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Transforming Graphs of Functions

Transforming Graphs of Functions. Introduction. This chapter focuses on multiple transformations of graphs It will introduce a new concept, ‘modulus’ We will also look at how to solve equations involving this…. Teachings for Exercise 5A. Transforming Graphs of Functions.

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Transforming Graphs of Functions

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  1. Transforming Graphs of Functions

  2. Introduction • This chapter focuses on multiple transformations of graphs • It will introduce a new concept, ‘modulus’ • We will also look at how to solve equations involving this…

  3. Teachings for Exercise 5A

  4. Transforming Graphs of Functions You need to be able to sketch the graph of the modulus function y = |f(x)| The modulus of a number is its positive numerical value. It is written like this: So you can effectively think of a modulus as swapping all negative values to positive ones… Sketch the graph of: y = x 1) Sketch the graph ignoring the modulus y = |x| 2) Reflect the negative part in the x-axis 5A

  5. Transforming Graphs of Functions You need to be able to sketch the graph of the modulus function y = |f(x)| The modulus of a number is its positive numerical value. It is written like this: So you can effectively think of a modulus as swapping all negative values to positive ones… Sketch the graph of: y = 3x - 2 1) Sketch the graph ignoring the modulus 2/3 -2 y = |3x - 2| 2) Reflect the negative part in the x-axis 2 2/3 When the modulus is applied to the whole equation, we are changing the output (y-values), hence the reflection in the x-axis… The red part of the graph is from the equation: y = 3x + 2 5A The blue part of the graph is from the equation: y = -(3x + 2)

  6. Transforming Graphs of Functions You need to be able to sketch the graph of the modulus function y = |f(x)| The modulus of a number is its positive numerical value. It is written like this: So you can effectively think of a modulus as swapping all negative values to positive ones… Sketch the graph of: y = x2 – 3x - 10 1) Sketch the graph ignoring the modulus -2 5 -10 y = |x2 – 3x - 10| 10 2) Reflect the negative part in the x-axis -2 5 The red part of the graph is from the equation: y = x2 – 3x - 10 The blue part of the graph is from the equation: y = -(x2 – 3x – 10) 5A

  7. Transforming Graphs of Functions You need to be able to sketch the graph of the modulus function y = |f(x)| The modulus of a number is its positive numerical value. It is written like this: So you can effectively think of a modulus as swapping all negative values to positive ones… Sketch the graph of: 1) Sketch the graph ignoring the modulus y = sinx 0 π/2 π 3π/2 2π y = |sinx| 2) Reflect the negative part in the x-axis 0 π/2 π 3π/2 2π The red part of the graph is from the equation: y = sinx The blue part of the graph is from the equation: y = -(sinx) 5A

  8. Teachings for Exercise 5B

  9. Transforming Graphs of Functions Sketch the graph of: You need to be able to sketch the graph of the modulus function y = f(|x|) The difference here is that we are changing the value of the inputs (x) We are not going to put any negative values into the function • The result is that the value we get at x = -3 will be the same as at x = 3 • The graph will be reflected in the y-axis… y = |x| - 2 1) Sketch the graph for x ≥ 0, ignoring the modulus… -2 2 2) Reflect the graph in the y-axis -2 5B

  10. Transforming Graphs of Functions Sketch the graph of: You need to be able to sketch the graph of the modulus function y = f(|x|) The difference here is that we are changing the value of the inputs (x) We are not going to put any negative values into the function • The result is that the value we get at x = -3 will be the same as at x = 3 • The graph will be reflected in the y-axis… 1) Sketch the graph for x ≥ 0, ignoring the modulus… -2 2 2) Reflect the graph in the y-axis y =4|x| - |x|3 5B

  11. Teachings for Exercise 5C

  12. Transforming Graphs of Functions Solve the Equation: You need to be able to solve equations involving a modulus Solutions to these equations are the places where the two graphs cross (if each side of the equation is plotted as a graph) You must pay careful attention to where they cross, on the original graph or the reflected part Try to keep sketches reasonably accurate by working out key points… y = 2x – 3/2 1) Draw both graphs, ignoring the modulus part. Work out key points if possible 3 y = 3 3/4 -3/2 y = |2x – 3/2| 2) Alter the graphs to take into account any modulus effects 3 y = 3 3/2 3/4 5C

  13. Transforming Graphs of Functions Solve the Equation: You need to be able to solve equations involving a modulus Solutions to these equations are the places where the two graphs cross (if each side of the equation is plotted as a graph) You must pay careful attention to where they cross, on the original graph or the reflected part Try to keep sketches reasonably accurate by working out key points… y = |2x – 3/2| 3) If a solution is on the reflected part, use –f(x) For example point A is on the original blue line, but the reflected red line… A B 3 y = 3 3/2 3/4 Solution A Solution B Solution B is on both original curves, so no modification needed… Using –f(x) for the equation of the red line 5C

  14. Transforming Graphs of Functions y = |5x – 2| y = |2x| 2 You need to be able to solve equations involving a modulus 2) Alter the graphs to take into account any modulus effects 2/5 0 Solve the Equation: A B Solution A (Reflected Red, Original Blue) y = 5x – 2 y = 2x 1) Draw both graphs, ignoring the modulus part. Work out key points if possible 3) If a solution is on the reflected part, use –f(x) 0 2/5 Solution B -2 (Original Red, Original Blue) 5C

  15. Transforming Graphs of Functions y = |x2 – 2x| You need to be able to solve equations involving a modulus 2) Alter the graphs to take into account any modulus effects 0 1/8 2 Solve the Equation: A y = 1/4 - 2x Solution A (Original Red, Original Blue) y = x2 – 2x 1) Draw both graphs, ignoring the modulus part. Work out key points if possible 3) If a solution is on the reflected part, use –f(x) 1/8 0 2 or y = 1/4 - 2x x < 0 at point A so the second solution is the correct one 5C

  16. Transforming Graphs of Functions y = |x2 – 2x| You need to be able to solve equations involving a modulus B 2) Alter the graphs to take into account any modulus effects 0 1/8 2 Solve the Equation: y = 1/4 - 2x Solution B (Reflected Red, Original Blue) y = x2 – 2x 1) Draw both graphs, ignoring the modulus part. Work out key points if possible 3) If a solution is on the reflected part, use –f(x) 1/8 0 2 y = 1/4 - 2x or x < 2 at point B so the second solution is the correct one 5C

  17. Teachings for Exercise 5D

  18. Transforming Graphs of Functions You need to be able to apply multiple transformations to the same curve f(x + a) is a horizontal translation of –a units f(x) + a is a vertical translation of a units f(ax) is a horizontal stretch of scale factor 1/a af(x) is a vertical stretch of scale factor a -f(x) is a reflection in the x-axis f(-x) is a reflection in the y-axis 5D

  19. Transforming Graphs of Functions y = (x – 2)2 + 3 y = (x – 2)2 y = x2 You need to be able to apply multiple transformations to the same curve Sketch the graph of:  Build the equation up from y = x2 7 Horizontal Translation, 2 units right y-intercept where x = 0 Vertical Translation, 3 units up 5D

  20. Transforming Graphs of Functions y = 2/x + 5 You need to be able to apply multiple transformations to the same curve Sketch the graph of:  Build the equation up from y = 1/x y = 1/x + 5 y = 1/x 2/5 Horizontal Translation, 5 units left At the y-intercept, x = 0 Vertical stretch, scale factor 2 5D

  21. Transforming Graphs of Functions You need to be able to apply multiple transformations to the same curve Sketch the graph of:  Build the equation up from y = cosx 2 y = cos2x 1 y = cosx 0 90 180 270 360 -1 -2 y = cos2x - 1 Horizontal ‘stretch’, scale factor 1/2 Vertical translation 1 unit down 5D

  22. Transforming Graphs of Functions y = 3|x – 1| y = x - 1 You need to be able to apply multiple transformations to the same curve Sketch the graph of:  Build the equation up from y = x - 1 3 y = |x – 1| 1 1/3 1 1 5/3 y = 3|x – 1| - 2 -1 Reflect negative values in the x-axis Vertical stretch, scale factor 3 You will need to do more than one sketch for these – do not do lots on the same diagram! Vertical translation 2 units down 5D

  23. Teachings for Exercise 5E

  24. Transforming Graphs of Functions B(6,8) B(6,7) When you are given a sketch of y = f(x), you need to be able to sketch transformations and show the final position of original co-ordinates To the right is the graph of y = f(x) Sketch the graph of: y = 2f(x) – 1 and state the new coordinates of O, A and B… B(6,4) O O(0,-1) A(2,-1) y = f(x) A(2,-2) A(2,-3) y = 2f(x) y = 2f(x) - 1 Vertical Stretch, scale factor 2  y-values double Vertical translation 1 unit down  y-values reduced by 1 5E

  25. Transforming Graphs of Functions y = f(x + 2) + 2 B(4,6) When you are given a sketch of y = f(x), you need to be able to sketch transformations and show the final position of original co-ordinates To the right is the graph of y = f(x) Sketch the graph of: y = f(x + 2) + 2 and state the new coordinates of O, A and B… B(6,4) B(4,4) A(0,1) O(-2,2) O O(-2,0) A(0,-1) A(2,-1) y = f(x) y = f(x + 2) Horizontal translation 2 units left  x-values reduced by 2 Vertical translation 2 units up  y-values increased by 2 5E

  26. Transforming Graphs of Functions y = 1/4f(2x) When you are given a sketch of y = f(x), you need to be able to sketch transformations and show the final position of original co-ordinates To the right is the graph of y = f(x) Sketch the graph of: y = 1/4f(2x) and state the new coordinates of O, A and B… B(3,4) B(6,4) B(3,1) A(1,-0.25) O A(1,-1) y = f(x) A(2,-1) y = f(2x) Horizontal stretch, scale factor 1/2  x-values divided by 2 Vertical stretch, scale factor 1/4  y-values divided by 4 5E

  27. Transforming Graphs of Functions y = f(x - 1) When you are given a sketch of y = f(x), you need to be able to sketch transformations and show the final position of original co-ordinates To the right is the graph of y = f(x) Sketch the graph of: y = -f(x - 1) and state the new coordinates of O, A and B… B(6,4) B(7,4) A(3,1) O(1,0) O A(3,-1) A(2,-1) y = f(x) B(7,-4) y = -f(x - 1) Horizontal translation 1 unit right  x-values increase by 1 Reflection in the x-axis  y-values ‘swap sign’ (times -1) 5E

  28. Summary • We have learnt about modulus graphs • We have seen how sketches help us solve equations involving a modulus • We have also practised multiple transformations and tracked given co-ordinates

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