further pure 1 lesson 7 – sketching graphs. wiltshire graphs and inequalities good diagrams...

Further Pure 1

Lesson 7 – Sketching Graphs

WiltshireGraphs and Inequalities

Good diagrams Communicate ideas efficiently. Helps discovery & understanding of

relationships. Essential Features include

Points where axes are crossed. Nature and position of turning points. Behaviour as x or y tend to infinity.

Need to be able to check using a calculator or computer but must be able to do it independently.

WiltshireRational Functions

A rational function is defined as a function

N(x)/D(x), where N(x) & D(x) are polynomials. Note that D(x) = 0 (This is similar to the notion of a rational

number, n/d) What does the graph of y = 1/x look like? What happens when x = 0? Now what would the graph of y = 1/(x-2)+3

look like.

WiltshireUsing Transformations

Now what would the graph of y = 1/(x-2)+3 look like.

This will look the same as y = 1/x but after two transformations.

The (x – 2) shifts the graph right 2.

Then the +3 shifts the graph up 3.

Notice where the asymptotes have moved too.

This would be the same as

y = (3x – 5)/(x - 2)

WiltshireUsing Transformations

How could we plot the graph

y = (2x+9)/(x+4)

y = 1/(x+4)+2

Or

2

x + 4 2x + 9

2x + 8

1 We can now use transformations to plot the graph.

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4

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4

1

4

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4

1

4

92

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x

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x

WiltshireQuestions

Plot the following graphs

i)

ii)

iii)4

133

x

xy

3

52

x

xy

2

3

x

xy

WiltshireVertical Asymptotes

An asymptote is a straight line which a curve approaches tangentially as x or y tends to infinity.

Asymptotes are usually shown on a graph by a dotted line.

In the graph y = 1/x there is an asymptote at x = 0. This is because as x gets smaller and smaller y tends

to infinity. (Note: it can help that 1/0 is undefined) What would the vertical asymptote be for the graph

y = 1/(x – 3)? x = 3. In general the asymptote for y = 1(x - a) will be x = a. and the asymptote for y = 1/(x + a) will be

x = -a.

WiltshireVertical Asymptotes

What are the vertical asymptotes for the following graphs?

i)

ii)

iii))9)(43(

162

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y

)3)(2(

9

xx

xy

7

23

x

xy

WiltshireSketching Graphs

We are now going to look at the steps for sketching the graph.

There are some steps to follow to ensure that the graph is plotted correctly.1) Find where the graph cuts the axes.2) Find the vertical asymptotes and examine the

behaviour near them.3) Examine the behaviour as x tends to infinity.4) Complete the sketch

)1)(3(

6

xx

xy


Step 1 - Find where the graph cuts the axes. The graph will cut the y-axis when x = 0, and the x-

axis when y = 0. So when x = 0 And when y = 0

The graph will cut the axes at the co-ordinates (0,2) and (6,0).

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)10)(30(

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)1)(3(

6

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6

60

)1)(3(

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)1)(3(

6

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The graph will cut the axes at the co-ordinates (0,2) and (6,0).

We can plot these two co-ordinates on the graph.


Step 2 - Find the vertical asymptotes and examine the behaviour near them.

The vertical limits will be when x = -3 & x = 1


There are 2 vertical asymptotes that can be approached from 2 directions.

We need to look at each of the 4 scenarios.


i) When x is slightly less than -3. (x-6) is negative, (x+3) is negative and very small and (x-1) is negative.

y

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vey

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))((

)1)(3(

6


ii) When x is slightly greater than -3. (x-6) is negative, (x+3) is positive and very small and (x-1) is negative.

y

vesmall

vey

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xy

))((

)1)(3(

6


i) When x is slightly less than 1. (x-6) is negative, (x+3) is positive and (x-1) is a very small negative .

y

vesmall

vey

vesmallve

vey

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xy

))((

)1)(3(

6


i) When x is slightly greater than 1. (x-6) is negative, (x+3) is positive (x-1) is very small positive.

y

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))((

)1)(3(

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Step 3 - Examine the behaviour as x tends to infinity. As x tends to infinity y tends to zero from above the x-axis. As x tends to negative infinity y tends to zero from below the x-

axis. So y = 0 is a horizontal asymptote.

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xy

x

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1

)1)(3(

6

2


Step 4 – Sketch the graph.


Notice that the graph is a smooth curve. There is a local minimum between x = -3 & 1. There is a local maximum just after x = 6. How do we know that there are only two turning points? We can find the

exact position of the maxima andminima by solving dy/dx = 0.

However you will not study this until C3.


An alternative way to look at it would be to consider the graph y equal to horizontal lines c.

If y = (x-6)/((x+3)(x-1)) = c Then (x-6)=c(x+3)(x-1) Unless c = 0 then

this is a quadratic equation with 3 possible outcomes.i) No solutionii) One solutioniii) Two solutions

As the quadratic cannot have more than 2 solutions, thenthe green line is impossible.

This proves that our original drawing was correct. Can you explain how this would also show that the local

maximum must be lower than the local minimum?


Sketch the following graphs?

i)

ii)

iii) )9)(43(

122

xx

y

)3)(2(

9

xx

xy

7

23

x

xy

WiltshireComplicated example

Sketch the following graph.

34

22

2

xx

xxy

WiltshireUsing Symmetry

Efficient knowledge of symmetrical curves can help you when sketching graphs.

If f(x) = f(-x) then f(x) is symmetrical about the y-axis.

Graphs of this type are known as even functions

Note if f(x) = -f(-x) then f(x) has rotational symmetry of order 2 about the origin.

Graphs of this type are known as odd functions.

Name an even and an odd function.


Sketch the graph

Step 1 – When x = 0, y = 5When y = 0 there is no x

Step 2 – There is no x that makes x2+3 = 0 therefore no vertical asymptotes exist.

Step 3 – As x tends to +ve/-ve infinity y tends to 2.So y = 2 is a horizontal asymptote.

Step 4 – As x only contains even powers of x, y is an even function, and so is symmetrical about the y-axis.

3

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2

x

xy


3

1522

2

x

xy

WiltshireUsing a Graphics Calculator

further pure 1 lesson 7 – sketching graphs. wiltshire graphs and inequalities good diagrams...

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