C3 Chapter 5: Transforming Graphs Dr J Frost (jfrost@tiffin.kingston.sch.uk) Last modified: 31st July 2014

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 .

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.

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

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.

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 ? ? ? ?

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

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 + 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.

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 2 . 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.

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

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.

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

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.

Test Your Understanding June 2012 Q4 Sketch > Sketch >

Exercises

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

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.

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− 2 3 +1 = 1 2 2 3 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=2 →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

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−2 𝒙=− 𝟒 𝟑 𝑥 = −4𝑥−5 𝒙=−𝟏, 𝒙=− 𝟓 𝟑 3𝑥= 𝑥 2 −4 𝒙=𝟏. 𝒙=𝟒 𝑥 −1=− 3𝑥 𝒙=± 𝟏 𝟒 24+2𝑥− 𝑥 2 = 5𝑥−4 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.