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Notes 3-7: Rational Functions
Warm Up
Find the zeros of each function.
1. f(x) = x2 + 2x – 15
–5, 3
2. f(x) = x2 – 49
±7
Simplify. Identify any x-values for which the
expression is undefined.
2 + 5x + 4
x
3.
x+4 x≠±1
2
x –1
x–1
2 – 8x + 12
x
4. 2
x – 12x + 36
x–2
x–6
x≠6
I. Rational Functions
A rational function is a function thatcan be written
as a ratio of two polynomials. The parent rational
1
function is 𝑓 𝑥 = .
𝑥
Like logarithmic and exponential
functions, rational functions may
have asymptotes. The function
𝑓 𝑥 =
1
𝑥
has a vertical asymptote
at
x = 0 and a horizontal asymptote
at y = 0.
II. Transformations of Rational Functions
The rational function 𝑓 𝑥 =
1
𝑥
can be transformed by
using methods similar to those used to transform other
types of functions.
𝟏
𝒙
Ex 1: Using the graph of 𝒇 𝒙 = as a guide, describe
the transformation and graph each function. Identify
the location of the vertical and horizontal asymptotes.
1
x+4
Because h = –4,
translate f 4 units left.
a. g(x) =
Vertical asymptote: x = -4
Horizontal asymptote: y = 0
b. g(x) = 1 + 1
x
Because k = 1,
translate f 1 unit up.
Vertical asymptote: x = 0
Horizontal asymptote: y = 1
III. Identifying Discontinuities for Other Types of Rational Functions:
𝟑𝒙−𝟏
For functions like f(x) =
, we can use the following strategies to
𝒙−𝟐
find discontinuities:
A. Vertical Asymptotes and Holes: To find vertical asymptotes, find the
zeros of the denominator. If there is the same zero in the numerator,
the discontinuity is a hole. If there is NOT the same zero in the
numerator, the discontinuity is a vertical asymptotes. Compare:
𝑥+1
𝑥−2
𝑓 𝑥 =
There is a vertical asymptote
at x = 2
(𝑥−2)(𝑥+1)
𝑥−2
g 𝑥 =
There
is a hole at x = 2
B. Horizontal Asymptotes
If the highest degree is in the denominator, the horizontal
asymptote is at y = 0. 𝑓 𝑥 = 𝑥 − 1 : horizontal asympotote: y = o
𝑥2 + 1
If the degrees in the numerator and denominator are the
same, the horizontal asymptote is the ratio of leading
coefficients.
3𝑥 − 1
3
𝑓 𝑥 =
2𝑥 + 1
: horizontal asympotote: y =
If the highest degree is in the numerator, there is no
horizontal asymptote.
𝑥2 − 1
𝑓 𝑥 =
𝑥+1
: no horizontal asymptote
2
Ex 1: Determine the discontinuities for the graph of
f(x) =(x2 + 7x + 6) .
f(x) = (x + 6)(x + 1)
x+ 3
x+ 3
Step 1 Vertical asymptotes/holes.
No Holes; Vertical
asymptote: x = –3
The denominator is 0 when x =
–3. (x + 3) is not in the
numerator, so it is a vertical
asymptote and not a hole.
Step 2 Horizontal asymptotes.
None: The exponent in the numerator is
the largest, so there is no horizontal
asymptote.
Ex 2: Determine the discontinuities for the graph of
x2 + x – 6
(x – 2)(x + 3)
f(x)
=
x– 2
x– 2
Step 1 Vertical asymptotes/holes.
f(x)=
No vertical asymptote.
A hole is at x = 2.
The denominator is 0
when x = 2. Since (x 2) is also in the
numerator, it is a hole,
not a vertical asymptote.
Step 2 Horizontal asymptotes.
None: The exponent in the numerator is
the largest.
Remember
This is the same as the graph of y = x + 3, except for
the hole at x = 2. On the graph, indicate the hole
with an open circle. The domain of f is
{x|x ≠ 2}.
Hole at x
=2
Ex 3: Determine the discontinuities for the graph of
x–2
x–2
f(x) =
.
f(x)
=
x2 + x
x(x + 1)
Step 1 Vertical asymptotes/holes.
No Holes; Vertical
asymptotes: x = -1
and x = 0.
The denominator is 0 when x = -1
or 0. Since neither of those factors
are also in the numerator, they are
vertical asymptotes and not a
holes.
Step 2 Horizontal asymptotes.
y = 0. The exponent in the denominator
is the largest.
Ex 4: Determine the discontinuities for the graph of
2 + 2x – 15
x
f(x) =
.
x–1
f(x) = (x – 3)(x + 5)
x–1
Step 1 Vertical asymptotes/holes.
No Holes; Vertical
asymptote: x = 1
The denominator is 0 when x =
1. (x - 1) is not in the numerator,
so it is a vertical asymptote and
not a hole.
Step 2 Horizontal asymptotes.
None: The exponent in the numerator is
the largest, so there is no horizontal
asymptote.
C. Slant Asymptotes.
The graph of a rational function has a slant
asymptote if the degree of the numerator is exactly
one more than the degree of the denominator. Long
division is used to find slant asymptotes.
The only time you have an oblique asymptote is
when there is no horizontal asymptote. You cannot
have both.
When doing long division, we do not care about the
remainder.
Example 1
• Determine if the following function has a slant
asymptote. If it does, find the equation for it.
n > d by exactly one, so no horizontal
asymptote, but there is an oblique
(slant) asymptote.
x2  2
f ( x) 
x 1
1
1
0
-2
1
1
1
1
1
The slant
asymptote is the
line y = x + 1
Example 2
• Determine if the following function has a slant
asymptote. If it does, find the equation for it.
x2  x  2
f  x 
x 1
1
n > d by exactly one, so no
horizontal asymptote, but there is
an oblique (slant) asymptote.
1
-1
-2
1
1
0
0
-2
The slant
asymptote is the
line y = x