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Have you ever noticed a function graph that approaches a line but never touches it? Well, that line is called an asymptote !
An asymptote is a straight line that is approached by a function graph when its variable value approaches infinity or approaches a certain value. Imagine like you're walking towards a wall but never actually touching it, that's the concept of an asymptote.
There are three types of asymptotes you need to know:
A vertical asymptote is a vertical line that the graph approaches when the function value approaches positive or negative infinity.
Definition: The line x = a x = a x = a
When x x x a a a f ( x ) → ± ∞ f(x) \to \pm\infty f ( x ) → ± ∞
When x x x a a a f ( x ) → ± ∞ f(x) \to \pm\infty f ( x ) → ± ∞
How to find: For rational functions, vertical asymptotes occur when denominator = 0 \text{denominator} = 0 denominator = 0 numerator ≠ 0 \text{numerator} \neq 0 numerator = 0 Q ( x ) = 0 Q(x) = 0 Q ( x ) = 0 P ( x ) ≠ 0 P(x) \neq 0 P ( x ) = 0
A horizontal asymptote is a horizontal line that the graph approaches when x x x
Definition: The line y = b y = b y = b
lim x → ∞ f ( x ) = b \lim_{x \to \infty} f(x) = b lim x → ∞ f ( x ) = b
lim x → − ∞ f ( x ) = b \lim_{x \to -\infty} f(x) = b lim x → − ∞ f ( x ) = b
An oblique asymptote is a slanted line that the graph approaches when x x x
Definition: The line y = m x + c y = mx + c y = m x + c
Let's focus on rational functions f ( x ) = P ( x ) Q ( x ) f(x) = \frac{P(x)}{Q(x)} f ( x ) = Q ( x ) P ( x ) P ( x ) P(x) P ( x ) Q ( x ) Q(x) Q ( x )
Find the value of x x x Q ( x ) = 0 Q(x) = 0 Q ( x ) = 0
Check if P ( x ) ≠ 0 P(x) \neq 0 P ( x ) = 0
If yes, then there is a vertical asymptote at x = a x = a x = a
Example: Determine the vertical asymptote of f ( x ) = x + 3 x − 2 f(x) = \frac{x + 3}{x - 2} f ( x ) = x − 2 x + 3
Denominator is zero when: x − 2 = 0 x - 2 = 0 x − 2 = 0 x = 2 x = 2 x = 2
When x = 2 x = 2 x = 2 2 + 3 = 5 ≠ 0 2 + 3 = 5 \neq 0 2 + 3 = 5 = 0
Therefore, vertical asymptote: x = 2 x = 2 x = 2
Let's look at the function behavior around the vertical asymptote:
x x x f ( x ) = x + 3 x − 2 f(x) = \frac{x + 3}{x - 2} f ( x ) = x − 2 x + 3 Description 1.9 1.9 1.9 1.9 + 3 1.9 − 2 = 4.9 − 0.1 = − 49 \frac{1.9 + 3}{1.9 - 2} = \frac{4.9}{-0.1} = -49 1.9 − 2 1.9 + 3 = − 0.1 4.9 = − 49
Rules for rational functions:
Let the degree of numerator is m m m n n n
If m < n m < n m < n y = 0 y = 0 y = 0
If m = n m = n m = n y = a b y = \frac{a}{b} y = b a
If m > n m > n m > n
Example: Determine the horizontal asymptote of:
f ( x ) = 2 x + 1 x 2 − 4 f(x) = \frac{2x + 1}{x^2 - 4} f ( x ) = x 2 − 4 2 x + 1 Solution:
The numerator degree is = 1 = 1 = 1 = 2 = 2 = 2 1 < 2 1 < 2 1 < 2 y = 0 y = 0 y = 0
g ( x ) = 3 x 2 − 1 2 x 2 + 5 g(x) = \frac{3x^2 - 1}{2x^2 + 5} g ( x ) = 2 x 2 + 5 3 x 2 − 1 Solution:
Let's see how the function approaches the horizontal asymptote:
x x x g ( x ) = 3 x 2 − 1 2 x 2 + 5 g(x) = \frac{3x^2 - 1}{2x^2 + 5} g ( x ) = 2 x 2 + 5 3 x 2 − 1 Approaches 10 10 10 3 ( 100 ) − 1 2 ( 100 ) + 5 = 299 205 ≈ 1.459 \frac{3(100) - 1}{2(100) + 5} = \frac{299}{205} \approx 1.459 2 ( 100 ) + 5 3 ( 100 ) − 1 = 205 299 ≈ 1.459
Oblique asymptotes appear when degree of numerator = degree of denominator + 1 \text{degree of numerator} = \text{degree of denominator} + 1 degree of numerator = degree of denominator + 1
How to find: Perform polynomial division.
Example: Determine the oblique asymptote of f ( x ) = x 2 + 2 x − 1 x − 1 f(x) = \frac{x^2 + 2x - 1}{x - 1} f ( x ) = x − 1 x 2 + 2 x − 1
Solution:
Using polynomial division:
When x → ± ∞ x \to \pm\infty x → ± ∞ 2 x − 1 → 0 \frac{2}{x - 1} \to 0 x − 1 2 → 0
Therefore, oblique asymptote: y = x + 3 y = x + 3 y = x + 3
Asymptotes are very helpful in drawing function graphs. Here are the steps:
Determine all asymptotes (vertical, horizontal, or oblique)Draw asymptotes with dashed lines Find intercepts with the axesDetermine some additional points Draw the curve that approaches the asymptotes Complete Example: Draw the graph of f ( x ) = x + 1 x − 2 f(x) = \frac{x + 1}{x - 2} f ( x ) = x − 2 x + 1
Step 1 1 1
Vertical asymptote: x = 2 x = 2 x = 2 denominator = 0 \text{denominator} = 0 denominator = 0
Horizontal asymptote: y = 1 y = 1 y = 1 = 1 / 1 = 1/1 = 1/1
y y y f ( 0 ) = 0 + 1 0 − 2 = − 1 2 f(0) = \frac{0 + 1}{0 - 2} = -\frac{1}{2} f ( 0 ) = 0 − 2 0 + 1 = − 2 1 x x x 0 = x + 1 x − 2 0 = \frac{x + 1}{x - 2} 0 = x − 2 x + 1 x = − 1 x = -1 x = − 1 Step 3 3 3
When x → 2 − x \to 2^- x → 2 − f ( x ) → − ∞ f(x) \to -\infty f ( x ) → − ∞
When x → 2 + x \to 2^+ x → 2 + f ( x ) → + ∞ f(x) \to +\infty f ( x ) → + ∞
When x → ± ∞ x \to \pm\infty x → ± ∞ f ( x ) → 1 f(x) \to 1 f ( x ) → 1
Step 4 4 4
x x x f ( x ) = x + 1 x − 2 f(x) = \frac{x + 1}{x - 2} f ( x ) = x − 2 x + 1 Description − 3 -3 − 3 − 3 + 1 − 3 − 2 = − 2 − 5 = 0.4 \frac{-3 + 1}{-3 - 2} = \frac{-2}{-5} = 0.4 − 3 − 2 − 3 + 1 = − 5 − 2 = 0.4
Determine all asymptotes of f ( x ) = 2 x 2 − 3 x + 1 x − 3 f(x) = \frac{2x^2 - 3x + 1}{x - 3} f ( x ) = x − 3 2 x 2 − 3 x + 1
Determine the asymptotes of g ( x ) = x 2 − 4 x 2 − 9 g(x) = \frac{x^2 - 4}{x^2 - 9} g ( x ) = x 2 − 9 x 2 − 4
The average cost function of a product is C ( x ) = 500 + 3 x x C(x) = \frac{500 + 3x}{x} C ( x ) = x 500 + 3 x
Draw a sketch of the graph h ( x ) = x x 2 − 1 h(x) = \frac{x}{x^2 - 1} h ( x ) = x 2 − 1 x
degree of numerator ( 2 ) = degree of denominator ( 1 ) + 1 \text{degree of numerator }(2) = \text{degree of denominator }(1) + 1 degree of numerator ( 2 ) = degree of denominator ( 1 ) + 1
There is an oblique asymptote. By division: f ( x ) = 2 x + 3 + 10 x − 3 f(x) = 2x + 3 + \frac{10}{x - 3} f ( x ) = 2 x + 3 + x − 3 10
Vertical asymptote: x = 3 x = 3 x = 3
Oblique asymptote: y = 2 x + 3 y = 2x + 3 y = 2 x + 3
Vertical asymptote: x 2 − 9 = 0 x^2 - 9 = 0 x 2 − 9 = 0 x = 3 x = 3 x = 3 x = − 3 x = -3 x = − 3
But when x = 2 x = 2 x = 2 numerator = 0 \text{numerator} = 0 numerator = 0 x = 2 x = 2 x = 2
When x = − 2 x = -2 x = − 2 numerator = 0 \text{numerator} = 0 numerator = 0 x = − 2 x = -2 x = − 2
Horizontal asymptote: y = 1 y = 1 y = 1 = 1 / 1 = 1/1 = 1/1
When x → ∞ x \to \infty x → ∞ 500 x → 0 \frac{500}{x} \to 0 x 500 → 0 = 3 = 3 = 3
Vertical asymptotes: x = 1 x = 1 x = 1 x = − 1 x = -1 x = − 1
Horizontal asymptote: y = 0 y = 0 y = 0
The graph has three separate parts due to two vertical asymptotes
Value table for h ( x ) = x x 2 − 1 h(x) = \frac{x}{x^2 - 1} h ( x ) = x 2 − 1 x
x x x h ( x ) h(x) h ( x ) Description − 2 -2 − 2 − 2 4 − 1 = − 2 3 \frac{-2}{4-1} = -\frac{2}{3} 4 − 1 − 2 = − 3 2 Left part − 0.5 -0.5 − 0.5 − 0.5 0.25 − 1 = 2 3 \frac{-0.5}{0.25-1} = \frac{2}{3} 0.25 − 1 − 0.5 = 3 2 0 0 0 0 0 − 1 = 0 \frac{0}{0-1} = 0 0 − 1 0 = 0 Intercept 0.5 0.5 0.5 0.5 0.25 − 1 = − 2 3 \frac{0.5}{0.25-1} = -\frac{2}{3} 0.25 − 1 0.5 = − 3 2 2 2 2 2 4 − 1 = 2 3 \frac{2}{4-1} = \frac{2}{3} 4 − 1 2 = 3 2
1.99 1.99 1.99 4.99 − 0.01 = − 499 \frac{4.99}{-0.01} = -499 − 0.01 4.99 = − 499 Getting more negative
2.01 2.01 2.01 5.01 0.01 = 501 \frac{5.01}{0.01} = 501 0.01 5.01 = 501 Approaches + ∞ +\infty + ∞
2.1 2.1 2.1 5.1 0.1 = 51 \frac{5.1}{0.1} = 51 0.1 5.1 = 51 Getting more positive
The numerator degree is = 2 = 2 = 2 = 2 = 2 = 2 y = 3 2 y = \frac{3}{2} y = 2 3
100 100 100 29999 20005 ≈ 1.4997 \frac{29999}{20005} \approx 1.4997 20005 29999 ≈ 1.4997 1.5 1.5 1.5
1000 1000 1000 2999999 2000005 ≈ 1.49997 \frac{2999999}{2000005} \approx 1.49997 2000005 2999999 ≈ 1.49997 1.5 1.5 1.5
− 1 -1 − 1 − 1 + 1 − 1 − 2 = 0 − 3 = 0 \frac{-1 + 1}{-1 - 2} = \frac{0}{-3} = 0 − 1 − 2 − 1 + 1 = − 3 0 = 0 x x x
0 0 0 0 + 1 0 − 2 = 1 − 2 = − 0.5 \frac{0 + 1}{0 - 2} = \frac{1}{-2} = -0.5 0 − 2 0 + 1 = − 2 1 = − 0.5 y y y
1 1 1 1 + 1 1 − 2 = 2 − 1 = − 2 \frac{1 + 1}{1 - 2} = \frac{2}{-1} = -2 1 − 2 1 + 1 = − 1 2 = − 2 Approaching vertical asymptote
3 3 3 3 + 1 3 − 2 = 4 1 = 4 \frac{3 + 1}{3 - 2} = \frac{4}{1} = 4 3 − 2 3 + 1 = 1 4 = 4 Right of asymptote
5 5 5 5 + 1 5 − 2 = 6 3 = 2 \frac{5 + 1}{5 - 2} = \frac{6}{3} = 2 5 − 2 5 + 1 = 3 6 = 2 Approaching horizontal asymptote
