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Domain, Range, Zeros,
Intercepts and
Asymptotes of
Rational Function
Objectives:
οƒ˜find the domain, range, zeroes
and intercepts of rational
functions
οƒ˜determine the vertical and
horizontal asymptotes of rational
function.
οƒ˜ The domain of a rational function 𝑓 π‘₯ =
𝑁(π‘₯)
𝐷(π‘₯)
is all the values of that π‘₯ will not make 𝐷(π‘₯)
equal to zero.
οƒ˜ To find the range of rational function is by
finding the domain of the inverse function.
οƒ˜ Another way to find the range of rational
function is to find the value of horizontal
asymptote.
Domain and Range of Rational Function
𝒇 𝒙 =
𝟐
𝒙 βˆ’ πŸ‘
EXAMPLE 1:
𝒇 𝒙 =
𝟐
𝒙 βˆ’ πŸ‘
𝒙 βˆ’ πŸ‘ = 𝟎
Focus on the
denominator
The domain of 𝒇(𝒙)
is the set of all real
numbers except πŸ‘.
EXAMPLE 1:
𝒙 = πŸ‘
To find the domain:
𝑫: 𝒙 𝒙 ∈ ℝ, 𝒙 β‰  πŸ‘
𝒇 𝒙 =
𝟐
𝒙 βˆ’ πŸ‘
οƒΌ Change 𝑓(π‘₯) into y
EXAMPLE 1:
To find the range:
π’š =
𝟐
𝒙 βˆ’ πŸ‘
οƒΌ Interchange the position
of x and y
𝒙 =
𝟐
π’š βˆ’ πŸ‘
οƒΌ Simplify the rational
expression
𝒙 π’š βˆ’ πŸ‘ = 𝟐
π’™π’š βˆ’ πŸ‘π’™ = 𝟐
οƒΌ Solve for y in terms of x π’™π’š = 𝟐 + πŸ‘π’™
π’™π’š
𝒙
=
𝟐 + πŸ‘π’™
𝒙
π’š =
𝟐 + πŸ‘π’™
𝒙
οƒΌ Equate the
denominator
to 0.
𝒙 = 𝟎
The range of 𝒇(𝒙) is the set
of all real numbers except 𝟎.
𝑹: π’š π’š ∈ ℝ, π’š β‰  𝟎
𝒇 𝒙 =
𝒙 βˆ’ πŸ“
𝒙 + 𝟐
EXAMPLE 2:
𝒇 𝒙 =
𝒙 βˆ’ πŸ“
𝒙 + 𝟐
𝒙 + 𝟐 = 𝟎
Focus on the
denominator
The domain of 𝒇(𝒙)
is the set of all real
numbers except βˆ’πŸ.
EXAMPLE 2:
𝒙 = βˆ’πŸ
To find the domain:
𝑫: 𝒙 𝒙 ∈ ℝ, 𝒙 β‰  βˆ’πŸ
𝒇 𝒙 =
𝒙 βˆ’ πŸ“
𝒙 + 𝟐
οƒΌ Change 𝑓(π‘₯) into y
EXAMPLE 2:
To find the range:
π’š =
𝒙 βˆ’ πŸ“
𝒙 + 𝟐
οƒΌ Interchange the position
of x and y
𝒙 =
π’š βˆ’ πŸ“
π’š + 𝟐
οƒΌ Simplify the rational
expression
𝒙 π’š + 𝟐 = π’š βˆ’ πŸ“
π’™π’š + πŸπ’™ = π’š βˆ’ πŸ“
οƒΌ Solve for y in terms of x π’™π’š βˆ’ π’š = βˆ’πŸ“ βˆ’ πŸπ’™
π’š(𝒙 βˆ’ 𝟏)
𝒙 βˆ’ 𝟏
=
βˆ’πŸ“ βˆ’ πŸπ’™
𝒙 βˆ’ 𝟏
π’š =
βˆ’πŸ“ βˆ’ πŸπ’™
𝒙 βˆ’ 𝟏
οƒΌ Equate the
denominator
to 0.
𝒙 βˆ’ 𝟏 = 𝟎
The range of 𝒇(𝒙) is
the set of all real
numbers except 𝟏.
𝑹: π’š π’š ∈ ℝ, π’š β‰  𝟏
π’š(𝒙 βˆ’ 𝟏) = βˆ’πŸ“ βˆ’ πŸπ’™
𝒙 = 𝟏
𝒇 𝒙 =
πŸ• + 𝒙
πŸπ’™ βˆ’ πŸ”
EXAMPLE 3:
𝒇 𝒙 =
πŸ• + 𝒙
πŸπ’™ βˆ’ πŸ”
πŸπ’™ βˆ’ πŸ” = 𝟎
Focus on the
denominator
The domain of 𝒇(𝒙)
is the set of all real
numbers except πŸ‘.
EXAMPLE 3:
πŸπ’™ = πŸ”
To find the domain:
𝑫: 𝒙 𝒙 ∈ ℝ, 𝒙 β‰  πŸ‘
𝒙 = πŸ‘
𝒇 𝒙 =
πŸ• + 𝒙
πŸπ’™ βˆ’ πŸ”
οƒΌ Change 𝑓(π‘₯) into y
EXAMPLE 3:
To find the range:
π’š =
πŸ• + 𝒙
πŸπ’™ βˆ’ πŸ”
οƒΌ Interchange the position
of x and y
𝒙 =
πŸ• + π’š
πŸπ’š βˆ’ πŸ”
οƒΌ Simplify the rational
expression
𝒙 πŸπ’š βˆ’ πŸ” = πŸ• + π’š
2π’™π’š βˆ’ πŸ”π’™ = πŸ• + π’š
οƒΌ Solve for y in terms of x πŸπ’™π’š βˆ’ π’š = πŸ• + πŸ”π’™
π’š(πŸπ’™ βˆ’ 𝟏)
πŸπ’™ βˆ’ 𝟏
=
πŸ• + πŸ”π’™
πŸπ’™ βˆ’ 𝟏
π’š =
βˆ’πŸ• + πŸ”π’™
πŸπ’™ βˆ’ 𝟏
οƒΌ Equate the
denominator
to 0.
πŸπ’™ βˆ’ 𝟏 = 𝟎
The range of 𝒇(𝒙) is the set of all
real numbers except
𝟏
𝟐
.
𝑹: π’š π’š ∈ ℝ, π’š β‰ 
𝟏
𝟐
π’š πŸπ’™ βˆ’ 𝟏 = πŸ• + πŸ”π’™
πŸπ’™ = 𝟏
𝒙 =
𝟏
𝟐
Vertical and
Horizontal
Asymptotes of
Rational Functions
They are the restrictions on the x
– values of a reduced rational
function. To find the restrictions,
equate the denominator to 0 and
solve for x.
Finding the Vertical Asymptotes
of Rational Functions
Let n be the degree of the numerator and m
be the degree of denominator:
β€’ If 𝒏 < π’Ž, π’š = 𝟎.
β€’ If 𝒏 = π’Ž, π’š =
𝒂
𝒃
, where 𝒂 is the leading
coefficient of the numerator and 𝒃 is the
leading coefficient of the denominator.
β€’ If 𝒏 > π’Ž , there is no horizontal
asymptote.
Finding the Horizontal Asymptotes
of Rational Functions
Find the Degree of Polynomial.
πŸ“π’™ 𝟏
π‘«π’†π’ˆπ’“π’†π’†
𝒙 βˆ’ πŸ’ 𝟏
πŸπ’™πŸ‘
βˆ’ 𝒙 βˆ’ πŸ’ πŸ‘
Find the Degree of Polynomial.
π’™πŸ
βˆ’ πŸπ’™πŸ“
βˆ’ 𝒙 πŸ“
π‘«π’†π’ˆπ’“π’†π’†
π’šπŸ
βˆ’ π’š + 𝟏 𝟐
πŸ— + πŸπ’™ βˆ’ π’™πŸ‘ πŸ‘
EXAMPLE 1:
𝒇 𝒙 =
πŸ‘
𝒙 βˆ’ πŸ“
𝒇 𝒙 =
πŸ‘
𝒙 βˆ’ πŸ“
To find the vertical
asymptote:
𝒙 βˆ’ πŸ“ = 𝟎
𝒙 = πŸ“
Focus on the
denominator
The vertical asymptote
is 𝒙 = πŸ“.
EXAMPLE 1:
𝒇 𝒙 =
πŸ‘
𝒙 βˆ’ πŸ“
To find the horizontal
asymptote:
𝒏 < π’Ž
Focus on the degree
of the numerator
and denominator
The horizontal asymptote
is π’š = 𝟎.
0
1
EXAMPLE 1:
𝒇 𝒙 =
πŸ’π’™ βˆ’ 𝟐
𝒙 + 𝟐
EXAMPLE 2:
𝒇 𝒙 =
πŸ’π’™ βˆ’ 𝟐
𝒙 + 𝟐
To find the vertical
asymptote:
𝒙 + 𝟐 = 𝟎
𝒙 = βˆ’πŸ
Focus on the
denominator
The vertical asymptote
is 𝒙 = βˆ’πŸ.
EXAMPLE 2:
𝒇 𝒙 =
πŸ’π’™ βˆ’ 𝟐
𝒙 + 𝟐
To find the horizontal
asymptote:
𝒏 = π’Ž
Focus on the degree
of the numerator
and denominator
The horizontal
asymptote is π’š = πŸ’.
1
1
EXAMPLE 2:
π’š =
𝒂
𝒃
=
πŸ’
𝟏
= πŸ’
a is the leading coefficient of 4x
b is the leading coefficient of x
𝒇 𝒙 =
πŸ‘π’™ + πŸ’
πŸπ’™πŸ + πŸ‘π’™ + 𝟏
EXAMPLE 3:
𝒇 𝒙 =
πŸ‘π’™ + πŸ’
πŸπ’™πŸ + πŸ‘π’™ + 𝟏
To find the vertical asymptote:
πŸπ’™πŸ + πŸ‘π’™ + 𝟏 = 𝟎
Focus on the
denominator
The vertical
asymptote are 𝒙 = βˆ’
𝟏
𝟐
and 𝒙 = βˆ’πŸ.
EXAMPLE 3:
πŸπ’™ + 𝟏 𝒙 + 𝟏 = 𝟎
πŸπ’™ + 𝟏 = 𝟎 𝒙 + 𝟏 = 𝟎
πŸπ’™ = βˆ’πŸ
𝒙 = βˆ’
𝟏
𝟐
𝒙 = βˆ’πŸ
1
2
EXAMPLE 3:
𝒇 𝒙 =
πŸ‘π’™ + πŸ’
πŸπ’™πŸ + πŸ‘π’™ + 𝟏
To find the horizontal
asymptote:
𝒏 < π’Ž
Focus on the degree
of the numerator
and denominator
The horizontal asymptote
is π’š = 𝟎.
𝒇 𝒙 =
πŸ’π’™πŸ‘
βˆ’ 𝟏
π’™πŸ + πŸ’π’™ βˆ’ πŸ“
EXAMPLE 4:
𝒇 𝒙 =
πŸ’π’™πŸ‘
βˆ’ 𝟏
π’™πŸ + πŸ’π’™ βˆ’ πŸ“
To find the vertical asymptote:
π’™πŸ + πŸπ’™ βˆ’ πŸ“ = 𝟎
Focus on the
denominator
The vertical
asymptote are 𝒙 = βˆ’πŸ“
and 𝒙 = 𝟏.
EXAMPLE 4:
𝒙 + πŸ“ 𝒙 βˆ’ 𝟏 = 𝟎
𝒙 + πŸ“ = 𝟎 𝒙 βˆ’ 𝟏 = 𝟎
𝒙 = βˆ’πŸ“ 𝒙 = 𝟏
𝒙 = βˆ’πŸ“
3
2
EXAMPLE 4:
To find the horizontal
asymptote:
𝒏 > π’Ž
Focus on the degree
of the numerator
and denominator
The rational function has
no horizontal asymptote.
𝒇 𝒙 =
πŸ’π’™πŸ‘
βˆ’ 𝟏
π’™πŸ + πŸ’π’™ βˆ’ πŸ“
Zeros of
Rational
Function
Finding the Zeros of Rational
Functions
Steps:
1. Factor the numerator and denominator.
2. Identify the restrictions.
3. Identify the values of x that make the
numerator equal to zero.
4. Identify the zero of f(x).
𝒇 𝒙 =
π’™πŸ
βˆ’ πŸ’π’™
𝒙 + 𝟏
EXAMPLE 1:
𝒇 𝒙 =
π’™πŸ
βˆ’ πŸ’π’™
𝒙 + 𝟏
οƒΌ Factor the numerator and
denominator
EXAMPLE 1:
𝒇(𝒙) =
𝒙(𝒙 βˆ’ πŸ’)
𝒙 + 𝟏
οƒΌ Identify the restrictions. 𝒙 + 𝟏 = 𝟎
𝒙 = βˆ’πŸ
οƒΌ Identify the values of x that
will make the numerator
equal to zero.
𝒙 𝒙 βˆ’ πŸ’ = 𝟎
𝒙 = 𝟎 𝒙 βˆ’ πŸ’ = 𝟎
𝒙 = πŸ’
οƒΌ Identify the zeroes of f(x).
𝒙 = 𝟎 𝒙 = πŸ’
𝒇 𝒙 =
(𝒙 βˆ’ πŸ’)(𝒙 + 𝟐)
(𝒙 βˆ’ πŸ‘)(𝒙 βˆ’ 𝟏)
EXAMPLE 2:
οƒΌ Factor the numerator and
denominator
EXAMPLE 2:
𝒇(𝒙) =
(𝒙 βˆ’ πŸ’)(𝒙 + 𝟐)
(𝒙 βˆ’ πŸ‘)(𝒙 βˆ’ 𝟏)
οƒΌ Identify the restrictions. 𝒙 βˆ’ πŸ‘ 𝒙 βˆ’ 𝟏 = 𝟎
𝒙 βˆ’ πŸ‘ = 𝟎
οƒΌ Identify the values of x
that will make the
numerator equal to
zero.
𝒙 βˆ’ πŸ’ 𝒙 + 𝟐 = 𝟎
𝒙 βˆ’ πŸ’ = 𝟎 𝒙 + 𝟐 = 𝟎
𝒙 = βˆ’πŸ
οƒΌ Identify the zeroes of f(x).
𝒙 = πŸ’ 𝒙 = βˆ’πŸ
𝒇 𝒙 =
(𝒙 βˆ’ πŸ’)(𝒙 + 𝟐)
(𝒙 βˆ’ πŸ‘)(𝒙 βˆ’ 𝟏)
𝒙 = πŸ‘
𝒙 βˆ’ 𝟏 = 𝟎
𝒙 = 𝟏
𝒙 = πŸ’
𝒇 𝒙 =
π’™πŸ
+ πŸ“π’™ + πŸ’
π’™πŸ βˆ’ πŸπ’™ βˆ’ πŸ‘
EXAMPLE 3:
οƒΌ Factor the numerator and
denominator
EXAMPLE 3:
𝒇(𝒙) =
(𝒙 + 𝟏)(𝒙 + πŸ’)
(𝒙 βˆ’ πŸ‘)(𝒙 + 𝟏)
οƒΌ Identify the restrictions. 𝒙 βˆ’ πŸ‘ 𝒙 + 𝟏 = 𝟎
𝒙 βˆ’ πŸ‘ = 𝟎
οƒΌ Identify the values of x
that will make the
numerator equal to
zero.
𝒙 + 𝟏 𝒙 + πŸ’ = 𝟎
𝒙 + 𝟏 = 𝟎 𝒙 + πŸ’ = 𝟎
𝒙 = βˆ’πŸ’
οƒΌ Identify the zeroes of f(x).
𝒙 = βˆ’πŸ’
𝒙 = πŸ‘
𝒙 + 𝟏 = 𝟎
𝒙 = βˆ’πŸ
𝒙 = βˆ’πŸ
𝒇 𝒙 =
π’™πŸ
+ πŸ“π’™ + πŸ’
π’™πŸ βˆ’ πŸπ’™ βˆ’ πŸ‘
Intercepts of
Rational
Functions
οƒ˜ Intercepts are x and y – coordinates of
the points at which a graph crosses the
x-axis or y-axis, respectively.
οƒ˜ y-intercept is the y-coordinate of the
point where the graph crosses the y-
axis.
οƒ˜ x-intercept is the x-coordinate of the
point where the graph crosses the x-
axis.
Note: Not all rational functions have both x and y intercepts. If the
rational function has no real solution, then it does not have intercepts.
Rule to find the Intercepts
1) To find the y-intercept, substitute 0
for x and solve for y or f(x).
2) To find the x-intercept, substitute 0
for y and solve for x.
𝒇 𝒙 =
𝒙 + πŸ’
𝒙 βˆ’ 𝟐
EXAMPLE 1:
𝒇 𝒙 =
𝒙 + πŸ’
𝒙 βˆ’ 𝟐
EXAMPLE 1:
y - intercept
𝒇(𝒙) =
𝒙 + πŸ’
𝒙 βˆ’ 𝟐
𝒙 = 𝟎
𝒇(𝒙) =
𝟎 + πŸ’
𝟎 βˆ’ 𝟐
𝒇(𝒙) =
πŸ’
βˆ’πŸ
𝒇(𝒙) = βˆ’πŸ
x - intercept
𝒇(𝒙) =
𝒙 + πŸ’
𝒙 βˆ’ 𝟐
π’š = 𝟎
𝟎 =
𝒙 + πŸ’
𝒙 βˆ’ 𝟐
𝟎 𝒙 βˆ’ 𝟐 = 𝒙 + πŸ’
𝟎 = 𝒙 + πŸ’
βˆ’πŸ’ = 𝒙
𝒇 𝒙 =
π’™πŸ
+ πŸ“π’™ + πŸ’
π’™πŸ βˆ’ πŸπ’™ βˆ’ πŸ‘
EXAMPLE 2:
EXAMPLE 1:
y - intercept
𝒙 = 𝟎
𝒇 𝒙 =
𝟎 + πŸ’
𝟎 βˆ’ πŸ‘
x - intercept
π’š = 𝟎
𝟎 = 𝒙 + πŸ’
βˆ’πŸ’ = 𝒙
𝒇 𝒙 =
π’™πŸ
+ πŸ“π’™ + πŸ’
π’™πŸ βˆ’ πŸπ’™ βˆ’ πŸ‘
𝒇 𝒙 =
(𝒙 + 𝟏)(𝒙 + πŸ’)
(𝒙 + 𝟏)(𝒙 βˆ’ πŸ‘)
𝒇 𝒙 =
𝒙 + πŸ’
𝒙 βˆ’ πŸ‘
𝒇 𝒙 = βˆ’
πŸ’
πŸ‘
𝒇 𝒙 =
𝒙 + πŸ’
𝒙 βˆ’ πŸ‘
𝟎 =
𝒙 + πŸ’
𝒙 βˆ’ πŸ‘
𝟎(𝒙 βˆ’ πŸ‘) = 𝒙 + πŸ’
Find each of the following:
a) Intercepts
1)𝑓 π‘₯ =
π‘₯
π‘₯+4
2) 𝑓 π‘₯ =
π‘₯ βˆ’7
π‘₯ βˆ’5
3) 𝑓 π‘₯ =
π‘₯2βˆ’5π‘₯βˆ’14
π‘₯2βˆ’4
Domain-Range-Intercepts-Zeros-and-Asymptotes-of-Rational-Function.pptx

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Domain-Range-Intercepts-Zeros-and-Asymptotes-of-Rational-Function.pptx

  • 1. Domain, Range, Zeros, Intercepts and Asymptotes of Rational Function
  • 2. Objectives: οƒ˜find the domain, range, zeroes and intercepts of rational functions οƒ˜determine the vertical and horizontal asymptotes of rational function.
  • 3. οƒ˜ The domain of a rational function 𝑓 π‘₯ = 𝑁(π‘₯) 𝐷(π‘₯) is all the values of that π‘₯ will not make 𝐷(π‘₯) equal to zero. οƒ˜ To find the range of rational function is by finding the domain of the inverse function. οƒ˜ Another way to find the range of rational function is to find the value of horizontal asymptote. Domain and Range of Rational Function
  • 4.
  • 5. 𝒇 𝒙 = 𝟐 𝒙 βˆ’ πŸ‘ EXAMPLE 1:
  • 6. 𝒇 𝒙 = 𝟐 𝒙 βˆ’ πŸ‘ 𝒙 βˆ’ πŸ‘ = 𝟎 Focus on the denominator The domain of 𝒇(𝒙) is the set of all real numbers except πŸ‘. EXAMPLE 1: 𝒙 = πŸ‘ To find the domain: 𝑫: 𝒙 𝒙 ∈ ℝ, 𝒙 β‰  πŸ‘
  • 7. 𝒇 𝒙 = 𝟐 𝒙 βˆ’ πŸ‘ οƒΌ Change 𝑓(π‘₯) into y EXAMPLE 1: To find the range: π’š = 𝟐 𝒙 βˆ’ πŸ‘ οƒΌ Interchange the position of x and y 𝒙 = 𝟐 π’š βˆ’ πŸ‘ οƒΌ Simplify the rational expression 𝒙 π’š βˆ’ πŸ‘ = 𝟐 π’™π’š βˆ’ πŸ‘π’™ = 𝟐 οƒΌ Solve for y in terms of x π’™π’š = 𝟐 + πŸ‘π’™ π’™π’š 𝒙 = 𝟐 + πŸ‘π’™ 𝒙 π’š = 𝟐 + πŸ‘π’™ 𝒙 οƒΌ Equate the denominator to 0. 𝒙 = 𝟎 The range of 𝒇(𝒙) is the set of all real numbers except 𝟎. 𝑹: π’š π’š ∈ ℝ, π’š β‰  𝟎
  • 8. 𝒇 𝒙 = 𝒙 βˆ’ πŸ“ 𝒙 + 𝟐 EXAMPLE 2:
  • 9. 𝒇 𝒙 = 𝒙 βˆ’ πŸ“ 𝒙 + 𝟐 𝒙 + 𝟐 = 𝟎 Focus on the denominator The domain of 𝒇(𝒙) is the set of all real numbers except βˆ’πŸ. EXAMPLE 2: 𝒙 = βˆ’πŸ To find the domain: 𝑫: 𝒙 𝒙 ∈ ℝ, 𝒙 β‰  βˆ’πŸ
  • 10. 𝒇 𝒙 = 𝒙 βˆ’ πŸ“ 𝒙 + 𝟐 οƒΌ Change 𝑓(π‘₯) into y EXAMPLE 2: To find the range: π’š = 𝒙 βˆ’ πŸ“ 𝒙 + 𝟐 οƒΌ Interchange the position of x and y 𝒙 = π’š βˆ’ πŸ“ π’š + 𝟐 οƒΌ Simplify the rational expression 𝒙 π’š + 𝟐 = π’š βˆ’ πŸ“ π’™π’š + πŸπ’™ = π’š βˆ’ πŸ“ οƒΌ Solve for y in terms of x π’™π’š βˆ’ π’š = βˆ’πŸ“ βˆ’ πŸπ’™ π’š(𝒙 βˆ’ 𝟏) 𝒙 βˆ’ 𝟏 = βˆ’πŸ“ βˆ’ πŸπ’™ 𝒙 βˆ’ 𝟏 π’š = βˆ’πŸ“ βˆ’ πŸπ’™ 𝒙 βˆ’ 𝟏 οƒΌ Equate the denominator to 0. 𝒙 βˆ’ 𝟏 = 𝟎 The range of 𝒇(𝒙) is the set of all real numbers except 𝟏. 𝑹: π’š π’š ∈ ℝ, π’š β‰  𝟏 π’š(𝒙 βˆ’ 𝟏) = βˆ’πŸ“ βˆ’ πŸπ’™ 𝒙 = 𝟏
  • 11. 𝒇 𝒙 = πŸ• + 𝒙 πŸπ’™ βˆ’ πŸ” EXAMPLE 3:
  • 12. 𝒇 𝒙 = πŸ• + 𝒙 πŸπ’™ βˆ’ πŸ” πŸπ’™ βˆ’ πŸ” = 𝟎 Focus on the denominator The domain of 𝒇(𝒙) is the set of all real numbers except πŸ‘. EXAMPLE 3: πŸπ’™ = πŸ” To find the domain: 𝑫: 𝒙 𝒙 ∈ ℝ, 𝒙 β‰  πŸ‘ 𝒙 = πŸ‘
  • 13. 𝒇 𝒙 = πŸ• + 𝒙 πŸπ’™ βˆ’ πŸ” οƒΌ Change 𝑓(π‘₯) into y EXAMPLE 3: To find the range: π’š = πŸ• + 𝒙 πŸπ’™ βˆ’ πŸ” οƒΌ Interchange the position of x and y 𝒙 = πŸ• + π’š πŸπ’š βˆ’ πŸ” οƒΌ Simplify the rational expression 𝒙 πŸπ’š βˆ’ πŸ” = πŸ• + π’š 2π’™π’š βˆ’ πŸ”π’™ = πŸ• + π’š οƒΌ Solve for y in terms of x πŸπ’™π’š βˆ’ π’š = πŸ• + πŸ”π’™ π’š(πŸπ’™ βˆ’ 𝟏) πŸπ’™ βˆ’ 𝟏 = πŸ• + πŸ”π’™ πŸπ’™ βˆ’ 𝟏 π’š = βˆ’πŸ• + πŸ”π’™ πŸπ’™ βˆ’ 𝟏 οƒΌ Equate the denominator to 0. πŸπ’™ βˆ’ 𝟏 = 𝟎 The range of 𝒇(𝒙) is the set of all real numbers except 𝟏 𝟐 . 𝑹: π’š π’š ∈ ℝ, π’š β‰  𝟏 𝟐 π’š πŸπ’™ βˆ’ 𝟏 = πŸ• + πŸ”π’™ πŸπ’™ = 𝟏 𝒙 = 𝟏 𝟐
  • 15. They are the restrictions on the x – values of a reduced rational function. To find the restrictions, equate the denominator to 0 and solve for x. Finding the Vertical Asymptotes of Rational Functions
  • 16. Let n be the degree of the numerator and m be the degree of denominator: β€’ If 𝒏 < π’Ž, π’š = 𝟎. β€’ If 𝒏 = π’Ž, π’š = 𝒂 𝒃 , where 𝒂 is the leading coefficient of the numerator and 𝒃 is the leading coefficient of the denominator. β€’ If 𝒏 > π’Ž , there is no horizontal asymptote. Finding the Horizontal Asymptotes of Rational Functions
  • 17. Find the Degree of Polynomial. πŸ“π’™ 𝟏 π‘«π’†π’ˆπ’“π’†π’† 𝒙 βˆ’ πŸ’ 𝟏 πŸπ’™πŸ‘ βˆ’ 𝒙 βˆ’ πŸ’ πŸ‘
  • 18. Find the Degree of Polynomial. π’™πŸ βˆ’ πŸπ’™πŸ“ βˆ’ 𝒙 πŸ“ π‘«π’†π’ˆπ’“π’†π’† π’šπŸ βˆ’ π’š + 𝟏 𝟐 πŸ— + πŸπ’™ βˆ’ π’™πŸ‘ πŸ‘
  • 19.
  • 20. EXAMPLE 1: 𝒇 𝒙 = πŸ‘ 𝒙 βˆ’ πŸ“
  • 21. 𝒇 𝒙 = πŸ‘ 𝒙 βˆ’ πŸ“ To find the vertical asymptote: 𝒙 βˆ’ πŸ“ = 𝟎 𝒙 = πŸ“ Focus on the denominator The vertical asymptote is 𝒙 = πŸ“. EXAMPLE 1:
  • 22. 𝒇 𝒙 = πŸ‘ 𝒙 βˆ’ πŸ“ To find the horizontal asymptote: 𝒏 < π’Ž Focus on the degree of the numerator and denominator The horizontal asymptote is π’š = 𝟎. 0 1 EXAMPLE 1:
  • 23. 𝒇 𝒙 = πŸ’π’™ βˆ’ 𝟐 𝒙 + 𝟐 EXAMPLE 2:
  • 24. 𝒇 𝒙 = πŸ’π’™ βˆ’ 𝟐 𝒙 + 𝟐 To find the vertical asymptote: 𝒙 + 𝟐 = 𝟎 𝒙 = βˆ’πŸ Focus on the denominator The vertical asymptote is 𝒙 = βˆ’πŸ. EXAMPLE 2:
  • 25. 𝒇 𝒙 = πŸ’π’™ βˆ’ 𝟐 𝒙 + 𝟐 To find the horizontal asymptote: 𝒏 = π’Ž Focus on the degree of the numerator and denominator The horizontal asymptote is π’š = πŸ’. 1 1 EXAMPLE 2: π’š = 𝒂 𝒃 = πŸ’ 𝟏 = πŸ’ a is the leading coefficient of 4x b is the leading coefficient of x
  • 26. 𝒇 𝒙 = πŸ‘π’™ + πŸ’ πŸπ’™πŸ + πŸ‘π’™ + 𝟏 EXAMPLE 3:
  • 27. 𝒇 𝒙 = πŸ‘π’™ + πŸ’ πŸπ’™πŸ + πŸ‘π’™ + 𝟏 To find the vertical asymptote: πŸπ’™πŸ + πŸ‘π’™ + 𝟏 = 𝟎 Focus on the denominator The vertical asymptote are 𝒙 = βˆ’ 𝟏 𝟐 and 𝒙 = βˆ’πŸ. EXAMPLE 3: πŸπ’™ + 𝟏 𝒙 + 𝟏 = 𝟎 πŸπ’™ + 𝟏 = 𝟎 𝒙 + 𝟏 = 𝟎 πŸπ’™ = βˆ’πŸ 𝒙 = βˆ’ 𝟏 𝟐 𝒙 = βˆ’πŸ
  • 28. 1 2 EXAMPLE 3: 𝒇 𝒙 = πŸ‘π’™ + πŸ’ πŸπ’™πŸ + πŸ‘π’™ + 𝟏 To find the horizontal asymptote: 𝒏 < π’Ž Focus on the degree of the numerator and denominator The horizontal asymptote is π’š = 𝟎.
  • 29. 𝒇 𝒙 = πŸ’π’™πŸ‘ βˆ’ 𝟏 π’™πŸ + πŸ’π’™ βˆ’ πŸ“ EXAMPLE 4:
  • 30. 𝒇 𝒙 = πŸ’π’™πŸ‘ βˆ’ 𝟏 π’™πŸ + πŸ’π’™ βˆ’ πŸ“ To find the vertical asymptote: π’™πŸ + πŸπ’™ βˆ’ πŸ“ = 𝟎 Focus on the denominator The vertical asymptote are 𝒙 = βˆ’πŸ“ and 𝒙 = 𝟏. EXAMPLE 4: 𝒙 + πŸ“ 𝒙 βˆ’ 𝟏 = 𝟎 𝒙 + πŸ“ = 𝟎 𝒙 βˆ’ 𝟏 = 𝟎 𝒙 = βˆ’πŸ“ 𝒙 = 𝟏 𝒙 = βˆ’πŸ“
  • 31. 3 2 EXAMPLE 4: To find the horizontal asymptote: 𝒏 > π’Ž Focus on the degree of the numerator and denominator The rational function has no horizontal asymptote. 𝒇 𝒙 = πŸ’π’™πŸ‘ βˆ’ 𝟏 π’™πŸ + πŸ’π’™ βˆ’ πŸ“
  • 33. Finding the Zeros of Rational Functions Steps: 1. Factor the numerator and denominator. 2. Identify the restrictions. 3. Identify the values of x that make the numerator equal to zero. 4. Identify the zero of f(x).
  • 34.
  • 35. 𝒇 𝒙 = π’™πŸ βˆ’ πŸ’π’™ 𝒙 + 𝟏 EXAMPLE 1:
  • 36. 𝒇 𝒙 = π’™πŸ βˆ’ πŸ’π’™ 𝒙 + 𝟏 οƒΌ Factor the numerator and denominator EXAMPLE 1: 𝒇(𝒙) = 𝒙(𝒙 βˆ’ πŸ’) 𝒙 + 𝟏 οƒΌ Identify the restrictions. 𝒙 + 𝟏 = 𝟎 𝒙 = βˆ’πŸ οƒΌ Identify the values of x that will make the numerator equal to zero. 𝒙 𝒙 βˆ’ πŸ’ = 𝟎 𝒙 = 𝟎 𝒙 βˆ’ πŸ’ = 𝟎 𝒙 = πŸ’ οƒΌ Identify the zeroes of f(x). 𝒙 = 𝟎 𝒙 = πŸ’
  • 37. 𝒇 𝒙 = (𝒙 βˆ’ πŸ’)(𝒙 + 𝟐) (𝒙 βˆ’ πŸ‘)(𝒙 βˆ’ 𝟏) EXAMPLE 2:
  • 38. οƒΌ Factor the numerator and denominator EXAMPLE 2: 𝒇(𝒙) = (𝒙 βˆ’ πŸ’)(𝒙 + 𝟐) (𝒙 βˆ’ πŸ‘)(𝒙 βˆ’ 𝟏) οƒΌ Identify the restrictions. 𝒙 βˆ’ πŸ‘ 𝒙 βˆ’ 𝟏 = 𝟎 𝒙 βˆ’ πŸ‘ = 𝟎 οƒΌ Identify the values of x that will make the numerator equal to zero. 𝒙 βˆ’ πŸ’ 𝒙 + 𝟐 = 𝟎 𝒙 βˆ’ πŸ’ = 𝟎 𝒙 + 𝟐 = 𝟎 𝒙 = βˆ’πŸ οƒΌ Identify the zeroes of f(x). 𝒙 = πŸ’ 𝒙 = βˆ’πŸ 𝒇 𝒙 = (𝒙 βˆ’ πŸ’)(𝒙 + 𝟐) (𝒙 βˆ’ πŸ‘)(𝒙 βˆ’ 𝟏) 𝒙 = πŸ‘ 𝒙 βˆ’ 𝟏 = 𝟎 𝒙 = 𝟏 𝒙 = πŸ’
  • 39. 𝒇 𝒙 = π’™πŸ + πŸ“π’™ + πŸ’ π’™πŸ βˆ’ πŸπ’™ βˆ’ πŸ‘ EXAMPLE 3:
  • 40. οƒΌ Factor the numerator and denominator EXAMPLE 3: 𝒇(𝒙) = (𝒙 + 𝟏)(𝒙 + πŸ’) (𝒙 βˆ’ πŸ‘)(𝒙 + 𝟏) οƒΌ Identify the restrictions. 𝒙 βˆ’ πŸ‘ 𝒙 + 𝟏 = 𝟎 𝒙 βˆ’ πŸ‘ = 𝟎 οƒΌ Identify the values of x that will make the numerator equal to zero. 𝒙 + 𝟏 𝒙 + πŸ’ = 𝟎 𝒙 + 𝟏 = 𝟎 𝒙 + πŸ’ = 𝟎 𝒙 = βˆ’πŸ’ οƒΌ Identify the zeroes of f(x). 𝒙 = βˆ’πŸ’ 𝒙 = πŸ‘ 𝒙 + 𝟏 = 𝟎 𝒙 = βˆ’πŸ 𝒙 = βˆ’πŸ 𝒇 𝒙 = π’™πŸ + πŸ“π’™ + πŸ’ π’™πŸ βˆ’ πŸπ’™ βˆ’ πŸ‘
  • 42. οƒ˜ Intercepts are x and y – coordinates of the points at which a graph crosses the x-axis or y-axis, respectively. οƒ˜ y-intercept is the y-coordinate of the point where the graph crosses the y- axis. οƒ˜ x-intercept is the x-coordinate of the point where the graph crosses the x- axis. Note: Not all rational functions have both x and y intercepts. If the rational function has no real solution, then it does not have intercepts.
  • 43. Rule to find the Intercepts 1) To find the y-intercept, substitute 0 for x and solve for y or f(x). 2) To find the x-intercept, substitute 0 for y and solve for x.
  • 44.
  • 45. 𝒇 𝒙 = 𝒙 + πŸ’ 𝒙 βˆ’ 𝟐 EXAMPLE 1:
  • 46. 𝒇 𝒙 = 𝒙 + πŸ’ 𝒙 βˆ’ 𝟐 EXAMPLE 1: y - intercept 𝒇(𝒙) = 𝒙 + πŸ’ 𝒙 βˆ’ 𝟐 𝒙 = 𝟎 𝒇(𝒙) = 𝟎 + πŸ’ 𝟎 βˆ’ 𝟐 𝒇(𝒙) = πŸ’ βˆ’πŸ 𝒇(𝒙) = βˆ’πŸ x - intercept 𝒇(𝒙) = 𝒙 + πŸ’ 𝒙 βˆ’ 𝟐 π’š = 𝟎 𝟎 = 𝒙 + πŸ’ 𝒙 βˆ’ 𝟐 𝟎 𝒙 βˆ’ 𝟐 = 𝒙 + πŸ’ 𝟎 = 𝒙 + πŸ’ βˆ’πŸ’ = 𝒙
  • 47. 𝒇 𝒙 = π’™πŸ + πŸ“π’™ + πŸ’ π’™πŸ βˆ’ πŸπ’™ βˆ’ πŸ‘ EXAMPLE 2:
  • 48. EXAMPLE 1: y - intercept 𝒙 = 𝟎 𝒇 𝒙 = 𝟎 + πŸ’ 𝟎 βˆ’ πŸ‘ x - intercept π’š = 𝟎 𝟎 = 𝒙 + πŸ’ βˆ’πŸ’ = 𝒙 𝒇 𝒙 = π’™πŸ + πŸ“π’™ + πŸ’ π’™πŸ βˆ’ πŸπ’™ βˆ’ πŸ‘ 𝒇 𝒙 = (𝒙 + 𝟏)(𝒙 + πŸ’) (𝒙 + 𝟏)(𝒙 βˆ’ πŸ‘) 𝒇 𝒙 = 𝒙 + πŸ’ 𝒙 βˆ’ πŸ‘ 𝒇 𝒙 = βˆ’ πŸ’ πŸ‘ 𝒇 𝒙 = 𝒙 + πŸ’ 𝒙 βˆ’ πŸ‘ 𝟎 = 𝒙 + πŸ’ 𝒙 βˆ’ πŸ‘ 𝟎(𝒙 βˆ’ πŸ‘) = 𝒙 + πŸ’
  • 49.
  • 50. Find each of the following: a) Intercepts 1)𝑓 π‘₯ = π‘₯ π‘₯+4 2) 𝑓 π‘₯ = π‘₯ βˆ’7 π‘₯ βˆ’5 3) 𝑓 π‘₯ = π‘₯2βˆ’5π‘₯βˆ’14 π‘₯2βˆ’4