number

Hole: None

## Exercise 8

y=5|x3| y 5 x3

### Solution

Vertical Asymptote: x=3x3 since |x3|=0 x3 0 , x3=0 x3 0 , x=3 x 3

Hole: None

## Exercise 9

y=4|x|4 y 4 x 4

### Solution

Vertical asymptotes: x=-4 x -4 and x=4 x 4 since |x|4=0 x 4 0 , |x|=4 x 4 , x=-4 x -4 and x=4 x 4

Hole: None

## Exercise 10

y=3(x2x6)4(x29) y 3 x 2 x 6 4 x 2 9

### Solution

Vertical Asymptote: x=-3x-3

Hole: (3, 5858) since 3(x2x6)4(x29)=3((x3)(x+2))4((x+3)(x3))=3((x+2))4((x+3)) 3 x 2 x 6 4 x 2 9 3 x 3 x 2 4 x 3 x 3 3 x 2 4 x 3 , (x-3) was cancelled, so the hole is at x=3. To find the y-coordinate, plug 3 into the reduced equation: 3((3+2))4((3+3))=3×54×6=1524=58 3 3 2 4 3 3 3 5 4 6 15 24 5 8

## Exercise 11

y=-2(x24)3(x2+4x+4) y -2 x 2 4 3 x 2 4 x 4

### Solution

-2(x24)3(x2+4x+4)=-2(x+2)(x2)3x+22=-2(x2)3(x+2) -2 x 2 4 3 x 2 4 x 4 -2 x 2 x 2 3 x 2 2 -2 x 2 3 x 2

Vertical Asymptote: x=-2x-2

Hole: None since the vertical asymptote takes care of the hole.

## Exercise 12

y=x24x+2 y x 2 4 x 2

### Solution

Vertical Asymptote: None

Hole: (-2,-4) since x24x+2=(x+2)(x2)x+2=x2 x 2 4 x 2 x 2 x 2 x 2 x 2 , (x+2) was cancelled, so the hole is at x = -2. To find the y-coordinate, plug -2 into the reduced equation: -22=-4 -2 2 -4

## Exercise 13

y=x2(x3)x23x y x 2 x 3 x 2 3 x

### Solution

Vertical Asymptotes: None

Holes: (3,3), (0,0) since x2(x3)x23x=x2(x3)x(x3)=x x 2 x 3 x 2 3 x x 2 x 3 x x 3 x , x and (x-3) were cancelled, so the holes are at x=0 and x=3. To find the y-coordinate, plug 0 and 3 into the reduced equation: 0, 3

## Exercise 14

y=x31x1 y x 3 1 x 1

### Solution

Vertical Asymptote: None

Hole: (1,3) since x31x1=(x1)(x2+x+1)x1=x2+x+1 x 3 1 x 1 x 1 x 2 x 1 x 1 x 2 x 1 , (x-1) was cancelled, so the hole is at x=1. To find the y-coordinate, plug 1 into the reduced equation: 12+1+1=3 1 2 1 1 3

## Exercise 15

y=2x23x5x21 y 2 x 2 3 x 5 x 2 1

### Solution

2x23x5x21=(2(x5))(x+1)(x+1)(x1)=2(x5)x1 2 x 2 3 x 5 x 2 1 2 x 5 x 1 x 1 x 1 2 x 5 x 1

Vertical asymptote: x=1 x 1 since x1=0 x 1 0

Hole: (-1, 72 7 2 ) Since (x+1) was cancelled, the hole is at x= -1. To find the y-coordinate, plug -1 into the reduced equation: 2×(-15)-11=72 2 -1 5 -1 1 72

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