1. C3 Chapter 5: Transforming Graphs
Dr J Frost
Last modified: 31st July 2014

2. Overview
If you did the Tiffin L6 Summer Programme, then you will have already covered half this Chapter! We will cover:
Already covered last year:
The modulus function, 𝑓 𝑥 =|𝑥|
Basic graph transformations, including 𝑦= 𝑓 𝑥 .
Building up a graph by starting with a basic graph and performing transformations on it, e.g. sketching 𝑦= 2 𝑥−1 by starting with 𝑦= 1 𝑥
Consider the effect on a specific point of transformation(s).
To cover:
Transformating graphs using 𝑓( 𝑥 )
Solving equations of the form 𝑓 𝑥 =𝑔 𝑥 or 𝑓 𝑥 = 𝑔 𝑥 .

3. f(x+3) f(x) + 3 f(3x) 3f(x) RECAP :: Basic Transformations  3 units ?
I initially sketch 𝑦=𝑓 𝑥 . What happens if I then sketch:
f(x+3) ?
 3 units ?
f(x) + 3
 3 units ?
f(3x)
 Squashed by factor of 3 on x-axis. ?
3f(x)
 Stretched by factor of 3 on y-axis.

4. 𝒂 𝒇 𝒃𝒙+𝒄 +𝒅 RECAP :: Basic Transformations Step 1: Step 3: Step 4:
Bro Tip: To get the order of transformations correct inside the f(..), think what you’d need to do to get from (𝑏𝑥+𝑐) back to 𝑥.
Step 1: ?
 c
Step 3: ?
↕ a
Step 4: ?
↑ d
Step 2: ?
↔  b

5. Quickfire Questions 2f(2x – 1) f(0.5x + 1) - 2 f(-x) -2f(-2x + 3) + 1
List the transformations required (in order).
2f(2x – 1)
f(0.5x + 1) - 2 ? ?
Shift right 1 unit.
Halve x values.
Double y values.
Shift left 1 unit.
Double x values.
Shift down 2 units.
f(-x)
-2f(-2x + 3) + 1 ? ?
Times x values by -1, i.e. reflect in y-axis.
Shift left 3 units.
Divide x values by -2 (i.e. Halve and reflect in y-axis.
Times y values by -2, i.e. Reflect in x axis and double y.
Shift up 1 unit.

6. RECAP :: Basic Transformations
Summary Table:
Affects which axis?
Does what we expect or opposite?
Inside function brackets x
Opposite
Outside function brackets y
What we expect ? ? ? ?

7. Click to Start Bromanimation
Sketching by transforming a simple function
Sketch 𝑦= 1 𝑥+3 −2
Start with 𝑓 𝑥 = 1 𝑥 . Then what is the function above? ?
y = f(x + 3) - 2
An alternative way to think about this is to start with 𝑦= 1 𝑥+3 as we saw earlier (knowing that we have an asymptote 𝑥=−3 as we can’t divide by 0), before shifting 2 down.
𝑥=−3
𝑦=−2
Click to Start Bromanimation

8. Sketching by transforming a simple function
By thinking about the transformations involved, we can now sketch a greater variety of functions.
𝑦=3− 1 𝑥 2
𝑦=2 sin 𝑥+ 𝜋 4
𝑦=2+ 1 𝑥 ? 𝑦
1 4 𝜋,2 ? ? 𝑦 𝑦 2
𝑦=3
− 1 4 𝜋
3 4 𝜋
7 4 𝜋
𝑦=2
− 1 3 𝑥 𝑥
5 4 𝜋,−2
− 1 2
Bro Tip: Start with 𝑦= 1 𝑥 2 , then think 𝑦=− 1 𝑥 2
Notice that unlike with 𝑦= 1 𝑥 we now have to work out the roots by setting 𝑦=0. You’ll lose marks otherwise!
Notice also that the horizontal asymptote now has to be explicitly identified.

9. Test Your Understanding
By thinking about the transformations involved, we can now sketch a greater variety of functions.
𝑦= 2 𝑥 2 −1
𝑦= sin 2 𝑥+ 𝜋 2
𝑦=1− 1 𝑥 ? 𝑦 ? ? 𝑦 𝑦 1 𝑥
𝜋 2
3 2 𝜋
𝑦=1 𝑥
− 2
+ 2 𝑥
𝑦=−1 1
Start with 𝑦= sin 𝑥+ 𝜋 Then square the 𝑦 values. Note that anything squared is always positive!
Again, you’d lose a mark if you didn’t have the root or the equation of the asymptote.

10. The Modulus Function
To draw 𝑦= 𝑓 𝑥 , first draw 𝑦=𝑓 𝑥 , then make the 𝑦 values positive wherever they were negative. Easy! 𝑦
𝑦=𝑥+1
Sketch >
𝑦= 𝑥+1
Sketch > 1 𝑥 −1

11. Examples
𝑦= 1 𝑥 +1
𝑦= 𝑥 2 −2𝑥−15 ? ?
= 𝑥+3 𝑥−5 𝑦 𝑦 15
𝑦=1 𝑥 -3 5 −1
−15
Bro Tip: The modulus function tends to lead to ‘sharp’ corners at the 𝑥-axis.

12. Test Your Understanding
𝑦= 1 𝑥 𝑥−2
𝑦= 2− 1 𝑥 ? ? 𝑦 𝑦
𝑦=𝑥 𝑥−2
𝑦=2
𝒚= 𝟏 𝒙 𝒙−𝟐 𝑥
1,1
1 2 𝑥 2
1,−1
𝑥=2
𝒚= 𝟏 𝒙 𝒙−𝟐

13. Further transformations involving |…|
We’ve seen how to draw 𝑦= 𝑓 𝑥 .
This made any 𝑦 values that were negative positive. This is because the |..| was outside the function brackets.
How do you think we might sketch 𝒚=𝒇 𝒙 then? 𝑦
𝑦=𝑓 𝑥
Sketch > 𝑥
Think about it: If 𝑥=−3, then we’d actually be using the 𝑦 value when 𝑥 was 3.
As usual, since the ‘transformation’ is inside the function brackets, it affects the 𝑥 values. Just imagine a mirror on the 𝑦-axis and looking from the right.

14. Test Your Understanding
June 2012 Q4
Sketch >
Sketch >

15. Exercises

16. Solving Equations of the Form 𝑓 𝑥 =𝑔(𝑥)
Solve the equation 𝑥 2 −2𝑥 = 1 4 −2𝑥
A helpful way to visualise what’s going on (but is not necessary in an exam) is to sketch each side of the equation.
Solution ?
𝒙 𝟐 −𝟐𝒙= 𝟏 𝟒 −𝟐𝒙
Solving gives 𝒙=− 𝟏 𝟐 𝒐𝒓 𝒙= 𝟏 𝟐
𝟐𝒙− 𝒙 𝟐 = 𝟏 𝟒 −𝟐𝒙
Solving gives 𝒙=𝟑.𝟗𝟒 𝒐𝒓 𝟎.𝟎𝟔
Checking each solution by substituting it into the original equation:
𝒙=𝟎.𝟎𝟔 works
𝒙=𝟑.𝟗𝟒 does not work
𝒙=− 𝟏 𝟐 works
𝒙= 𝟏 𝟐 does not work
So 𝒙=− 𝟏 𝟐 𝒐𝒓 𝒙=𝟎.𝟎𝟔
Sketch ? 𝑦
The key is to solve two equations, where 𝑥 2 −2𝑥 was not reflected (i.e. use 𝑥 2 −2𝑥) and where it was reflected (use − 𝑥 2 −2𝑥 )
𝒚=| 𝒙 𝟐 −𝟐𝒙| 𝑥
𝒚= 𝟏 𝟒 −𝟐𝒙

17. Why did we have to check if solutions worked?𝑦
The problem is that when we use the reflected graph 𝑦=2𝑥− 𝑥 2 , the intersection may have occurred where the graph wasn’t actually reflected.
𝒚=| 𝒙 𝟐 −𝟐𝒙| 𝑥 2
𝒚= 𝟏 𝟒 −𝟐𝒙
The textbook gets you to reject solutions by considering from the diagram if the intersection occurred on the actual graphs.
A much easier way is to not bother with the graphs and just sub the values into the original equation to check.

18. Test Your Understanding
June 2008 Q3 b ?
2− 𝑥+1 = 1 2 𝑥
When 𝑥+1 is not reflected:
2−𝑥−1= 1 2 𝑥 𝑥= 2 3
When 𝑥+1 is reflected:
2+𝑥+1= 1 2 𝑥 𝑥=−6
Check:
2− = works
2− −6+1 = 1 2 −6 works a ?
Q: When 𝑥=0, 𝑓 𝑥 =2− 0+1 =1
𝑄 0,1
R: When 𝑦=0, 2− 𝑥+1 =0 𝑥+1 =2
Either 𝑥+1=2 → 𝑥=1
Or −𝑥−1= →𝑥=−3
𝑅 1,0
P: Graph is at its maximum when 𝑥+1 =0
Thus 𝑥=−1 (alternative by symmetry, -1 is halfway between -3 and 1)
𝑃 −1,2

19. Exercise 5C
Solve the following. Then sketch the graphs to demonstrate the solutions.
Note: solving 𝑓 𝑥 =|𝑔 𝑥 | uses the same technique as solving 𝑓 𝑥 =𝑔(𝑥). We don’t need to try 𝑓 𝑥 =−𝑔(𝑥) if we’ve already tried −𝑓 𝑥 =𝑔(𝑥) and we don’t need to try −𝑓 𝑥 =−𝑔(𝑥) if we’ve already tried 𝑓 𝑥 =𝑔 𝑥 .
−2𝑥= 1 2 𝑥− 𝒙=− 𝟒 𝟑 𝑥 = −4𝑥− 𝒙=−𝟏, 𝒙=− 𝟓 𝟑 3𝑥= 𝑥 2 − 𝒙=𝟏. 𝒙=𝟒
𝑥 −1=− 3𝑥 𝒙=± 𝟏 𝟒 24+2𝑥− 𝑥 2 = 5𝑥− 𝒙=−𝟐.𝟏𝟖, 𝒙=𝟒 ? 1 ? 2 3 ? ? 4
We can simplify! Noting that 𝑎𝑥 =𝑎|𝑥|, we get:
𝑥 + 3𝑥 = 𝑥 +3 𝑥 =4|𝑥|
So 𝑥 = 1 4 ? 5
Bro Tip: For the first two questions you could also square both sides and solve, because 𝑥 2 = 𝑥 2 , since squaring gives the same result whether the value was negative or positive. You still need to check your values at the end, because squaring introduces false solutions (e.g. if 𝑥=2 then 𝑥 2 =4, which introduces the false “solution” of 𝑥=−2)
For Q3/5 the method is not helpful as you’ll have a quartic. And for Q4 the |..| is not isolated.