1. Topic 3 Introduction to Functions II

2.  Distinction between functions and relations (covered in Topic 1)  Distinction between continuous functions, discontinuous functions and discrete functions.  Practical applications of quadratic functions, the reciprocal function and inverse variation  Relationship between the graph of f(x) and the graphs of f(x)+a, f(x+a), af(x), f(ax) for both positive and negative values of the constant a  General shapes of graphs of absolute value functions, the reciprocal function and polynomial functions up to and including the fourth degree  Algebraic and graphical solution of two simultaneous equations in two variables (to be applied to only linear and quadratic functions (covered in Topic 1)  Concept of the inverse of a function  Composition of two functions

3. Continuous functions and discontinuous functions With a continuous function, you do not need to lift your pencil in order to draw the graph:

4. Continuous functions and discontinuous functions With a continuous function, you do not need to lift your pencil in order to draw the graph:

5. With a continuous function, you need to lift your pencil With a discontinuous function, you need to lift your pencil in order to draw the graph: ● ●

6. Discrete Functions Consider the table below which shows the maximum temperature in Brisbane for the first week of July. Date in July 1234567 Maximum Temperature (° C) 16181213172011 This is a finite sequence. Also there is no relationship between successive terms

7. Discrete Functions Date in July 1234567 Maximum Temperature (° C) 16181213172011 This is a finite sequence. Also there is no relationship between successive terms

8.  On the other hand, consider this sequence 1, 3, 5, 7, 9, …  It represents the odd numbers and is an infinite series. In this sequence it is easy to work out the pattern.  Each number in this sequence is called a term. Rather than using function notation, the terms are shown using subscript notation.  e.g. t 3 = 5

9.  Each number in this sequence is called a term. Rather than using function notation, the terms are shown using subscript notation.  e.g. t 3 = 5  The n th term of the sequence, t n, is refered to as the general term. In this example, the general term is… t n = 2n – 1

10. Model: For the sequence t n = 3n + 4 : (a) write down the first 4 terms (b) which term is equal to 49? (a) t n = 3n + 4 t 1 = 3 × 1 + 4 = 7 t 2 = 3 × 2 + 4 = 10 t 3 = 3 × 3 + 4 = 13 t 4 = 3 × 4 + 4 = 16

11. Model: For the sequence t n = 3n + 4 : (a) write down the first 4 terms (b) which term is equal to 49? (b) t n = 3n + 4 49 = 3n + 4 n = (49 – 4) ÷ 3 n = (49 – 4) ÷ 3 n = 15 n = 15 Therefore 49 is the 15 th term.

12. Exercise Page 132 Ex 4.5 1 – 4

13. Quadratic Functions Model: Use a graph to find values of x such that (a) x 2 – x – 6 = 0 (b) x 2 + 5x + 2 = 0 (b) x 2 + 5x + 2 = 0 How can we do this? Where a graph cuts the x-axis is called a root or a zero of the equation.

14. A quadratic equation may have 2 roots 2 roots or 1 root or no roots

15. Model: Use a GC to draw a graph and find approximate roots for the equation 4x 2 – 3x = 10 Show how to find roots by using the “table” on the GC and changing the increments to smaller and smaller sizes.

16. Exercise Page 237 Ex 7.1 1 2 3 a,b,h,j

17. Review Factorising  Exercise Page 240 Ex 7.2 1-6 (a couple from each)

18. Completing the Square Solve x 2 – 4x -12 = 0 by completing the square. x 2 – 4x -12 = 0 x 2 – 4x = 12 x 2 – 4x + (-2) 2 = 12 + 4 (x-2) 2 = 16 x-2 =  4 x-2 =  4 x = 2  4 x = 2  4 x = 6 or x = -2 x = 6 or x = -2

19. Solve x 2 + 6x – 2 = 0 by completing the square. x 2 + 6x = 2 x 2 + 6x + 9 = 2 + 9 (x+3) 2 = 11 x+3 =  11 x+3 =  11 x = -3  11 x = -3  11 x =.317 or x = -6.317 (3dp) x =.317 or x = -6.317 (3dp)

20. Solve x 2 + 5x – 2 = 0 by completing the square. x 2 + 5x = 2 x 2 + 5x + 6.25 = 2 + 6.25 (x+2.5) 2 = 8.25 x+2.5 =  8.25 x+2.5 =  8.25 x = -2.5  8.25 x = -2.5  8.25 x =.372 or x = -5.372 (3dp) x =.372 or x = -5.372 (3dp)

21. Solve 2x 2 + 12x – 5 = 0 by completing the square. 2x 2 + 12x = 5 x 2 + 6x = 2.5 x 2 + 6x + 9 = 2.5 + 9 (x+3) 2 = 11.5 x+3 =  11.5 x+3 =  11.5 x = -3  11.5 x = -3  11.5 x =.391 or x = -6.391 (3dp) x =.391 or x = -6.391 (3dp)

22. Does this look familiar ?

23. Exercise Page 240 Ex 7.2 1-6 (a couple from each) 7-10 Page 242 Ex 7.3 3(a-g completing the square), 4-8

24. Your GC can be used to look at a family of curves: e.g.y = x 2, y = x 2 + 3, y = x 2 – 3 e.g. y = x 2, y = (x-3) 2, y = (x+3) 2

25. Model: Find the turning point of the following curves by completing the square: (a) y = x 2 – 4x + 7 (b) y = 2x 2 + 6x - 5 (a) y = x 2 – 4x + 7 = x 2 – 4x + (-2) 2 + 7 – 4 = x 2 – 4x + (-2) 2 + 7 – 4 = (x-2) 2 + 3 = (x-2) 2 + 3 Now y will be as small as y can be when x = 2 When x = 2, y = 3 ∴ (2,3) is a turning point

26. Model: Find the turning point of the following curves by completing the square: (a) y = x 2 – 4x + 7 (b) y = 2x 2 + 12x - 5 (b) y = 2x 2 - 12x - 5 = 2(x 2 - 6x) – 5 = 2(x 2 - 6x) – 5 = 2[x 2 - 3x + (-3) 2 ] – 5 – 2 × 9 = 2[x 2 - 3x + (-3) 2 ] – 5 – 2 × 9 = 2(x - 3) 2 – 23 = 2(x - 3) 2 – 23 Now y will be as small as y can be when x = 3 When x = 3, y = -23 ∴ (3,-23) is a turning point

27. y = -3x 2 + 12x + 7

28. xtxt

29. i.e. Turning points occur when x = -b 2a 2a Check that this is so in previous model

30. Exercise Page 246 Ex 7.4 No. 2, 3, 4 Page 251 Ex 7.5 7-9

31. Solving simultaneous equations Solve y = x 2 – 4x + 6 and 2x – y = 3 y = x 2 – 4x + 6 ….(1) 2x – y = 3 ….(2) (2)  y = 2x – 3 ….(3) Equating (1) and (3) x 2 – 4x + 6 = 2x – 3 x 2 - 6x + 9 = 0 (x-3) 2 = 0  x = 3  y = 3 (Check graphically)

32. Exercise Page 254 Ex 7.6

33. Inverse Variation  As one quantity increases the other decreases.  Consider 100 papers to be picked up from off the school oval. The more students involved, the less each picks up. No. of Students 124510202550 Papers each picks up

34. Inverse Variation  As one quantity increases the other decreases.  Consider 100 papers to be picked up from off the school oval. The more students involved, the less each picks up. No. of Students 124510202550 Papers each picks up 10050252010542

35. No. of Students 124510202550 Papers each picks up 10050252010542 X times Y 100100100100100100100100

36.  Inverse variations always fit the pattern  These graphs will always have the general shape…

37.  Our example obviously only relates to x > 0  The graph above (y=1/x) is a discontinuous graph. It has a discontinuity (does not exist) at x=0.

38. Exercise Year 10 Set 12 E & F

39. Model: For the function (a)Graph the function (b)State the domain and range There will be a discontinuity when t-3=0  t = 3 is an asymptote Horizontal asymptote at y = -4 As t ↑ 3, h(t) → ∞ As t ↓ 3, h(t) → - ∞ When t = 0, h(t) = -3⅓ When h(t) = 0, t = 2½ -3⅓ 2½2½ Domain: all real numbers except 3 Range: all real numbers except -4

40. Attempt to graph Domain: any real, except -2Range: y > 3

41. Exercise Photocopied sheets of “Graphing a Function”

42. Inverse of a Function  Take a function and swap the x and y pronumerals. e.g. y = 3x + 2becomesx = 3y + 2 y = 3x + 2becomesx = 3y + 2 rearrange(x – 2)/3 = y y = ⅓x - ⅔  Graph both functions

43. These lines are a reflection in the line y = x

44. Exercise Photocopied sheets of “Inverse of a Function”

45. Composition of Two Functions  Consider f(x) = x 2 – 3x + 2 f(3)= 3 2 – 3 × 3 + 2 = 2 = 2 f(a) = a 2 – 3a + 2  If g(x) = 2x + 1 f [g(x)] = f( 2x + 1) = (2x+1) 2 – 3(2x+1) + 2 = 4x 2 + 4x + 1 – 6x – 3 + 2 = 4x 2 – 2x

46. Try These … 1. f(x) = x 2 + 5x – 2 g(x) = 3x – 2  Find 1. f [g(x)] 2. g [f(x)] 3. g [g(x)] 2. f(x) = 2x 2 + 5x – 4 g(x) = 5 – 3x  Find 1. f [g(x)] 2. g [f(x)] 3. g [g(x)]

47. Try These … 1. f(x) = x 2 + 5x – 2 g(x) = 3x – 2  Find 1. f [g(x)] = 9x 2 + 3x – 8 2. g [f(x)] = 3x 2 + 15x – 8 3. g [g(x)] = 9x – 8 2. f(x) = 2x 2 + 5x – 4 g(x) = 5 – 3x  Find 1. f [g(x)] = 18x 2 – 75x + 71 2. g [f(x)] = 17 – 15x – 6x 2 3. g [g(x)] = 9x – 10

48. When functions are contained within other functions, they can be simplified as follows. y = (x 2 - 5) 2 - 5(x 2 - 5) + 6 x 2 - 5 is a function,  let u = x 2 - 5 y = u 2 - 5u + 6

49. When functions are contained within other functions, they can be simplified as follows. y = (x 2 - 5) 2 - 5(x 2 - 5) + 6 x 2 - 5 is a function,  let u = x 2 - 5 y = u 2 - 5u + 6  solve (x 2 - 5) 2 - 5(x 2 - 5) + 6 = 0

50. Function of a Function handout sheets