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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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