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Section: 5.4
Objective(s): Graph rational functions
: Transform rational functions by changing parameters
Rational function
A function that can be written as the quotient of two polynomials with
integral coefficients, plus a constant term. The graph of a rational function is
π(π₯)
usually a hyperbola. π(π₯) = π(π₯) + πΆ, where π(π₯) and π(π₯) are polynomials
and C is a constant, is the generic form of a rational function.
Continuous function
A function which contains no discontinuities. In mathematical terms, a
function is continuous when βπ β π·πππππ, lim π(π₯) = π(π).
π₯βπ
Discontinuity
Parent function of the family of
rational functions.
A point for which a function is undefined, yet the same function is defined as
it approaches the discontinuity from either side. Discontinuities may be
vertical asymptotes or βholesβ.
π(π₯) =
1
π₯
The y-axis is a vertical asymptote. The x-axis is a horizontal asymptote.
The general graphing form of
the function:
π(π₯) =
π
+ π;
π₯ββ
a is the vertical stretch or compression
a < 0 is a reflection over the y-axis
h is the horizontal translation
k is the vertical translation
Describe the transformation
from the parent graph of the
following:
π(π₯) =
1
π₯+2
The graph is translated left 2 units.
π(π₯) =
1
β3
π₯
The graph is translated 3 units down.
β(π₯) =
5
+6
π₯β4
The graph is stretched by a factor of 5.
The graph is translated right 4 units.
The graph is translated up 6 units.
Identifying asymptotes of
rational functions.
Vertical asymptotes
Vertical asymptotes occur at the x-values which cause the value of the
denominator to be 0. Since division by 0 is undefined, the function is
undefined at these x-values.
π(π₯)
If π(π₯) = π(π₯) + πΆ , then the vertical asymptotes exist at the x-values where
π(π₯) = 0.
Horizontal asymptotes
π(π₯)
If π(π₯) = π(π₯) + πΆ, consider the largest exponent to appear in π(π₯) and the
largest exponent to appear in π(π₯).
If the largest exponent in π(π₯) is higher than the largest exponent in π(π₯),
there are no horizontal asymptotes.
If the largest exponent in π(π₯) is higher than the largest exponent in π(π₯),
there is a horizontal asymptote at π₯ = πΆ.
If the largest exponent in the denominator is equal to the largest exponent in
the numerator, then a horizontal asymptote exists at the following ratio:
π₯=
πππππππππππ‘ ππ π£πππππππ π€ππ‘β βππβππ π‘ ππ₯ππππππ‘ π(π₯)
+πΆ
πππππππππππ‘ ππ π£πππππππ π€ππ‘β βππβππ π‘ ππ₯ππππππ‘ ππ π(π₯)
Identify the vertical and
horizontal asymptotes:
π(π₯) =
3π₯ 3 + 4
β2π₯ 2
Identify vertical asymptotes:
Determine what values of x make the function undefined:
β2π₯ 2 = 0
π₯=0
β΄ a vertical asymptote exists at π₯ = 0.
Identify horizontal asymptotes:
Since the highest exponent in the numerator is greater than the highest
exponent in the denominator (the value of C is irrelevant), no horizontal
asymptotes exist.
π(π₯) =
5π₯ 2
2π₯ 2 β 5π₯ + 2
Identify vertical asymptotes:
Determine what values of x make the function undefined.
2π₯ 2 β 5π₯ + 2 = 0
(2π₯ β 1)(π₯ β 2) = 0
2π₯ β 1 = 0 ππ π₯ β 2 = 0
1
π₯ = ππ π₯ = 2
2
1
β΄ vertical asymptotes exist at π₯ = 2 and π₯ = 2
Identify horizontal asymptotes:
Since the highest exponent in the numerator equals the highest power in the
denominator and C = 0, a horizontal asymptote exists at the ratio of the
coefficients of those terms.
5
β΄ a horizontal asymptote exists at π¦ = 2
π(π₯) =
π₯+2
π₯ 2 β 3π₯
Identify vertical asymptotes:
Determine what values of x make the function undefined.
π₯ 2 β 3π₯ = 0
π₯(π₯ β 3) = 0
π₯ = 0 ππ π₯ β 3 = 0
π₯ = 0 ππ π₯ = 3
β΄ vertical asymptotes exist at π₯ = 0 and π₯ = 3
Identify the horizontal asymptotes:
Since the highest exponent in the numerator is less than the highest exponent
in the denominator and C = 0, a horizontal asymptote exists at π¦ = 0.
Domain
The domain of a rational function is the set of all real numbers except the xvalues of the discontinuities (asymptotes or holes).
x-intercepts (roots)(zeros)
If π(π₯) = π(π₯) , the x-intercepts (also called the βrootsβ or βzerosβ of the
π(π₯)
function) exist at the x-values where π(π₯) = 0.
If π(π₯) is a rational function, the y-intercepts are identified at π(0).
y-intercepts
βholesβ
0.
Identify the discontinuities,
vertical asymptotes, horizontal
asymptotes, zeros, holes, yintercepts and domain, for the
following:
π(π₯) =
π(π₯)
If π(π₯) = π(π₯) , a βholeβ exists at x-values where both π(π₯) = 0 and π(π₯) =
2
π₯β3
Discontinuities: 3
Vertical asymptote: π₯ = 3
Horizontal asymptote: π¦ = 0
Zeros: ππππ
Holes: None
2
y-intercepts: (0, β 3)
Domain: π
β 3
π(π₯) =
π₯ 2 β 7π₯ + 12
π₯2 β 4
Discontinuities: 2, -2
Vertical asymptote: π₯ = ±2
Horizontal asymptote: π¦ = 1
Zeros: {3,4} or (3,0) and (4,0) (note the different notations)
y-intercepts: (0, β3)
Domain: π
β ±2
π(π₯) =
π₯ 3 + π₯ 2 β 12π₯
+4
π₯ 3 β 9π₯
Vertical asymptote: π₯(π₯ β 3)(π₯ + 3) = 0
π₯ = β3 (not 3 or 0 though because they also make the
numerator 0 β΄ π₯ = 3 πππ π₯ = 0 are holes, not
asymptotes)
Holes: π₯ = 0 πππ π₯ = 3
1
Horizontal asymptote: π¦ = 1 + 4; π¦ = 5
Zeros: Zeros get really messy here because the C value is not 0 (itβs 4). So
itβs not as simple as just setting the numerator equal to 0, because that will
make the y-value 4, not 0. Weβll save finding zeros on something where C
isnβt zero until pre-calc.
y-intercepts: None (because π₯ = 0 is a hole)
Domain: π
β ±3,0
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