Lesson 2 Asymptotes - Lecture notes 3 PDF

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ASYMPTOTES Learning Outcomes As a result of studying this topic, students will be able to: 1. Explain the concept of limits to determine the value and behaviour of a function. 2. Explain under what conditions asymptote(s) of a rational function exists. 3. Define the terms ‘vertical asymptote’, ‘horizontal asymptote’ and ‘slant asymptote’. 4. Identify a function’s asymptotes and sketch the graph of the function. 5. Perform long division in determining the slant asymptote.

1. Introduction An asymptote of a curve is a line such that the distance between the curve and the line approaches zero as one or both of the x or y coordinates tends to infinity. Asymptotes convey information about the behavior of curves in the large, and determining the asymptotes of a function is an important step in sketching its graph. The study of asymptotes of functions, construed in a broad sense, forms a part of the subject of asymptotic analysis.

2. Asymptotes An asymptote is a line that is approached by the curve of a function as the function approaches infinity or certain points of discontinuity. There are three types of asymptotes: vertical asymptotes, horizontal asymptotes and oblique asymptotes. Such a function will give a break in the graph when there is an attempt to divide by zero.

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a) Vertical Asymptotes If lim− 𝑓 (𝑥 ) = ±∞ or lim+ 𝑓 (𝑥 ) = ±∞ then x=C is a vertical asymptote of 𝑥→𝐶

𝑥→𝐶

the curve (graph) of y = f(x). Examples 1. Determine the vertical asymptote of 𝑓 (𝑥 ) =

2𝑥 2 +4

𝑥 2 +6𝑥+8

.

Solution For a vertical asymptote equate the denominator to zero: 𝑥 2 + 6𝑥 + 8 = 0 => (𝑥 + 4)(𝑥 + 2) = 0 => x = −4 or x = −2

Thus x = −4 and x = −2 are the required vertical asymptotes.

2. Consider the graph of (𝑥 ) =

2𝑥−6 𝑥+4

.

For a vertical asymptote equate the denominator to zero: 𝑥 + 4 = 0 => x = −4

Thus x = −4 is the required vertical asymptote. b) Horizontal Asymptotes If line y=L is a horizontal asymptote of the function y = f(x), then lim 𝑓 (𝑥 ) = 𝐿 or lim 𝑓 (𝑥 ) = 𝐿 .

𝑥→+∞

𝑥→−∞

Examples i).Determine the horizontal asymptote of =

2𝑥 2 +4

𝑥 2 +6𝑥+8

Solution

.

4 2 + 𝑥2 2+0 2𝑥 2 + 4 = lim 2 = lim =2 6 8 𝑥→∞ 𝑥 + 6𝑥 + 8 𝑥→∞ 1 + 0 + 0 + 1+𝑥 𝑥2 Thus y=2 is the horizontal asymptote. 𝑥+3

ii). Determine the horizontal asymptote of = 3𝑥−4 . Solution

2

lim 𝑥 + 3 = lim 3𝑥 4 1 + 0 = 13 13 + 3𝑥 − 4 − =3−0 𝑥→∞ 𝑥→∞

𝑥 Thus 𝑦 = is the horizontal asymptote. 1

3

Worked Examples. 𝑥+3

1. Determine any horizontal or vertical asymptote of =2𝑥−4 . Solution Horizontal asymptote:

3 1+ 𝑥 1+0 1 𝑥+3 = = lim . = lim 4 2−0 2 𝑥→∞ 𝑥→∞ 2𝑥 − 4 2−𝑥 1

Thus 𝑦 = is the horizontal asymptote. 2

Vertical asymptote: For a vertical asymptote equate the denominator to zero: 2𝑥 − 4 = 0 => x = 2

Thus 𝑥 = 2 is the vertical asymptote.

2. Determine any horizontal or vertical asymptote of 𝑦 =

𝑥 2 −1

4𝑥+1

Solution Horizontal asymptote: 1 1− 2 1−0 1 𝑥2 − 1 𝑥 = =∞ lim = lim 2 1 = 𝑥→∞ 0 + 0 𝑥→∞ 4𝑥 + 1 0 + 𝑥 𝑥2

Thus there is no horizontal asymptote. Vertical asymptote:

For a vertical asymptote equate the denominator to zero: 4𝑥 + 1 = 0 => x = −

1

4

3

.

1

Thus x = − 4 is the vertical asymptote.

3. Determine any horizontal or vertical asymptote of 𝑦 =

𝑥

√𝑥 2 −9

.

Solution Horizontal asymptote: 𝑥 𝑥 = lim lim 𝑥→∞ 𝑥→∞ √𝑥 2 − 9 √𝑥 2 (1 − 9 2 ) 𝑥 = lim

𝑥→∞

𝑥

= lim

9 𝑥 √(1 − 2 ) 𝑥

𝑥→∞

1

√(1 − 92 ) 𝑥

= ±1

Thus the horizontal asymptotes are 𝑦 = 1 and 𝑦 = −1. Vertical asymptote:

For a vertical asymptote equate the denominator to zero: √𝑥 2 − 9 = 0

=> 𝑥 2 − 9 = 0

=> x = ±3

Thus x = 3 and x = −3 are the vertical asymptotes.

4. Sketch the graph of 𝑦 =

𝑥+3

2𝑥−4

Solution Vertical asymptote: For a vertical asymptote equate the denominator to zero: 2𝑥 − 4 = 0

=> 𝑥 = 2

Thus x = 3 and x = −3 are the vertical asymptotes. Horizontal asymptote:

3 1+ 𝑥+3 𝑥+3 𝑥 = 1+0= 1 lim = lim = lim 4 2−0 2 𝑥→∞ 𝑥→∞ 2𝑥 − 4 𝑥→∞ 2𝑥 − 4 2− 𝑥 1

Thus y = is the vertical asymptote. 2

4

𝑦 = 0.5 x-axis

0

-3

-3/4

x=2

Behaviour of the curve near the asymptotes: As 𝑥 → 2+ ,

𝑦 → +∞ ;

As 𝑥 → −∞,

As 𝑥 → 2− ,

𝑦 → 0.5− ;

As 𝑥 → +∞,

𝑦 → −∞

𝑦 → 0.5+

2 5. Sketch of y  22 x , stating the x- and y-intercepts, horizontal and vertical

x 1

asymptotes. Solution Horizontal asymptote lim

𝑥→∞

2𝑥 2

𝑥 2 −1

= lim

𝑥→∞

2

1 1− 𝑥2

=

2

1−0

=2

Horizontal asymptote is therefore y=2. Vertical asymptotes 𝑥 2 − 1 = 0 => 𝑥 = ±1 Thus the vertical asymptotes are x = 1 and x = -1.

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Also x = 0 when y=0 So taking these points into consideration, the sketch should appear as:

y=2 x-axis

x=-1

x=1 y-axis

Behaviour of the curve near the asymptotes: As 𝑥 → −1+ ,

𝑦 → −∞;

As 𝑥 → 1+ ,

𝑦 → +∞;

As 𝑥 → +∞,

6. Sketch of 𝑦 =

As 𝑥 → −1− ,

As 𝑥 → 1− ,

𝑦 → 2+ ;

1

As 𝑥 → −∞,

Solution Horizontal asymptote lim

𝑥→∞ 𝑥 2 −3𝑥

= lim

𝑥→∞

𝑦 → −∞ 𝑦 → 2+

stating the x- and y-intercepts, horizontal and vertical

𝑥 2 −3𝑥

asymptotes.

1

𝑦 → +∞

1 𝑥2

3 1−𝑥

=

0

1−0

=0

Horizontal asymptote is therefore y=2. Vertical asymptotes 𝑥 2 − 3𝑥 => 𝑥 (𝑥 − 3) = 0

=> 𝑥 = 0 𝑜𝑟 𝑥 = 3

Thus the vertical asymptotes are x = 0 and x = 3. Thus both axes are asymptotes.

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Turning point: 𝑦′ =

(𝑥 2 −3𝑥).0−1(2𝑥−3) 𝑥 2 −3𝑥

When x=1.5, 𝑦 ′ =

=> (2𝑥 − 3) = 0

=0 1

1.52 −3(1.5)

=> 𝑥 = 1.5

≈ −0.43. So the turning point is (1.5, -0.43).

1.5 -0.43

x-axis

0

x=3

y-axis

Behaviour of the curve near the asymptotes: As 𝑥 → 3+ ,

As 𝑥 → 0+ ,

As 𝑥 → +∞,

𝑦 → +∞ ;

As 𝑥 → 3− ,

𝑦 → −∞;

As 𝑥 → 0− ,

𝑦 → 0+ ;

As 𝑥 → −∞,

𝑦 → −∞

𝑦 → +∞ 𝑦 → 0+

Exercise Sketch the following graphs, and of state the x- and y-intercepts, horizontal and vertical asymptotes. 𝑎) 𝑦 =

2x + 1 𝑥−1

𝑏) 𝑦 =

3 𝑥−1

𝑐)

7

𝑦=

1 1−𝑥

Answers a) Vertical asymptote is x = 1 Horizontal asymptote is y=2 When x = 0, y =1 When y = 0, x = - 0.5 Behaviour of the curve near the asymptotes: As 𝑥 → 1+ ,

𝑦 → +∞

As 𝑥 → +∞,

𝑦 → 2+

As 𝑥 → 1− ,

𝑦 → −∞

As 𝑥 → −∞,

𝑦 → 2−

𝑦=2

-

x

0 -1

The sketch:

x=

b) Vertical asymptote is x = 1 Horizontal asymptote is y=0 When x = 0, y = -3 Also y = 0 => 0 = 3. This is impossible. So the curve does not cut the x-axis. Behaviour of the curve near the asymptotes: As 𝑥 → 1+ ,

𝑦 → +∞

As 𝑥 → 1− ,

𝑦 → −∞

As 𝑥 → −∞,

𝑦 → 0−

As 𝑥 → +∞,

𝑦 → 0+

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The sketch.

0

-3

x=1

c) Vertical asymptote is x = 1 Horizontal asymptote is y = 0 When x = 0, y = 1 Also y = 0 => 0 = 1. This is impossible.

1

So the curve does not cut the x-axis. Behaviour of the curve near the asymptotes: As 𝑥 → 1+ ,

𝑦 → −∞

As 𝑥 → 1− ,

𝑦 → +∞

As 𝑥 → −∞,

𝑦→0

As 𝑥 → +∞,

0

𝑦 → 0−

x=1

+

c) Oblique Asymptotes In a rational function of y against x, when the degree of x in the numerator is greater than in the denominator, an oblique asymptote occurs in the graph of

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y against x. To obtain the oblique asymptote, start by performing long division. Examples i). Determine the oblique asymptote of 𝑦 =

𝑥 2 +3𝑥+2 𝑥−2

.

Solution 𝑥 2 + 3𝑥 + 2 12 =𝑥+5+ 𝑥−2 𝑥−2

Thus 𝑦 = 𝑥 + 5 is the oblique asymptote. Note also that 𝑥 = 2 is a vertical

asymptote. ii). Consider the graph of 𝑦 =

2𝑥3 +4𝑥2 −9 3−𝑥2

.

6x + 3 2𝑥 3 + 4𝑥 2 − 9 = −2𝑥 − 4 + 3 − 𝑥2 3 − 𝑥2

Thus 𝑦 = −2𝑥 − 4 is the oblique asymptote. Also that 𝑥 = √3 and

𝑥 = −√3 are two vertical asymptotes of the curve of y against x. iii). Find the asymptotes of 𝑦 =

𝑥 3 −x−2 x−2

.

Solution 0 𝑥3 − x − 2 = 𝑥+1+ x−2 x−2

Thus 𝑦 = 𝑥 + 1 is the oblique asymptote. Also that 𝑥 = −2 is the vertical asymptote. To get horizontal asymptote, find:

1 2 1− 2− 3 𝑥3 − x − 2 𝑥 = 1−0−0= 1 =∞ 𝑥 = lim lim 2 1 𝑥→∞ 𝑥→∞ 0 x−2 0−0 − 𝑥3 𝑥3 So there is no horizontal asymptote.

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iv). Sketch the graph of 𝑦 =

𝑥 2 +3𝑥+2 discussed 𝑥−2

in (1) above.

Solution 0 𝑥3 − x − 2 =𝑥+1+ x−2 x−2 When y = 0, the curve cuts the x-axis

5

at x = -1, and x = -2. -5

When x=0, y = -1

-2

-1

0 -1

Behaviour of the curve near the asymptotes: As 𝑥 → 2 ,

x=2

𝑦 → +∞

+

As 𝑥 → 2− ,

𝑦 → −∞

𝑦 → (𝑥 + 5)−

As 𝑥 → −∞,

𝑦 → (𝑥 + 5)+

As 𝑥 → +∞,

Exercise Sketch the following graphs, and of state the x- and y-intercepts, horizontal and vertical asymptotes. 2x2 + 1 𝑎) 𝑦 = 𝑥−1

𝑏) 𝑦 =

2x3 + 4x 2 − 9 = 3 − 𝑥2

3x2 + 𝑥 𝑥 +1

Answers a) 𝑦 = 2𝑥 + 2 +

3

𝑥−1

Vertical asymptote is x = 1 Oblique asymptote is 𝑦 = 2𝑥 + 2 For the line:

11

𝑐)

𝑦

When x = 0, y = 2 When y = 0, x = - 1

The curve does not cut the x-axis. -1

0 -1

Behaviour of the curve near the asymptotes: As 𝑥 → 1 , +

As 𝑥 → −∞,

𝑦 → +∞;

As 𝑥 → 1 , −

𝑦 → (2𝑥 + 2)− ;

As 𝑥 → +∞,

x=1

𝑦 → −∞

𝑦 → (2𝑥 + 2)+

2

b) 𝑦 = 3𝑥 − 2 + 𝑥+1 Vertical asymptote is x = -1 Oblique asymptote is 𝑦 = 3𝑥 − 2. For the line: When x = 0, y = - 2

-1

-2

When y = 0, x = 2/3

For the curve: x=-1

When y = 0, x = 0 and x = -1/3 When x = 0, y = 0 Behaviour of the curve near the asymptotes: As 𝑥 → 1+ ,

As 𝑥 → 1− ,

As 𝑥 → −∞,

2/3

0

𝑦 → +∞

𝑦 → −∞

𝑦 → (3𝑥 − 2)−

12

As 𝑥 → +∞, c) 𝑦 = −2𝑥 − 4 +

𝑦 → (3𝑥 − 2)+ 6𝑥+3 3−𝑥2

Vertical asymptotes are is 𝑥 = √3 and 𝑥 = −√3. Oblique asymptote is 𝑦 = −2𝑥 − 4. For the line: When x = 0, y = - 4

-1.73

0

1.73

-4

When y = 0, x = -2 For the curve: When x = 0, y = -3. Behaviour of the curve near the asymptotes: As 𝑥 → 1+ ,

As 𝑥 → −∞,

𝑦 → +∞;

As 𝑥 → 1− ,

𝑦 → (3𝑥 − 2)− ;

As 𝑥 → +∞,

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𝑦 → −∞

𝑦 → (3𝑥 − 2)+...