c3 chapter 5 modulus and transformations

Chapter 5 Modulus and Transformations

What is the modulus?

• The modulus of the no. z is written |z|• |z| is the absolute value of z (the positive

numerical value)|1| = 1 |-1| = 1

• When x>0, |x|=x• When x<0, |x|=-x

|x-3|

• When x>3, |x-3| = x-3• HOWEVER, when x<3, |x-3|=-(x-3)=-x+3

By the end of this lesson you should be able to…

1. Sketch the graph of the modulus function y = |f(x)|2. Sketch the graph of the function y = f(|x|)3. Solve equations involving the modulus function4. Apply a combination of two (or more)

transformations to the same curve5. Sketch transformations of the graph y = f(x)

Sketching modulus graphs

• Lets look at y = x• Lets draw y=|x|

• When x>0, y = |x|• All the negative values of x

become positive• When x<0, |x|=-x

OR there is a reflection in the x axis

How to sketch y = |x|

1. Draw the graph without the modulus2. Reflect in the x-axis so that no negative values of y remain

Another example

• y = |2x – 4|– Step 1 -> draw without modulus y = 2x - 4

• y = |2x – 4|– Step 2 -> Reflect in the x-axis

A harder example

• y = |x2 - 4x + 3|– Step 1 -> draw without modulus

• y = |x2 - 4x + 3|– Step 2 -> Reflect in the x-axis

How to sketch y = |x| + a• Method 1 (to draw y = |x| + 3)

1. Sketch without modulus

• Method 1 (to draw y = |x| + 3)2. Reflect in the y-axis (before you always reflected in the x-axis – don’t mix them up!)

• Method 2 (to draw y = |x| + 31. Sketch y = |x|

• Method 2 (to draw y = |x| + 32. Vertical translation of “ + 3 “

A harder example

• Sketch y = |x|3-|x|2

– Step 1 -> Sketch the graph without modulus

• Sketch y = |x|3-|x|2

– Step 2 -> Reflect in the y-axis

Key Points

• The Modulus function |x| is defined by

• When working with entire equations in the modulus (e.g. |x-5|), sketch out the graph without modulus, and then reflect in the x-axis

• When working with equations with only x within the modulus symbols (e.g. 3|x|-5), then sketch graph without modulus and then reflect in the y-axis

Exercise

Exercises

Solving Equations with Modulus

• These will look like |2x – 5| = 3 (with only one modulus equation) and |x+4|=|2x-3| (with two modulus equations

• Best way to calculate x values is using a mixture of graphical and algebraic calculations

Solving |2x-5|=3

• For simple calculations like this, sketch y = |2x-5| and y = 3

This part of the modulus was the original line, so we know the equation of this part of the line is y = 2x-5

The reflected part of the modulus is different as we have found the absolute values of x. To find the equation of this line, you simply do -1(original eq), so it is y = -(2x-5) = -2x + 5

When f(x) < 0, |f(x)|= -f(x)

y = -2x + 5 y = 2x - 5

y = 3

We can see there are two intersections. Now all we do is solve simultaneously.

First intersection (reflected part)y = 3, y= -2x + 5

3 = -2x + 52x = 5 - 3

2x = 2x = 1

Second intersection (original part)y = 3, y = 2x - 5

2x - 5 = 32x = 3 + 5

2x = 8x = 4

Therefore the solutions to |2x-5|= 3 are:x=1 and x=4

You can see that you don’t specifically need to draw the graph, but it is

important that you do to double check your

solutions and to avoid making silly mistakes

Solving |2x-5|=3 by squaring both sides

(2 𝑥−5 )2=32𝑥2−5 𝑥+4=0

A harder example -> |2x-7|=|x-6|• Sketch!• You have 2 modulus equations here

y = |2x-7| and y = |x-6|Hence, you can create 4 non-modulus equations

y = 2x – 7

y = -2x + 7

y = x - 6

y = -x + 6

We have 2 intersections

Rememberwhen f(x) < 0 |f(x)| = -f(x)

We can see that the first solution occurs with the reflected portion of both equationsy = -2x + 7 and y = -x + 6

The second solution occurs with the an original eq. and the reflected portion of the other eq.y = 2x – 7 and y = -x + 6

y = -2x + 7 & y = -x + 6-2x + 7 = -x + 67 – 6 = -x + 2x

x = 1

y = 2x - 7 & y = -x + 62x - 7 = -x + 62x + x = 6 + 7

3x = 13x = or 4. (2s.f.)

y = 2x – 7

y = -2x + 7

y = x - 6

y = -x + 6

Rememberwhen f(x) < 0 |f(x)| = -f(x)

We have 2 intersections

We can see that the first solution occurs with the reflected portion of both equationsy = -2x + 7 and y = -x + 6

The second solution occurs with the an original eq. and the reflected portion of the other eq.y = 2x – 7 and y = -x + 6

y = -2x + 7 & y = -x + 6-2x + 7 = -x + 67 – 6 = -x + 2x

x = 1

y = 2x - 7 & y = -x + 62x - 7 = -x + 62x + x = 6 + 7

3x = 13x = or 4. (2s.f.)

y = 2x – 7

y = -2x + 7

y = -x + 6

y = -x + 6

If asked for a coordinate, then insert the relevant x-coord into the corresponding equation and solve for y.

Eg. If finding the coordinates for

x = 1 insert it only into y = -2x + 7 or y = -x + 6

A common mistake here is to insert the coordinate into the wrong equations

|2x-7|=|x-6| by Squaring both sides

(2 𝑥−7 )2=(𝑥−6 )2

𝑥=− (−16 )± √(−16 )2−4 (3 ) (13 )

2 (3 )Again we get the correct answer, but this method is not fool proof.

Maybe another one -> |x2+4|=|x+3|

y = x2 + 4 y = -x2 - 4

y = x + 3(y = -x – 3)We can ignore this

4 Intersections betweeny = x2 + 4 & y = x + 3 x2y = -x2 – 4 & y = x + 3 x2

x2 + 4 = x + 3x2 + 4 - x - 3 = 0

x2 - x + 1 = 0

-x2 - 4 = x + 30 = x + 3 + x2 + 4

x2 + x + 7 = 0

And then factorise to find 2 solutions for x for each

quadratic

This example is impossible to factorise with real roots but you

get the point.

Why you should solve using graphs

• Using 100% algebraic methods to solve these sorts of questions can come up with problems.

• Questions can look like – |x+3| = 4 or |x+3| = |x-2|– They can also get a lot more complicated

• The next two questions will show you the problem of solving without graphs

|x-2| = -2

• Using no graphs, and the method we call squaring both sides we find the roots by combining the two equations:

(x-2)2 = (-3)2 x2 -4x +4 = 9x2 -4x -5 = 0

(x-5)(x+1) = 0 x=5 & -1

However, this is wrong. This would get you no marks. And here’s why.

Sketching |x-2|=-2

This equation has no solutions.

|x2 – 1| = 6xFor one solution

For the other solution

So we have 4 solutions:


This is only partially correct.

From the graph we can see that we have

only 2 solutions

• We substitute the x values in the original equation.


This is only partially correct.

For each of the four values of x, the LHS is always positive. But, two of the values has their RHS as negative and hence these two values are incorrect.

Exam Question (June 2010)

Exam Question (June 2008)

C1 Recap Transformations

• Is a horizontal stretch of scale factor

• Is a vertical translation of +a

• Is a vertical stretch of scale factor a

• Is a horizontal translation of -a

• f(x+a)

• f(x) + a

• f(ax)

• af(x)

Sketching more than 1 transformation on the same graph

y = ( x – 3)2 + 2• Always start from y = x2, working from the inside

to the out• y = x2 -> translation of f( x – 3 ) -> y = ( x – 3 )2

• y = ( x – 3 )2 -> translation of f( x ) +2 -> y = ( x – 3 )2 + 2

Sketch y = ( x – 3)2 + 2• y = x2 translation of f( x – 3 ) y = ( x – 3 )2

• y = ( x – 3 )2 translation of f( x ) +2 y = ( x – 3 )2 + 2

Sketch Here, start from • translation of f( x – 3 )

• vertical stretch of scale factor 3 / transformation of 3f( x )

Sketch Here, start from • horizonatal stretch of sf / transformation of f( 2x )

• translation of f(x) -1

Sketch Here, start from translation of f( x - 4 )

• vertical stretch of sf 2 / transformation of 2f( x )

• translation of f( x ) + 3

This is just an extension of the previous sub-topic

You will be given a sketch of f(x) and your job will be to apply transformations to it.

This is y = f(x). The curve passes through the origin, A(-2,7) and B(1,5).

Sketch y = 2f(x-2)

Work from the inside of the brackets to the

outside

Remember to show the new position of O, A and B.

Exam Question (Jan 2010)

Exam Q

uestion (Jan 2008)

Exam Question June 2007

c3 chapter 5 modulus and transformations

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