1. Graph Sketching: Asymptotes and Rational Functions
Unit 6 Lesson #2 Asymptotes
Graph Sketching: Asymptotes and Rational Functions
OBJECTIVES
Find limits involving infinity.
Determine the asymptotes of a function’s graph.

2. Unit 6 Lesson #2 Asymptotes
DEFINITION:
A rational function is a function f that can be described by
where P(x) and Q(x) are polynomials, with Q(x) not the zero polynomial. The domain of f consists of all inputs x for which Q(x) ≠ 0.

3. Unit 6 Lesson #2 Asymptotes
A vertical asymptote is a vertical line that a function approaches, but never reaches.
If a is a value for x that makes the denominator = 0, the line x = a may be a vertical asymptote.
The graph will never cross a vertical asymptote

4. Unit 6 Lesson #2 Asymptotes
The line x = a is a vertical asymptote if any of the following limit statements are true: ∞
𝐥𝐢𝐦 𝒙→ 𝒂 − 𝒇 𝒙 =∞
𝐥𝐢𝐦 𝒙→ 𝒂 + 𝒇 𝒙 =∞ ∞
𝐥𝐢𝐦 𝒙→ 𝒂 + 𝒇 𝒙 =∞
x = a
x = a -∞
𝐥𝐢𝐦 𝒙→ 𝒂 − 𝒇 𝒙 =−∞

5. Unit 6 Lesson #2 Asymptotes
The line x = a is a vertical asymptote if any of the following limit statements are true:
x = a
x = a ∞
𝐥𝐢𝐦 𝒙→ 𝒂 − 𝒇 𝒙 =∞
𝐥𝐢𝐦 𝒙→ 𝒂 − 𝒇 𝒙 =−∞
𝐥𝐢𝐦 𝒙→ 𝒂 + 𝒇 𝒙 =−∞ -∞ -∞
𝐥𝐢𝐦 𝒙→ 𝒂 + 𝒇 𝒙 =−∞

6. Unit 6 Lesson #2 Asymptotes
A horizontal asymptote behaves differently.
We determine what happens to the function as x approaches positive or negative infinity.
The graph of the function can cross the horizontal asymptote.
y = 0

7. Unit 6 Lesson #2 Asymptotes
𝒇 𝒙 = 𝟑 𝒙−𝟏
EXAMPLE 1: Graph the function
x-intercept
y-intercept
𝟑 𝒙−𝟏 =𝟎
𝒇 𝟎 = 𝟑 𝟎−𝟏 =−𝟑
Does not exist
𝟎, −𝟑
𝒇 𝟏 = 𝟑 𝟏−𝟏
𝐥𝐢𝐦 𝒙→𝟏 𝟑 𝒙−𝟏
does not exist
does not exist
There is a vertical asymptote at x = 1

8. Unit 6 Lesson #2 Asymptotes
𝒇 𝒙 = 𝟑 𝒙−𝟏
EXAMPLE 1: Graph the function
Now lets look at the one-sided limits around x = 1 1
lim 𝑥→ 𝑥−1 x
1.0001
1.001
1.01
1.2
1.5 2 3 y
30 000
3000
300 15 6 ∞ 1
lim 𝑥→ 1 − 3 𝑥−1 x -3 -1
0.9
0.99
0.999
0.9999 y
-0.75
-1.5
-30
-300
-3000 -∞

9. Unit 6 Lesson #2 Asymptotes
𝒇 𝒙 = 𝟑 𝒙−𝟏
EXAMPLE 1: Graph the function
lim 𝑥→∞ 3 𝑥−1 ∞
𝐥𝐢𝐦 𝒙→ 𝟏 + 𝒇 𝒙 =∞
lim 𝑥→∞ 3 𝑥 𝑥 𝑥 − 1 𝑥
y = 0
0 1−0 =0
(0, -3)
x = 1 -∞
There is a horizontal
asymptote at y = 0
𝐥𝐢𝐦 𝒙→ 𝒂 − 𝒇 𝒙 =∞

10. Unit 6 Lesson #2 Asymptotes
𝒚= 𝟐𝒙+𝟏 𝒙−𝟑
EXAMPLE 2: Graph the function
x-intercept
y-intercept
𝟐𝒙+𝟏 𝒙−𝟑 =𝟎
𝒚= 𝟐(𝟎)+𝟏 𝟎−𝟑 =− 𝟏 𝟑
𝟐𝒙+𝟏=𝟎
𝒙=− 𝟏 𝟐
− 𝟏 𝟐 , 𝟎
𝟎, − 𝟏 𝟑

11. Unit 6 Lesson #2 Asymptotes
EXAMPLE 2: Graph the function
𝒚= 𝟐(𝟑)+𝟏 𝟑−𝟑
𝐥𝐢𝐦 𝒙→𝟑 𝟐𝒙+𝟏 𝒙−𝟑
does not exist
does not exist
There is a vertical asymptote at x = 3
x = 3

12. Unit 6 Lesson #2 Asymptotes
Now lets look at the one-sided limits around x = 3
𝐥𝐢𝐦 𝒙→ 𝟑 + 𝟐𝒙+𝟏 𝒙−𝟑
𝐥𝐢𝐦 𝒙→ 𝟑 − 𝟐𝒙+𝟏 𝒙−𝟑
𝟐 𝟐.𝟗𝟗𝟗𝟗 +𝟏 𝟐.𝟗𝟗𝟗𝟗−𝟑 =−𝟔𝟗 𝟗𝟗𝟖
𝟐 𝟑.𝟎𝟎𝟎𝟏 +𝟏 𝟑.𝟎𝟎𝟎𝟏−𝟑 =𝟕𝟎 𝟎𝟎𝟐 ∞ -∞

13. Unit 6 Lesson #2 Asymptotes
𝒚= 𝟐𝒙+𝟏 𝒙−𝟑
Look for Horizontal Asymptotes ∞
𝐥𝐢𝐦 𝒙→∞ 𝟐𝒙+𝟏 𝒙−𝟑
y = 2
𝐥𝐢𝐦 𝒙→∞ 𝟐𝒙 𝒙 + 𝟏 𝒙 𝒙 𝒙 − 𝟑 𝒙
− 𝟏 𝟐 , 𝟎
𝟎, − 𝟏 𝟑
x = 3
𝟐+𝟎 𝟏+𝟎 =𝟐
There is a horizontal
asymptote at y = 2 -∞

14. EXAMPLE 3: Graph the function
𝒇 𝒙 = 𝒙 𝟐 −𝒙−𝟐 𝒙 𝟐 −𝟐𝒙−𝟑
EXAMPLE 3: Graph the function
𝒇 𝒙 = (𝒙−𝟐)(𝒙+𝟏) (𝒙−𝟑)(𝒙+𝟏)
Domain:
𝒙 ≠ −𝟏, 𝟑
x- intercept
y-intercept
𝟎 𝟐 −𝟎−𝟐 𝟎 𝟐 −𝟐 𝟎 −𝟑 = 𝟐 𝟑
(𝒙−𝟐)(𝒙+𝟏) (𝒙−𝟑)(𝒙+𝟏) =𝟎
𝟎, 𝟐 𝟑
𝒙=𝟐, 𝒙 ≠−𝟏
𝟐, 𝟎

15. Unit 6 Lesson #2 Asymptotes
𝒇 𝒙 = 𝒙 𝟐 −𝒙−𝟐 𝒙 𝟐 −𝟐𝒙−𝟑 = (𝒙−𝟐)(𝒙+𝟏) (𝒙−𝟑)(𝒙+𝟏)
EXAMPLE 3: cont
𝐥𝐢𝐦 𝒙→𝟏 (𝒙−𝟐)(𝒙+𝟏) (𝒙−𝟑)(𝒙+𝟏)
−𝟏, 𝟑 𝟒
𝟎, 𝟐 𝟑
𝐥𝐢𝐦 𝒙→𝟏 (𝒙−𝟐)(𝒙+𝟏) (𝒙−𝟑)(𝒙+𝟏)
𝟐, 𝟎
−𝟏−𝟐 −𝟏−𝟑 = 𝟑 𝟒
There is a 'hole' at −𝟏, 𝟑 𝟒

16. Unit 6 Lesson #2 Asymptotes
𝒇 𝒙 = 𝒙 𝟐 −𝒙−𝟐 𝒙 𝟐 −𝟐𝒙−𝟑 = (𝒙−𝟐)(𝒙+𝟏) (𝒙−𝟑)(𝒙+𝟏)
EXAMPLE 3: cont
𝐥𝐢𝐦 𝒙→𝟑 (𝒙−𝟐)(𝒙+𝟏) (𝒙−𝟑)(𝒙+𝟏)
−𝟏, 𝟑 𝟒
𝟎, 𝟐 𝟑
𝟑−𝟐 𝟑−𝟑 = 𝟏 𝟎
𝟐, 𝟎
DNE
x = 3
There is a vertical asymptote at x = 3

17. Unit 6 Lesson #2 Asymptotes
𝒇 𝒙 = 𝒙 𝟐 −𝒙−𝟐 𝒙 𝟐 −𝟐𝒙−𝟑 = (𝒙−𝟐)(𝒙+𝟏) (𝒙−𝟑)(𝒙+𝟏)
Look at one-sided limits
around x = 3
From the left
From the right
f(2.999) = ∞
f(3.001) = ∞ ∞
x = 3 -∞

18. Unit 6 Lesson #2 Asymptotes
Is there a horizontal asymptote?
lim 𝒙→∞ 𝒙 𝟐 𝒙 𝟐 − 𝒙 𝒙 𝟐 − 𝟐 𝒙 𝟐 𝒙 𝟐 𝒙 𝟐 − 𝟐𝒙 𝒙 𝟐 − 𝟑 𝒙 𝟐
lim 𝒙→∞ 𝒙 𝟐 −𝒙−𝟐 𝒙 𝟐 −𝟐𝒙−𝟑
𝟏−𝟎−𝟎 𝟏−𝟎−𝟎 =𝟏
There is a horizontal asymptote at y = 1 ∞
𝟎, 𝟐 𝟑
y = 1
−𝟏, 𝟑 𝟒
𝟐, 𝟎
x = 3 -∞