Date: Click here to enter text. Section: 5.4 Objective(s): Graph

Date: Click here to enter text.
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
𝑝(π‘₯)
usually a hyperbola. 𝑓(π‘₯) = π‘ž(π‘₯) + 𝐢, where 𝑝(π‘₯) and π‘ž(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 βˆ€π‘Ž ∈ π·π‘œπ‘šπ‘Žπ‘–π‘›, lim 𝑓(π‘₯) = 𝑓(π‘Ž).
π‘₯β†’π‘Ž
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 β€œholes”.
𝑓(π‘₯) =
1
π‘₯
The y-axis is a vertical asymptote. The x-axis is a horizontal asymptote.
The general graphing form of
the function:
𝑓(π‘₯) =
π‘Ž
+ π‘˜;
π‘₯βˆ’β„Ž
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:
𝑔(π‘₯) =
1
π‘₯+2
The graph is translated left 2 units.
𝑓(π‘₯) =
1
βˆ’3
π‘₯
The graph is translated 3 units down.
β„Ž(π‘₯) =
5
+6
π‘₯βˆ’4
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.
𝑝(π‘₯)
If 𝑓(π‘₯) = π‘ž(π‘₯) + 𝐢 , then the vertical asymptotes exist at the x-values where
π‘ž(π‘₯) = 0.
Horizontal asymptotes
𝑝(π‘₯)
If 𝑓(π‘₯) = π‘ž(π‘₯) + 𝐢, consider the largest exponent to appear in 𝑝(π‘₯) and the
largest exponent to appear in π‘ž(π‘₯).
If the largest exponent in 𝑝(π‘₯) is higher than the largest exponent in π‘ž(π‘₯),
there are no horizontal asymptotes.
If the largest exponent in 𝑝(π‘₯) is higher than the largest exponent in π‘ž(π‘₯),
there is a horizontal asymptote at π‘₯ = 𝐢.
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‘₯=
π‘π‘œπ‘’π‘“π‘“π‘–π‘π‘–π‘’π‘›π‘‘ π‘œπ‘“ π‘£π‘Žπ‘Ÿπ‘–π‘Žπ‘π‘™π‘’ π‘€π‘–π‘‘β„Ž β„Žπ‘–π‘”β„Žπ‘’π‘ π‘‘ 𝑒π‘₯π‘π‘œπ‘›π‘’π‘›π‘‘ 𝑝(π‘₯)
+𝐢
π‘π‘œπ‘’π‘“π‘“π‘–π‘π‘–π‘’π‘›π‘‘ π‘œπ‘“ π‘£π‘Žπ‘Ÿπ‘–π‘Žπ‘π‘™π‘’ π‘€π‘–π‘‘β„Ž β„Žπ‘–π‘”β„Žπ‘’π‘ π‘‘ 𝑒π‘₯π‘π‘œπ‘›π‘’π‘›π‘‘ 𝑖𝑛 π‘ž(π‘₯)
Identify the vertical and
horizontal asymptotes:
𝑓(π‘₯) =
3π‘₯ 3 + 4
βˆ’2π‘₯ 2
Identify vertical asymptotes:
Determine what values of x make the function undefined:
βˆ’2π‘₯ 2 = 0
π‘₯=0
∴ 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.
𝑓(π‘₯) =
5π‘₯ 2
2π‘₯ 2 βˆ’ 5π‘₯ + 2
Identify vertical asymptotes:
Determine what values of x make the function undefined.
2π‘₯ 2 βˆ’ 5π‘₯ + 2 = 0
(2π‘₯ βˆ’ 1)(π‘₯ βˆ’ 2) = 0
2π‘₯ βˆ’ 1 = 0 π‘œπ‘Ÿ π‘₯ βˆ’ 2 = 0
1
π‘₯ = π‘œπ‘Ÿ π‘₯ = 2
2
1
∴ 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
∴ a horizontal asymptote exists at 𝑦 = 2
𝑓(π‘₯) =
π‘₯+2
π‘₯ 2 βˆ’ 3π‘₯
Identify vertical asymptotes:
Determine what values of x make the function undefined.
π‘₯ 2 βˆ’ 3π‘₯ = 0
π‘₯(π‘₯ βˆ’ 3) = 0
π‘₯ = 0 π‘œπ‘Ÿ π‘₯ βˆ’ 3 = 0
π‘₯ = 0 π‘œπ‘Ÿ π‘₯ = 3
∴ 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 𝑦 = 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 𝑓(π‘₯) = π‘ž(π‘₯) , the x-intercepts (also called the β€œroots” or β€œzeros” of the
𝑝(π‘₯)
function) exist at the x-values where 𝑝(π‘₯) = 0.
If 𝑓(π‘₯) is a rational function, the y-intercepts are identified at 𝑓(0).
y-intercepts
β€œholes”
0.
Identify the discontinuities,
vertical asymptotes, horizontal
asymptotes, zeros, holes, yintercepts and domain, for the
following:
𝑓(π‘₯) =
𝑝(π‘₯)
If 𝑓(π‘₯) = π‘ž(π‘₯) , a β€œhole” exists at x-values where both 𝑝(π‘₯) = 0 and π‘ž(π‘₯) =
2
π‘₯βˆ’3
Discontinuities: 3
Vertical asymptote: π‘₯ = 3
Horizontal asymptote: 𝑦 = 0
Zeros: π‘π‘œπ‘›π‘’
Holes: None
2
y-intercepts: (0, βˆ’ 3)
Domain: 𝑅 β‰  3
𝑓(π‘₯) =
π‘₯ 2 βˆ’ 7π‘₯ + 12
π‘₯2 βˆ’ 4
Discontinuities: 2, -2
Vertical asymptote: π‘₯ = ±2
Horizontal asymptote: 𝑦 = 1
Zeros: {3,4} or (3,0) and (4,0) (note the different notations)
y-intercepts: (0, βˆ’3)
Domain: 𝑅 β‰  ±2
𝑓(π‘₯) =
π‘₯ 3 + π‘₯ 2 βˆ’ 12π‘₯
+4
π‘₯ 3 βˆ’ 9π‘₯
Vertical asymptote: π‘₯(π‘₯ βˆ’ 3)(π‘₯ + 3) = 0
π‘₯ = βˆ’3 (not 3 or 0 though because they also make the
numerator 0 ∴ π‘₯ = 3 π‘Žπ‘›π‘‘ π‘₯ = 0 are holes, not
asymptotes)
Holes: π‘₯ = 0 π‘Žπ‘›π‘‘ π‘₯ = 3
1
Horizontal asymptote: 𝑦 = 1 + 4; 𝑦 = 5
Zeros: Zeros get really messy here because the C value is not 0 (it’s 4). So
it’s not as simple as just setting the numerator equal to 0, because that will
make the y-value 4, not 0. We’ll save finding zeros on something where C
isn’t zero until pre-calc.
y-intercepts: None (because π‘₯ = 0 is a hole)
Domain: 𝑅 β‰  ±3,0