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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

© Department of Education Western Australia 2020 1

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



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



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.



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



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



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



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.



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



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.



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



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.



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



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.



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).



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)



(b)

(c)



(d)

(e)



3. On the axes provided draw a sketch of each of the following functions.

(a) 𝑦 =

1

4 − 𝑥

(b) 𝑦 =

1

𝑥2 − 9



(c) 𝑦 =

1

𝑥(𝑥 − 4)(𝑥 + 6)



Lesson 2

Rational functions with polynomials of a lower degree

in the numerator than in the denominator


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.



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.



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.



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



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.



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.



Our final sketch of 𝑓(𝑥) on the right.


Draw a sketch of each of the following functions

1. 𝑦 =𝑥−7

𝑥2



2. 𝑦 =𝑥+2

(𝑥+7)(𝑥−3)

3. 𝑦 =𝑥2+3𝑥−18

𝑥4−16



Lesson 3

Rational Functions with higher degree polynomials in

the numerator


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.



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:



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.



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



When we sketch the curve it will

tend towards each asymptote

as it extends towards the edges

of the plane.



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.



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)




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



2. Sketch the graphs of each of the following functions.

(a) y =

3x − 6

x − 3



(b) 𝑦 =

x2 − 11

(x − 4)(x + 3)



(c) y =

x2 − x − 5

x + 2



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



Solutions

year 12 mathematics specialist - doe-ict-homeschool-prd ... · 3.2.8 sketch the graphs of simple...

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