Year 12 Mathematics Specialist Rational Functions
Year 12 Mathematics Specialist
Rational Functions
Year 12 | Mathematics Specialist | Rational Functions | © Department of Education WA 2020
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Year 12 Maths Specialist Rational Functions
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Contents
Signposts… ........................................................................................................... 2
Overview ................................................................................................................ 3
Lesson 1 – Reciprocal Functions ................................................................... 5
Lesson 2 – Rational functions with lower degree numerators ........... 18
Lesson 3 – Rational Functions with higher degree numerators ........ 26
Summary ............................................................................................................... 37
Solutions .............................................................................................................. 38
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Signposts
Each symbol is a sign to help you.
Here is what each one means.
Important Information
Mark and Correct your work
You write an answer or response
Use your CAS calculator
A point of emphasis
Refer to a text book
Contact your school teacher (if you can)
Check with your school about Assessment submission
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Overview
This booklet contains approximately 8 hours of work. Some students may need some additional time to complete all activities.
To guide the pace at which you work through the booklet and submit your assessments refer to the content page.
Space is provided for you to write your solutions in this PDF booklet. If you need more space, then attach a page to the page you are working on.
Answers are given to all questions: it is assumed you will use them responsibly, to maximise your learning. You should check your day to day lesson work.
Assessments
All of your assessments are provided for you separately by your school.
Assessments will be either response or investigative. Weightings for assessments are provided by your school.
Calculator
This course assumes the use of a CAS calculator. Screen displays will appear throughout the booklets to help you with your understanding of the lessons. Further support documents are available.
Textbook
You are encouraged to use a text for this course. A text will further explain some topics and can provide you with extra practice questions.
Online Support
Search for a range of online support.
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Content covered in this booklet
The syllabus content focused on in this booklet includes:
Sketching graphs
3.2.7 examine the relationship between the graph of
( )y f x and the graphs of 1
, ( )( )
y y f xf x
and ( )y f x
3.2.8 sketch the graphs of simple rational functions where the numerator and
denominator are polynomials of low degree
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Lesson 1
Reciprocal Functions
By the end of this lesson you should be able to:
Write the reciprocal of any polynomial function
Identify vertical and horizontal asymptotes of a reciprocal function
Sketch the graph of a reciprocal function.
What is a reciprocal function?
Any non-zero number x has the reciprocal 1
x
So the reciprocal of function f(x) is 1
𝑓(𝑥)
For example, if 𝑦 = 3𝑥 − 5, its reciprocal is 𝑦 =1
3𝑥−5
Finding vertical asymptotes
Soon we will be drawing sketches of reciprocal functions and the best first step is to find any
vertical asymptotes.
What is a vertical asymptote?
On the Cartesian plane a vertical asymptote represents values of 𝑥 for which the function 𝑓(𝑥) is
not defined.
For reciprocal functions we can’t let the denominator of any fraction = 0 as any number
divided by 0 is undefined. To find our vertical asymptotes simply find the values of x that
make the denominator of any fraction equal to 0.
Example: This graphs of 𝑦 =1
𝑥−2 shows the
dotted line for the asymptote 𝑥 = 2
Notice that we use a dotted line as this isn’t
actually part of the function
In fact when we draw the function the curve
cannot touch this asymptote
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Example
Find the vertical asymptotes for:
a. 𝑦 =1
𝑥−7
Solution
Make an equation of the
denominator = 0
𝑥 − 7 = 0
𝑥 = 7
Asymptote: 𝑥 ≠ 7
b. 𝑦 =1
7𝑥+2
7𝑥 + 2 = 0
7𝑥 = −2
𝑥 = −2
7
Asymptote 𝑥 ≠ −2
7
c. 𝑦 =1
𝑥2−2𝑥−8
𝑥2 − 2𝑥 − 8 = 0
(𝑥 − 4)(𝑥 + 2) = 0
𝑥 = 4 𝑜𝑟 − 2
Asymptotes:
𝑥 = −2 𝑎𝑛𝑑 𝑥 = 4
Finding Horizontal asymptotes
A horizontal asymptote represents the y-values that will never be the range (output) of the
function 𝑦 = 𝑓(𝑥).
Consider a fraction of the form 𝑐
𝑓(𝑥) where c is a non-zero constant.
The denominator is the only variable part. No matter what value that denominator takes,
the fraction cannot equal 0.
So if y = 𝑐
𝑓(𝑥) then 𝑦 ≠ 0.
Further if 𝑦 = 𝑐
𝑓(𝑥)+ 𝑘, where k is another constant, then 𝑦 ≠ 𝑘 since the fraction part of
the expression cannot equal 0.
Example
Find the horizontal asymptotes for:
a. 𝑦 =1
𝑥2+7𝑥+12
Solution
asymptote: 𝑦 = 0
b. 𝑦 =1
2𝑥+9− 4
asymptote: 𝑦 = −4
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How the curve approaches the asymptotes.
Let’s use as an example, the function
𝑦 =1
𝑥−3
Firstly, we would find its asymptotes:
𝑥 = 3 𝑎𝑛𝑑 𝑦 = 0
and plot them on a graph.
Remember that our curve cannot cross
these lines.
Next we should find a point on the
curve on either side of the vertical
asymptote.
Substituting 𝑥 = 2 we get 𝑦 =1
2−3= 1
So (2, −1) is a point on the curve.
Substituting x = 4 we get 𝑦 =1
4−3= 1
So (4,1) is another point on the curve.
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Now let us explore how the function
behaves as 𝑥 gets very large.
We write this as 𝑥 → ∞, which means
‘as 𝑥 approaches infinity’.
We could substitute ∞ into the function
and get 𝑦 =1
∞−3 but it is more practical
to consider a really large value for 𝑥,
such as 𝑥 = 1000
𝑦 =1
1000 − 3
𝑦 =1
997
This gives a very small, positive 𝑦-
value. As 𝑥 gets larger, 𝑦 will get
smaller but still be positive. So we say
as 𝑥 → ∞, 𝑦 → 0+.
That is, as 𝑥 approaches infinity, 𝑦
approaches 0 from above. We plot this
as the arrow to the right getting close to
𝑦 = 0 from above as we go to the right.
By considering large negative values of
𝑥 we find:
As 𝑥 → −∞, 𝑦 → 0−
We plot this as the arrow getting close
to 𝑦 = 0 from below as we go to the left
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So how does the curve approach the
vertical asymptote? Here we will look at
how 𝑦 behaves as 𝑥 approaches 3 from
above and from below,
i.e. as 𝑥 → 3+
Try 𝑥 = 3.1
𝑦 =1
3.1 − 3=
1
0.1= 10
Therefore, 𝑦 gets large and positive
and it will continue get larger as 𝑥 gets
closer to 3.
We represent this by drawing an
upwards vertical arrow as we approach
the vertical asymptote from the right.
Repeating this process from the left,
i.e. as 𝑥 → 3−
Try 𝑥 = 2.9
𝑦 =1
2.9−3=
1
−0.1= −10
𝑦 gets large and negative as we
approach from the left. We draw a
downwards vertical arrow as we
approach the vertical asymptote from
the left.
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We now have enough to sketch the
whole curve. It must go through both
the points we plotted and join to the
arrows that go off the sides of the
plane.
Our sketch is now complete. Notice
that the curve will get closer and closer
to the asymptotes as it goes off the
plane but it won’t actually touch them.
Drawing 𝒚 =𝟏
𝒇(𝒙) from the graph of 𝒚 = 𝒇(𝒙)
Often a quicker method is to consider the values 𝑓(𝑥) takes and draw the reciprocal from
that. Follow these general rules:
When 𝒇(𝒙) is: 𝟏
𝒇(𝒙) is
0 Undefined (vertical asymptote)
1 1
-1 -1
Positive and small Positive and big
Positive and big Positive and small
Negative and small Negative and big
Negative and big Negative and small
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Example
Below is a graph of 𝑦 = 𝑥 + 2. On the same set of axes draw a sketch of 𝑦 =1
𝑥+2.
Solution
We will plot 𝑦 =1
𝑥+2 in blue
Firstly, draw the vertical asymptote. It is located where 𝑓(𝑥) = 0.
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Example
Plot points where 𝑓(𝑥) = 1 𝑎𝑛𝑑/𝑜𝑟 − 1 as our reciprocal curve must pass through these
points.
Solution
Finally, we sketch the curve
Left of the
asymptote at
𝑥 = −2, 𝑓(𝑥)
is negative. It
is close to the
asymptote so 1
𝑓(𝑥) will be
large as it
approaches
the
asymptote.
Further from
the
asymptote
𝑓(𝑥) gets
larger so 1
𝑓(𝑥)
will get
smaller
Right of the
asymptote
𝑓(𝑥) is
positive. It
starts small
then gets
bigger so our
reciprocal will
start off very
large then get
smaller but
never cross 0
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Example
(Below is the graph of 𝑦 = 𝑥2 − 4. On the same set of axes sketch 𝑦 =1
𝑥2−4
Solution
Firstly we will sketch in our asymptotes and plot points where 𝑓(𝑥) = 1 𝑎𝑛𝑑/𝑜𝑟 − 1
The asymptotes split the plane into 3 regions: left middle and right
Next we will sketch the curve in each of the 3 regions
In the left region the 𝑓(𝑥) is positive. It’s smallest close to the asymptote and gets bigger
as we move away to the left. Our reciprocal will be biggest close to the asymptote and get
smaller further away.
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The region right of 𝑥 = 2 is simply a reflection on the left region in this case
In the central region between 𝑥 = −2 and 𝑥 = 2, 𝑓(𝑥) is negative and therefore, so is1
𝑓(𝑥).
It is smallest close to the asymptotes so 1
𝑓(𝑥) will be largest there.
𝑓(𝑥)has a local minimum at (0, −4). This means that our reciprocal function will have a
local maximum at (0, −1
4).
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Skills Development 1.1
1. For each function write the equation(s) of the vertical asymptote(s).
(a) 𝑦 =
1
2𝑥 − 11
(b) 𝑦 =
1
(𝑥 − 8)(𝑥 + 6)
(c) 𝑦 =
1
𝑥2 + 3𝑥 − 40
(d) 𝑦 =
1
𝑥2 + 4
2. For each of the graphs below, 𝑓(𝑥) is drawn. On the same set of axes, draw the graph of 1
𝑓(𝑥)
(a)
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(b)
(c)
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(d)
(e)
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3. On the axes provided draw a sketch of each of the following functions.
(a) 𝑦 =
1
4 − 𝑥
(b) 𝑦 =
1
𝑥2 − 9
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(c) 𝑦 =
1
𝑥(𝑥 − 4)(𝑥 + 6)
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Lesson 2
Rational functions with polynomials of a lower degree
in the numerator than in the denominator
By the end of this lesson you should be able to:
Identify the degree of the numerator and denominator of a rational function
Spot vertical asymptotes of a rational function
Spot 0 values of the numerator in a rational function
Find the sign of a rational function for all values of x in its natural domain
Sketch a graph of any rational function 𝑦 =𝑓(𝑥)
𝑔(𝑥) where 𝑓(𝑥) has a lower degree
than 𝑔(𝑥)
Rational functions A rational number can be expressed as
𝑎
𝑏 where a and b are integers.
A rational function can be expressed as 𝑓(𝑥)
𝑔(𝑥) where 𝑓(𝑥) and 𝑔(𝑥) are polynomial functions.
The degree of a polynomial function is equal to the highest power of x in that function.
For example, 𝑦 =𝑥+2
(𝑥2−8𝑥+9) is a rational function with a degree 1 polynomial in the
numerator and a degree 2 function in the denominator.
For this lesson we will focus on rational functions that have a lower degree polynomial in
the numerator.
The horizontal asymptote First observe how the rational function behaves as 𝑥 → ∞
Since we are studying only rational functions with a lower degree polynomial in the
numerator than the denominator in this lesson, let us use the example
𝑦 =𝑥 + 9
𝑥2 − 2𝑥 − 15
For small values of 𝑥, every part of this function is relevant.
For example, if 𝑥 = 1, 𝑦 =1+9
12−2×1−15=
10
−16
All of the terms on the numerator and the denominator had a significant effect on the y
value.
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What happens if we use a much larger 𝑥 value?
If 𝑥 = 1000, 𝑦 =1000+9
10002−2×1000−15=
1009
1000000−2000−15=
1009
997985 = 0.001011
Notice that the terms with the greatest influence on the output are the 𝑥 on the numerator
and the 𝑥2 on the denominator. This will be even more pronounced as 𝑥 gets larger.
The larger power of x on the denominator means that as 𝑥 → ∞, 𝑦 → 0.
So it would appear that this type of rational function has an asymptote 𝑦 = 0.
However, there is one exception. Since the numerator is 0 when 𝑥 = −9 then 𝑦 = 0
This is the only time the curve can go through the line 𝑦 = 0.
We’ll call this a zero point.
Vertical Asymptotes.
These work in much the same way as with reciprocal functions. We look for any 𝑥 values
that would make the denominator function equal 0.
From our previous example
𝑦 =𝑥 + 9
𝑥2 − 2𝑥 − 15
We would solve 𝑥2 − 2𝑥 − 15 = 0
(𝑥 − 5)(𝑥 + 3) = 0
𝑥 = 5 𝑜𝑟 − 3
So our vertical asymptotes are
𝑥 = 5 𝑎𝑛𝑑 𝑥 = −3
Unlike our horizontal asymptote there are no exceptions here. The curve cannot cross
these lines and will extend beyond the top and bottom of the graph.
Number lines and regions.
Now is a good time to set up a visual guide to show how the function behaves in the
regions defined by the important x values.
For this we’ll use 3 number lines. Each line will keep track of how part of the function
behaves. One for the numerator, the denominator and the entire function.
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Plot the point on the
numerator number line where
the numerator equals 0. This
is at 𝑥 = −9
Then mark the two points on
the denominator number line
where the denominator equals
0. This is at 𝑥 = −3 𝑎𝑛𝑑 5
Below both of these we’ll place
a number line to represent
𝑓(𝑥).
Now draw a vertical line though all of our number lines at each of these three points.
At 𝑥 = −9 we have 0 in the
numerator which means the
𝑓(𝑥) will have a zero point
which we indicate with a Z.
At 𝑥 = −3 𝑎𝑛𝑑 𝑥 = 5 we have
0 in the denominator which
means 𝑓(𝑥) has vertical
asymptotes at these locations.
We identify these on the
number line for 𝑓(𝑥) with ‘A’.
Our vertical lines have spit the function line into 4 regions, a through d. Our next step is to
determine the sign for each of the numerator, denominator and 𝑓(𝑥) within each region.
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The numerator is negative
when 𝑥 < −9 and positive
otherwise
The denominator is negative
when −3 < 𝑥 < 5 and positive
otherwise.
We can get the sign of 𝑓(𝑥)
within each region by reading
vertically down through the
regions. For example, in
region a, the numerator is
negative and the denominator
is positive. Therefore 𝑓(𝑥) is −𝑣𝑒
+𝑣𝑒 which is negative. In region
b, both the numerator and
denominator are positive so
we have 𝑓(𝑥) is +𝑣𝑒
+𝑣𝑒 which is
positive, and so on.
Sketch the graph using what we have deduced above.
Firstly, plot the zero point and the asymptotes. This splits the graph into 4 regions.
Then we add the horizontal asymptote, with its one exception at (−9,0).
Next, highlight the regions based on whether the function is positive or negative in that
region. For example we’ve determined in region a (where 𝑥 < −10) the function must be
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negative so the highlight is drawn below the x axis to indicate that is the feasible region for
the curve to appear. Similarly region b is positive and so on.
Now, sketch the curve in each of the regions. Note that as the curve approaches an
asymptote it will get closer and closer to it but never touch it.
Notice that there are 2 stationary points in this graph.
The one in the middle region is easy to spot but the one on the left is harder to identify.
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The curve must pass through the zero point, but, as 𝑥 → ∞ , 𝑦 → 0, which means it must
turn at some point. It is usually not necessary to find these stationary points, but if we had
to, we could use differentiation or a CAS calculator.
It is a good idea to draw the graph on a graphics calculator after you have sketched yours
to see how it compares.
Example
1. Sketch a graph of 𝑦 =𝑥2−4
𝑥3
Solution
We will go through the steps a bit faster this time.
Here are the number lines split into regions based on solving 𝑥2 − 4 = 0 and 𝑥3 = 0:
𝑥2 = 0 has solutions at 𝑥 = ±2 and 𝑥3 = 0 has solutions at 𝑥 = 0.
Using this information we will set up the zero points, asymptotes and regions on the graph.
Finally, we sketch the curve guided by our shading. 𝑓(𝑥) must pass through the zero
points and tend towards, but not touch the asymptotes.
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Our final sketch of 𝑓(𝑥) on the right.
Skills Development 2.1
Draw a sketch of each of the following functions
1. 𝑦 =𝑥−7
𝑥2
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2. 𝑦 =𝑥+2
(𝑥+7)(𝑥−3)
3. 𝑦 =𝑥2+3𝑥−18
𝑥4−16
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Lesson 3
Rational Functions with higher degree polynomials in
the numerator
By the end of this lesson you should be able to:
Sketch a graph of any rational function 𝑦 =𝑓(𝑥)
𝑔(𝑥) where 𝑓(𝑥) has a higher or equal degree
to 𝑔(𝑥)
Rational functions with the same degree on the
numerator and denominator
Our rational functions in the previous lesson all have a horizontal asymptote at 𝑦 = 0. This was due to
there being a higher degree polynomial function in the denominator. However, what if the degrees of
the polynomials in the numerator and denominator are the same?
Example
Find the limit for 𝑦 as 𝑥 → ∞ for the function.
𝑦 =𝑥+2
𝑥+8
Solution
Method 1: substitute a large value for x
𝑥 = 1000, 𝑦 =1000 + 2
1000 + 8=
1002
1008= 0.994
Clearly y is approaching 1 from below since the numerator will always be smaller than the
denominator.
So as 𝑥 → ∞ 𝑦 → 1−
Method 2: Express the rational function as a mixed function
Since the numerator has a degree equal to the denominator we can divide the functions and get a
remainder like so:
𝑦 =𝑥 + 2
𝑥 + 8=
𝑥 + 8 − 6
𝑥 + 8=
𝑥 + 8
𝑥 + 8−
6
𝑥 + 8= 1 −
6
𝑥 + 8
𝑦 = 1 −6
𝑥 + 8
Here we can see the function rewritten as a simpler rational function (lower degree on the
numerator) with a vertical translation of 1 unit provided by the 1 when graphed.
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The fraction −6
𝑥+8→ 0− as 𝑥 → ∞ and the 1 will be constant. Similarly −
6
𝑥+8→ 0+ 𝑎𝑠 𝑥 → −∞
The overall function must therefore tend to 1 from below as 𝑥 → ∞ and above as 𝑥 → −∞
So as 𝑥 → ∞ 𝑦 → 1− and as 𝑥 → −∞ 𝑦 → 1+
The second method is quite helpful here as we can rewrite a rational number with the same degree
on the numerator and denominator as the sum of a polynomial function together with a simpler
rational function.
Example
Draw a sketch of the function 𝑦 =𝑥+2
𝑥−3.
Solution
Express the rational function as a mixed function:
𝑦 =𝑥 − 3 + 5
𝑥 − 3=
𝑥 − 3
𝑥 − 3+
5
𝑥 − 3= 1 +
5
𝑥 − 3
𝑦 = 1 +5
𝑥 − 3
This gives a horizontal asymptote is 𝑦 = 1
Now let’s consider the fraction part of the function. It’s a reciprocal with the vertical asymptote 𝑥 =
3
So far our graph looks like this:
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Since the fraction can never have a 0 in the
numerator, there are no gaps in the horizontal
asymptote.
It’s easy to see the fraction part of 𝑓(𝑥)will be
positive for 𝑥 > 3 and negative for 𝑥 < 3 so
this gives us a guide to the position of 𝑓(𝑥) in
relation to the asymptotes.
Notice that when the fraction part is negative
the overall function can still be positive, i.e.
when 𝑥 ≤ 2.
Finally, sketch our curve including a couple of
key points calculated from the function.
𝑓(0) = −2
3 and 𝑓(4) = 6, giving the points
(0, −2
3) 𝑎𝑛𝑑 (4,6)
Sloping asymptotes
Let us look at the situation where variables appear in all parts of the function.
For example, 𝑦 = 𝑥 +1
𝑥
The vertical asymptote is easy to spot because 𝑥 ≠ 0 the fraction part 1
𝑥.
In this case, there is no constant that 𝑦 approaches as 𝑥 → ∞, but the fraction part of the function
will still tend towards 0.
This means the graph of 𝑓(𝑥) will get closer to, but never touch the line 𝑦 = 𝑥
So as 𝑥 → ∞, 𝑦 → 𝑥
This results a sloping asymptote.
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On the graph of 𝑦 = 𝑥 +1
𝑥 the
asymptotes would look like this.
To determine the location of the
graph of 𝑓(𝑥)with respect to the
asymptotes, consider the
fraction part of the function
When 𝑥 > 0 then 1
𝑥> 0 so the
curve will be above the
asymptote 𝑦 = 𝑥
When 𝑥 < 0 then 1
𝑥< 0 so the
curve will be below the
asymptote 𝑦 = 𝑥
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When we sketch the curve it will
tend towards each asymptote
as it extends towards the edges
of the plane.
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Functions with a higher degree polynomial in the numerator than in the denominator.
Consider 𝑦 =𝑓(𝑥)
𝑔(𝑥) where the degree of 𝑓(𝑥) is greater than the degree of 𝑔(𝑥)
For these functions we need to combine all of the techniques learned so far in this topic. Follow the
worked solution to Example 3 to see these in action.
Example
Sketch the graph of 𝑦 =𝑥2+5𝑥−6
𝑥+1.
Solution
Firstly, express the fraction as a mixed number:
𝑦 =𝑥2 + 5𝑥 − 6
𝑥 + 1=
𝑥(𝑥 + 1) + 4𝑥 − 6
𝑥 + 1= 𝑥 +
4𝑥 − 6
𝑥 + 1= 𝑥 +
4(𝑥 + 1) − 10
𝑥 + 1= 𝑥 + 4 −
10
𝑥 + 1
i.e,
𝑦 = 𝑥 + 4 −10
𝑥 + 1
With f(x) rewritten in this form
we can see our asymptotes will
be 𝑥 = −1 and 𝑦 = 𝑥 + 4.
These are shown on the right.
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The fraction part of the function
will determine if the curve is
above or below the sloping
asymptote.
When 𝑥 < −1 𝑡ℎ𝑒𝑛 −10
𝑥+1> 0
When 𝑥 > −1 𝑡ℎ𝑒𝑛 −10
𝑥+1< 0
Now we sketch the curve in the
correct regions. It’s a good idea
to plot a point on each part of
the curve. In this case finding
the 𝑥- intercepts is appropriate.
0 = 𝑥 + 4 −10
𝑥 + 1
Has the solutions
𝑥 = −6 𝑎𝑛𝑑 1
The curve passes through
(−6,0) and (1,0)
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Skills Development 3.1
1. Express the following rational functions as a mixed function. (Whole part and a fraction part)
(a) 𝑦 =
𝑥 + 5
𝑥 + 3
(b) 𝑦 =
𝑥2 − 7𝑥 + 10
𝑥 − 4
(c) 𝑦 =
𝑥2 + 13𝑥 + 9
𝑥2 + 9𝑥 − 2
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2. Sketch the graphs of each of the following functions.
(a) y =
3x − 6
x − 3
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(b) 𝑦 =
x2 − 11
(x − 4)(x + 3)
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(c) y =
x2 − x − 5
x + 2
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Summary
Reciprocal
function A reciprocal function is of the form
1
𝑓(𝑥) where 𝑓(𝑥) is a non-zero function
Rational function A rational function is a function such that
g xf x
h x where g x and h x
are polynomials. Usually g x and h x are chosen so as to have no
common factor of degree greater than or equal to 1, and the domain of f is
usually taken to be \ : 0R x h x
Asymptote A line that the curve approaches as it heads towards positive or negative
infinity but never actually touches. Asymptotes are indicated with a dotted
line
Vertical
asymptote
Comes from a restriction that x can’t equal a constant. Often to prevent the
denominator of a fraction from being 0
Horizontal
asymptote
Comes from a restriction that y can’t equal a constant. Often due to the
fact that a fraction with a non-zero numerator cannot equal 0
Sloping or
oblique
asymptote
Comes from a restriction that y cannot equal a function of x.
For example 𝑦 ≠ 𝑥 + 3
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Solutions
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