sketching rational functions

Sketching Rational Sketching Rational FunctionsFunctions

CIE Centre A-level CIE Centre A-level Further Pure MathsFurther Pure Maths

Definition of rational functionDefinition of rational function

In this section we will be looking at functions ofthe form:

( )( )p xq x

where p and q are polynomials in x. Almost all of our examples will be restricted to degree 1 and degree 2.

Keep in mind that we will continue to need our basic theory (from P3) about division of polynomials and partial fractions.

http://en.wikipedia.org/wiki/Image:Descartes.jpg

Curve sketchingCurve sketching

Let’s review some basic points to remember about sketching the curve of a function.•Sketching is not plotting•Gather all information before starting the sketch•Mark on the intercepts - points where the curve crosses the coordinate axes (as best you can)•Mark on the asymptotes (I will show you how to find them)•Find the turning points and determine their nature (where possible)•Draw smooth curves between these key features.


Type 1: Linear/LinearType 1: Linear/Linear

We start with the simplest kind of rational function:

Let’s try an example:

The curve cuts the y-axis at …

Ax BCx D

1( )3

xy f xx

-1/3

The curve cuts the x-axis at …

-1


AsymptotesAsymptotes

What is an asymptote?“An asymptote is a line which becomes a tangent to a curve as x or y tends to infinity” (p. 130)

What are the asymptotes of these functions:-xy e

1/y x

cosy x

0y

0, 0x y

None!

We will restrict our discussion to linear asymptotes


Type 1: Linear/linearType 1: Linear/linear

Going back to our example:

Finding the asymptotes.

1( )3

xy f xx

There are three types of asymptote:-•Horizontal – of the form y = k, found by observing the behaviour as •Vertical – of the form x = k, occur at points where the function is undefined. •Oblique – of the form y = kx + c, found by observing the behaviour as .

x

x

What kind of asymptotes does f(x) have?



1( )3

xy f xx

A: One vertical asymptote at x=3, one horizontal asymptote at y=1.

Finding the vertical asymptotes is trivial for rational functions; they lie at the roots of the denominator.Find the horizontal asymptotes and oblique asymptotes sometimes requires more thought.


Technique: finding non-vertical Technique: finding non-vertical asymptotesasymptotes

1

3

11( )3 1

Let ; ( ) 1Let ; ( ) 1 also.

x

x

xy f xx

x f xx f x

This method of argument is fine for type 1 (linear/linear), but it will trick us into a mistake with quadratic functions. Note this alternative approach:

1 4( ) 13 3

xf xx x

By splitting into partial fractions, the asymptotic behaviour is observed directly.



Let’s sum up what we’ve learnt so farabout this function:

1( )3

xy f xx

•Intercepts the coordinate axes at: (0,-1/3) and (-1,0)•Has vertical asymptote at x = 3•Has horizontal asymptote at y = 1

You can easily prove that there are no stationarypoints (how?).

Let’s mark those details onto a graph…



x

y

3

1

-113



Note! -> Type 1 rational functions never cross their asymptotes. This can help identifying the curves.Let 3 . 0 1

2( ) 0

Let 32( ) 0

x

f x

x

f x

This more rigorous approachwill help when studying the asymptotes of more complicated functions.It’s not necessary here.

As x-> inf. , does the curve approach the horizontalasymptote from above or below?



A: Recall that the partial fractions form is:1 4( ) 13 3

xf xx x

If x is greater than 3, f(x) > 1.

By the same reason, the curve approaches theasymptote from below for x -> - inf.

We are now ready to give the full sketch of the curve.



-8

-6

-4

-2

0

2

4

6

8

-10 -5 0 5 10 15


Type 2: Quadratic/linearType 2: Quadratic/linear

(2 3)( 1)( )3

x xy f xx

Find the points where the curve crosses the axes, and identify the vertical asymptote.

Finding the vertical asymptotes is trivial for rational functions; they lie at the roots of the denominator.Find the horizontal asymptotes and oblique asymptotes sometimes requires more thought.

A: (0,-1), (1,0), (-3/2,0). The vertical asymptoteis x=-3.



2(2 3)( 1) 2 3( )3 3

2 1 3/ 2 1 as 1 3/

x x x xy f xx x

x x x xx

Of course, y is “about” 2x for large x, but we can make a better linear approximation. There are two ways to prove it.

First, we will use a binomial expansion.



2 2

2

1 1

3 9 3 3 9

(2 3)( 1) 2 3( )3 3

2 1 3/1 3/

(2 1)(1 3/ ) (3 / )(1 3/ )(2 1)(1 ...) (1 ...) , 3

2 6 1 as x x xx x

x x x xy f xx x

x xx

x x x xx x

x x

According to this analysis y=2x-5 is an oblique asymptote.Now we will use a different method.



(2 3)( 1)( )3 3

(2 3)( 1) ( )( 3)2, 5, 12

12( ) 2 53

2 5 as

x x Cy f x Ax Bx x

x x Ax B x CA B C

f x xx

f x x

We have again obtained the same oblique asymptote.I hope you notice that polynomial division and partialfractions is the most natural way to analyse asymptoticbehaviour. It also makes it easier to find stationary points.



2

122 53

122 0( 3)

y xx

dydx x

It is easier to differentiate the function in partial fractions form than the original polynomial form (needs quotient rule).Here you should find two stationary points have x-coordinates 3 6

Task: Use the information we have gathered tosketch the “skeleton” of the graph.Also deduce the behaviour of the graph near theasymptotes to sketch the complete curve. Don’tforget to identify the stationary points and theirnature.

Intercepts: (0,-1), (1,0), (-3/2,0). The vertical asymptoteis x=-3. The oblique asymptote is y=2x-5. Stat. points as above.



-50

-40

-30

-20

-10

0

10

20

30

40

50

-10 -8 -6 -4 -2 0 2 4 6

2 5y x

Turning points 3 6

(2 3)( 1)( )3

x xf xx


Type 3: Quadratic/QuadraticType 3: Quadratic/Quadratic

( 5)( 2)( )( 3)( 4)x xy f xx x

Find the points where the curve crosses the axes, and identify the vertical and horizontal asymptotes.

Note! We have a horizontal asymptote, not an oblique asymptote, just because the degree of the denominator and numerator are the same.

A: (0,-5/6), (5,0), (-2,0). The vertical asymptotesare x=3, x=4. The horizontal asymptote is at y=1.

Q: What’s the easy way to find the stationary points?


Type 3: Quadratic/quadraticType 3: Quadratic/quadratic

A: Differentiation can be much simpler if we expressthe function in partial fractions form:

( 5)( 2)( )( 3)( 4) 3 4

10 6( ) 13 4

x x B Cf x Ax x x x

f xx x

2 2

2 2

12

10 6'( ) 03 4

5( 8 16) 3( 6 9)

(11 15)

f xx x

x x x x

x

Often you won’t be asked to do this.

Using this technique to find the range of f(x) isoften a troublesome and slow process. Is there another way?


Which values of y correspond to values of x?

2

2

2 2

2

( 5)( 2)If ( 3)( 4)

( 3)( 4) ( 5)( 2)

( 1) (3 7 ) 10 12 0For real values, we require 0

(3 7 ) 4( 1)(10 12 ) 0

9 42 49 4(10 12 10 12 ) 0

34 49 0

17 4 15 , 17 4 15

x xyx x

y x x x x

x y x y yx

y y y

y y y y y

y y

y y


Does the curve cross any of its asymptotes?

A: Yes. The type 3 curves will usually cross the horizontal asymptote once. So, you should alwaysfind out where this point is:

2 2

( 5)( 2)1( 3)( 4)

7 12 3 1011/ 2

x xyx x

x x x xx

So, at (5.5,1) the curve crosses the asymptote.


An accurate sketch can be constructed now. Let mesummarize all the information for you:

1. Axis crossings at (5,0), (-2,0), (0,-5/6).2. Vertical asymptotes at x = 3, x = 43. Horizontal asymptote at y = 1.4. Curve crosses asymptote at x = 11/2.5. Turning points at 6. Stationary values corresponding are:

12 (11 15)

17 4 15

Draw the curve and mark on all the key features.


-30

-20

-10

0

10

20

30

40

50

60

70

-9 -4 1 6 11

Large Scale view

( 5)( 2)( 3)( 4)x xyx x

-6

-4

-2

0

2

4

6

-9 -4 1 6 11


Small Scale view

( 5)( 2)( 3)( 4)x xyx x


Bear in mind:A type 3 (quadratic/quadratic) rational function normally has two vertical asymptotes; the asymptotic behaviour is usually opposite on each side of each asymptote (+inf vs. –inf.).However if the roots of the denominator are repeated, these two asymptotes merge and so this rule is no longer obeyed.Also, if the denominator does not have real roots, then there are no vertical asymptotes at all.

sketching rational functions

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