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2
x f(x)
2 0.5
1 1
0.5 2
0.1 10
0.01 100
0.001 1000
x f(x)
-2 -0.5
-1 -1
-0.5 -2
-0.1 -10
-0.01 -100
-0.001 -1000
As x → 0–, f(x) → -∞.As x → 0+, f(x) → +∞.
A rational function is a function of the form f(x) = ,
where P(x) and Q(x) are polynomials and Q(x) = 0.)(
)(
xQ
xP
f(x) =x
1
Example: f (x) = is defined for all real numbers except x = 0.x
1
3
x
x = a
as x → a –
f(x) → + ∞
x
x = a
as x → a –
f(x) → – ∞
x
x = a
as x → a +
f(x) → + ∞
x
x = a
as x → a +
f(x) → – ∞
The line x = a is a vertical asymptote of the graph of y = f(x), if and only if f(x) → + ∞ or f(x) → – ∞ as x → a + or as x → a –.
4
Example: Show that the line x = 2 is a vertical asymptote of the
graph of f(x) = .2)2(
4
x
x f(x)
1.5 16
1.9 400
1.99 40000
2 -
2.01 40000
2.1 400
2.5 16
Observe that:x→2–, f (x) → – ∞
x→2+, f (x) → + ∞
This shows that x = 2 is a vertical asymptote.
y
x100
0.5
f (x) = 2)2(
4
x
x = 2
5
Set the denominator equal to zero and solve. Solve the quadratic equation x2 + 4x – 5. (x – 1)(x + 5) = 0
Therefore, x = 1 and x = -5 are the values of x for which f may have a vertical asymptote.
As x →1– , f(x) → – ∞.
As x →1+, f(x) → + ∞.
As x → -5–, f(x) → + ∞.
As x →-5+, f(x) → – ∞.
x = -5 is a vertical asymptote.x = 1 is a vertical asymptote.
A rational function may have a vertical asymptote at
x = a for any value of a such that Q(a) = 0.)(
)(
xQ
xP
Example:Find the vertical asymptotes of the graph of f(x) = .
)54(
12 xx
6
1. Find the roots of the denominator. 0 = x2 – 4 = (x + 2)(x – 2) Possible vertical asymptotes are x = -2 and x = +2.
2. Calculate the values approaching -2 and +2 from both sides. x → -2, f(x) → -0.25; so x = -2 is not a vertical asymptote.
x → +2–, f(x) → – ∞ and x →+2+, f(x) → + ∞. So, x = 2 is a vertical asymptote.
f is undefined at -2
A hole in the graph of f at (-2, -0.25) shows a removable singularity.
x = 2
Example: Find the vertical asymptotes of the graph of f(x) = .
)4(
)2(2
x
x
x
y
(-2, -0.25)
7
y
y = b
as x → + ∞ f(x) → b –
y
y = b
as x → – ∞ f(x) → b –
y
y = b
as x → + ∞ f(x) → b +
y
y = b
as x → – ∞ f(x) → b +
The line y = b is a horizontal asymptote of the graph of y = f(x) if and only if f(x) → b + or f(x) → b – as x → + ∞ or as x → – ∞.
8
x f(x)
10 0.1
100 0.01
1000 0.001
0 –-10 -0.1
-100 -0.01
-1000 -0.001
As x becomes unbounded positively, f(x) approaches zero from above; therefore, the line y = 0 is a horizontal asymptote of the graph of f. As f(x) → – ∞, x → 0 –.
Example: Show that the line y = 0 is a horizontal asymptote of the graph of the function f(x) = .
x
1
x
y
f(x) =x
1
y = 0
9
y
x
Similarly, as x → – ∞, f(x) →1–.
Therefore, the graph of f has y = 1 as a horizontal asymptote.
Example: Determine the horizontal asymptotes of the graph of
f(x) = .)1( 2
2
x
x
Divide x2 + 1 into x2. f(x) = 1 – )1(
12 x
As x → +∞, → 0– ; so, f(x) = 1 – →1 –.)1(
12 x )1(
12 x
y = 1
10
Finding Asymptotes for Rational Functions
• If c is a real number which is a root of both P(x) and Q(x), then there is a removable singularity at c.
• If c is a root of Q(x) but not a root of P(x), then x = c is a vertical asymptote.
• If m > n, then there are no horizontal asymptotes.
• If m < n, then y = 0 is a horizontal asymptote.
• If m = n, then y = am is a horizontal asymptote.bn
Given a rational function: f (x) = P(x) am xm + lower degree terms
Q(x) bn xn + lower degree terms=
11
Factor the numerator and denominator.
The only root of the numerator is x = -1. The roots of the denominator are x = -1 and x = 2 .
Since -1 is a common root of both, there is a hole in the graph at -1 .
Since 2 is a root of the denominator but not the numerator, x = 2 will be a vertical asymptote.
Since the polynomials have the same degree, y = 3 will be a horizontal asymptote.
Example: Find all horizontal and vertical asymptotes of f (x) = .
2
3632
2
xx
xx
y = 3
x = 2
x
y
12
A slant asymptote is an asymptote which is not vertical or horizontal.
The slant asymptote is y = 2x – 5.
As x → + ∞, → 0+. 3
14
x
Example: Find the slant asymptote for f(x) = .3
12 2
x
xx
x
yx = -3
y = 2x - 5
Divide:3
12 2
x
xx3
1452
xx
Therefore as x → ∞, f(x) is more like the line y = 2x – 5.
3
14
xAs x → – ∞, → 0–.