1. Objectives: Be able to….. Distinguish between rational and irrational numbers. Understand what is meant by a surd. Simplify expressions involving surds. Perform arithmetic with surds.

2. Objectives: Be able to….. Use language of coordinate geometry. Find the distance between two given points. Find the mid points of a line segment joining two given points. Find, use and interpret the gradient of a line segment. Know the relationship between the gradients for parallel and perpendicular lines. Rearrange equation to show line in for Ax + By + C = 0 and vice versa. Find the equations of straight lines given – (a) the gradient and a point – (b) two points Verify, given their coordinates, that points lie on a line. Find the coordinates of a point of intersection of two lines. Find the fourth vertex of a parallelogram given the other three.

3. Chapter 4: Quadratic functions and there graphs. Learning objectives: To be able to… Solve quadratic equations. Complete the square of a quadratic expression. Understand the effect of a translation on a quadratic graph. Find the equation of a parabola when translated by a given vector. Understand the term discriminant.

4. What do these have in common?

5. Quadratic expressions A quadratic expression is an expression in which the highest power of the variable is 2. For example: x 2 – 2 w 2 + 3 w + 14 – 5 g 2 t2t2 2 x is a variable. a is the coefficient of x 2. b is the coefficient of x. c is a constant term. ax 2 + bx + c (where a ≠ 0) The general form of a quadratic expression in x is:

6. Factorising quadratic expressions Factorising an expression is the inverse of expanding it. Expanding or multiplying out factorising When we expand an expression we multiply out the brackets. ( x + 1)( x + 2) x 2 + 3 x + 2 When we factorise an expression we write it with brackets.

7. Factorising Quadratics – The Process 3x 2 - 7x + 2 Step 1: Find factor pairs of 3 x 2 that sum -7 6 -1&-6 3x 2 -1x -6x + 2 Step 2: Rewrite expression with factor pairs Step 3: Factorise the two halves x(3x -1) -2(3x-1) Step 4: Put Into brackets (3x-1) (x-2)

8. Factorising Quadratics – The Process 4x 2 + x - 3 Step 1: Find factor pairs of 4 x -3 that sum +1 -12 -3&4 4x 2 -3x +4x -3 Step 2: Rewrite expression with factor pairs Step 3: Factorise the two halves x(4x -3) +1(4x-3) Step 4: Put Into brackets (4x-3) (x+1)

9. Factorising Quadratics – The Process -8x 2 -14 x +15 Step 1: Find factor pairs of-8 x 15 that sum -14 -120 -6&20 -8x 2 +6x -20x+15 Step 2: Rewrite expression with factor pairs Step 3: Factorise the two halves -2x(4x -3) -5(4x-3) Step 4: Put Into brackets (4x-3) (-2x-5)

10. x-2012345 x2x2 4101491625 x 2 -3x1040-2 0410 x 2 -3x – 460-4-6 -406 y60 -6 -406 Significant points on a Quadratic Curve that crosses the x axis. Sketch the graph y = x 2 -3x – 4 The Y INTERCEPT = C /constant. (Where X = 0) Find from equation. The ROOTS = the x intercepts. (Where Y = 0) Find from table and or graph. The VERTEX = the maximum or minimum point on the graph. Find by substituting the X value, that is the  way between the roots, to find Y.

11. The quadratic curve y = -x 2 + 1

12. Perfect squares Some quadratic expressions can be written as perfect squares. For example: x 2 + 2 x + 1 = ( x + 1) 2 x 2 + 4 x + 4 = ( x + 2) 2 x 2 + 6 x + 9 = ( x + 3) 2 x 2 – 2 x + 1 = ( x – 1) 2 x 2 – 4 x + 4 = ( x – 2) 2 x 2 – 6 x + 9 = ( x – 3) 2

13. Completing the square Adding 16 to the expression x 2 + 8 x to make it into a perfect square is called completing the square. x 2 + 8 x = x 2 + 8 x + 16 – 16We can write If we add 16 we then have to subtract 16 so that both sides are still equal. By writing x 2 + 8 x + 16 we have completed the square and so we can write this as x 2 + 8 x = ( x + 4) 2 – 16 In general:

14. Completing the square Complete the square for x 2 – 10 x. Compare this expression to ( x – 5) 2 = x 2 – 10 x + 25 = ( x – 5) 2 – 25 x 2 – 10 x = x 2 – 10 x + 25 – 25 Complete the square for x 2 + 3 x. Compare this expression to

15. Completing the square How can we complete the square for x 2 – 8 x + 7? = ( x – 4) 2 – 9 x 2 – 8 x + 7 = x 2 – 8 x + 16 – 16 + 7 Look at the coefficient of x. This is –8 so compare the expression to ( x – 4) 2 = x 2 – 8 x + 16. In general:

16. Completing the square When the coefficient of x 2 is not 1, quadratic equations in the form ax 2 + bx + c can be rewritten in the form a ( x + p ) 2 + q by completing the square. Complete the square for 2 x 2 + 8 x + 3. 2 x 2 + 8 x + 3 = 2( x 2 + 4 x ) + 3 By completing the square, x 2 + 4 x = ( x + 2) 2 – 4 so 2 x 2 + 8 x + 3 = 2(( x + 2) 2 – 4) + 3 = 2( x + 2) 2 – 8 + 3 = 2( x + 2) 2 – 5 Take out the coefficient of x 2 as a factor from the terms in x :

17. Solving quadratics by completing the square Quadratic equations that cannot be solved by factorisation can be solved by completing the square. For example, the quadratic equation x 2 – 4 x – 3 = 0 can be solved by completing the square as follows: x = 4.65 x = –0.646 (to 3 s.f.) ( x – 2) 2 – 7 = 0 ( x – 2) 2 = 7 x – 2 = x = 2 +or x = 2 –

18. Sketching graphs by completing the square In general, when the quadratic function y = ax 2 + bx + c is written in completed square form as The coordinates of the vertex will be (– p, q ). The axis of symmetry will have the equation x = – p. Also: If a > 0 (– p, q ) will be the minimum point. If a < 0 (– p, q ) will be the maximum point. Plotting the y -intercept, (0, c ) will allow the curve to be sketched using symmetry.

19. When y = –5, we have,( x + 2) 2 – 5 = –5 ( x + 2) 2 = 0 x = –2 The coordinates of the vertex are therefore (–2, –5). The equation of the axis of symmetry is x = –2. Also, when x = 0 we have y = –1 So the curve cuts the y-axis at the point (0, -1). Using symmetry we can now sketch the graph. y x 0 (0, -1) x = –2 y = x 2 + 4 x – 1 (–2, –5) Sketch the graph of y = x 2 + 4 x – 1 by writing it in completed square form. x 2 + 4 x – 1 = ( x + 2) 2 – 5 The least value that ( x + 2) 2 can have is 0 because the square of a number cannot be negative. ( x + 2) 2 ≥ 0 ( x + 2) 2 – 5 ≥ – 5 Therefore The minimum value of the function y = x 2 + 4 x – 1 is therefore y = –5. Solving (x + 2) 2 – 5 = 0 gives the roots as x = ±√5 – 2.

20. Translations of Parabola (Movement of Curve) Vector [] Movement of graph

21. Translations of Parabola (Movement of Curve) Vector [] Movement of graph

22. Quadratic Graphs y = x 2 y = x 2 +1 y=x 2 -1 x-2012 x2x2 41014 y41014 x-2012 x2x2 41014 x 2 +152125 y52125 x-2012 x2x2 41014 x 2 -13003 y30 03 y = x 2 y = x 2 +1 y = x 2 - 1 What do you notice about the curves? What do you notice about where the lines cross the y-axis?

23. Quadratic Graphs y = x 2 y =  x 2 y=2x 2 x-2012 x2x2 41014 y41014 x-2012 x2x2 41014 x2x2 2  0  2 y2  0  2 x-2012 x2x2 41014 2x 2 82028 y82028 y = x 2 y = 2x 2 y =  x 2 What do you notice curves? What do you notice about where the lines cross the y-axis? 6 7 8

24. Sketching a curve - process

25. Objective Practise and consolidate graph sketching. Be able to translate cubic graphs. Sketch curves of functions showing points of intersection.

26. Key points in sketching quadratic graphs ax + bx + c Shape a Positive Pothole / Minus Mountain Y Intercept = c X Intercept (Roots) from Discriminant / Solving – b 2 – 4ac > 0 two roots (cuts x axis) factorise/formula/square – b 2 – 4ac = 0 one repeated root (vertex touches x axis) – b 2 – 4ac < 0 no real roots (vertex above/below x axis) Vertex (turning point → positive/min or negative/max) – Symmetry / half way between roots for x then substitute for y. – Completing the square

27. The quadratic formula

28. Learning Objectives: Be able to… Understand the vocabulary of polynomials Add subtract and multiply polynomials. Find, compare and equate coefficients of polynomials.

29. Vocabulary, definitions and conventions. An expression comprising constants, variables and powers / indices. The powers can only be positive integers : x 2 − 4x + 7 is a polynomial, but x 2 − 4/x + 7x  is not. The degree or order of the polynomial is given by the highest power of the variable, so; CONSTANT (Polynomial of degree 0)= 5 LINEAR (Polynomial in m of order 1)= 2m + 3 QUADRATIC (Polynomial in x of order 2)= 3x 2 + 4x – 4 CUBIC (Polynomial in g of order 3) = 5g 3 – 8g – 3 QUARTIC (Polynomial in f of order 4) = f 4 + 3f + 7 The normal convention is that the polynomial is written in descending powers of the variable. Polynomials can be added or subtracted by collecting like terms. They can be multiplied using a variety of methods. Coefficient means the constant preceding a variable.

30. Adding and subtracting polynomials. Simply follow normal rules of column addition / subtraction and simplify by collecting like terms. Make sure that polynomials are lined up in correct order in columns. Be careful when handling negative numbers. X 4 + 3x + 7 + 3x 2 + 4x – 4 X 4 +3x 2 + 7x + 3 X 4 + 3x + 7 - 3x 2 + 4x – 4 X 4 -3x 2 - x + 11

31. x²x²3x3x-2 2x²2x² -x-x 4 Multiply x² + 3x 3x – 2 by 2x² – x + 4 Multiplying polynomials – using a grid.

32. x²x²3x3x-2 2x²2x² -x-x 4 Multiply x² by 2x²2x² Multiplying polynomials – using a grid.

33. x²x²3x3x-2 2x²2x²2x42x4 -x-x 4 Multiply x² by 2x² Multiplying polynomials – using a grid.

34. x²x²3x3x-2 2x²2x²2x42x4 -x-x 4 Multiply 3x 3x by 2x²2x² Multiplying polynomials – using a grid.

35. x²x²3x3x-2 2x²2x²2x42x4 6x³6x³ -x-x 4 Multiply 3x by 2x² Multiplying polynomials – using a grid.

36. x²x²3x3x-2 2x²2x²2x42x4 6x³6x³ -x-x 4 Fill in the rest of the table in the same way Multiplying polynomials – using a grid.

37. x²x²3x3x-2 2x²2x²2x42x4 6x³6x³-4x² -x-x 4 Fill in the rest of the table in the same way Multiplying polynomials – using a grid.

38. x²x²3x3x-2 2x²2x²2x42x4 6x³6x³-4x² -x-x-x³-x³ 4 Fill in the rest of the table in the same way Multiplying polynomials – using a grid.

39. x²x²3x3x-2 2x²2x²2x42x4 6x³6x³-4x² -x-x-x³-x³-3x² 4 Fill in the rest of the table in the same way Multiplying polynomials – using a grid.

40. x²x²3x3x-2 2x²2x²2x42x4 6x³6x³-4x² -x-x-x³-x³-3x²2x2x 4 Fill in the rest of the table in the same way Multiplying polynomials – using a grid.

41. x²x²3x3x-2 2x²2x²2x42x4 6x³6x³-4x² -x-x-x³-x³-3x²2x2x 44x²4x² Fill in the rest of the table in the same way Multiplying polynomials – using a grid.

42. x²x²3x3x-2 2x²2x²2x42x4 6x³6x³-4x² -x-x-x³-x³-3x²2x2x 44x²4x²12x Fill in the rest of the table in the same way Multiplying polynomials – using a grid.

43. x²x²3x3x-2 2x²2x²2x42x4 6x³6x³-4x² -x-x-x³-x³-3x²2x2x 44x²4x²12x-8 Fill in the rest of the table in the same way Multiplying polynomials – using a grid.

44. x²x²3x3x-2 2x²2x²2x42x4 6x³6x³-4x² -x-x-x³-x³-3x²2x2x 44x²4x²12x-8 Now add up all the terms in the table First the term in x4x4 (x² + 3x 3x – 2)(2x² - x + 4) = 2x42x4 Multiplying polynomials – using a grid.

45. x²x²3x3x-2 2x²2x²2x42x4 6x³6x³-4x² -x-x-x³-x³-3x²2x2x 44x²4x²12x-8 Now add up all the terms in the table then the terms in x³x³ (x² + 3x – 2)(2x² - x + 4) = 2x 4 + 5x³5x³ Multiplying polynomials – using a grid.

46. x²x²3x3x-2 2x²2x²2x42x4 6x³6x³-4x² -x-x-x³-x³-3x²2x2x 44x²4x²12x-8 Now add up all the terms in the table then the terms in x²x² (x² + 3x – 2)(2x² - x + 4) = 2x 4 + 5x³- 3x² Multiplying polynomials – using a grid.

47. x²x²3x3x-2 2x²2x²2x42x4 6x³6x³-4x² -x-x-x³-x³-3x²2x2x 44x²4x²12x-8 Now add up all the terms in the table (x² + 3x – 2)(2x² - x + 4) = 2x 4 then the terms in x + 5x³- 3x² + 14x Multiplying polynomials – using a grid.

48. x²x²3x3x-2 2x²2x²2x42x4 6x³6x³-4x² -x-x-x³-x³-3x²2x2x 44x²4x²12x-8 Now add up all the terms in the table (x² + 3x – 2)(2x² - x + 4) = 2x 4 and finally the constant term + 5x³- 3x²+ 14x - 8 Multiplying polynomials – using a grid.

49. x²x²3x3x-2 2x²2x²2x42x4 6x³6x³-4x² -x-x-x³-x³-3x²2x2x 44x²4x²12x-8 (x² + 3x – 2)(2x² - x + 4) = 2x 4 + 5x³- 3x²+ 14x- 8 Multiplying polynomials – using a grid.

50. x²x²2x2x-2 6x²6x²-12x² -3x-6x² 44x²4x² - 14x² Using a grid to find a specific coefficient in a term. Eg: Find the coefficient of x 2 in (x² + 2x – 2)(6x² -3x + 4)

51. Equating coefficients: used to find either a specific coefficient in a term or a unknown polynomial. (x + 6)P(x) = x 3 + 12x 2 + 34x – 12 (x + 6)(ax 2 + bx + c)= x 3 + 12x 2 + 34x – 12 ax 3 + (6a +b)x 2 + (6b + c)x – 6c= x 3 + 12x 2 + 34x – 12 → (equate coefficients of x 3 )a = 1 → (equate coefficients of x 2 )6a + b = 12 because a = 1 → b = 6 → (equate coefficients of x)6b + c = 34 because b = 6 → c = -2 → (equate to constant)6c= -12→ c = -2

52. The Factor Theorem If ( x – a ) is a factor of a polynomial f ( x ) then f ( a ) = 0. The converse is also true: If f ( a ) = 0 then ( x – a ) is a factor of a polynomial f ( x ). This is the Factor Theorem:

53. The Factor Theorem ( x + 2) is a factor of 3 x 2 + 5 x – 2 if f (–2) = 0 f (–2) = 3(–2) 2 + 5(–2) – 2 = 12 – 10 – 2 = 0as required. We can write 3 x 2 + 5 x – 2 = ( x + 2)( ax + b ) Use the factor theorem to show that ( x + 2) is a factor of f ( x ) = 3 x 2 + 5 x – 2. Hence or otherwise, factorize f ( x ). By inspection a = 3 and b = –1 And so3 x 2 + 5 x – 2 = ( x + 2)(3 x – 1)

54. Factorizing polynomials The factor theorem can be used to factorize polynomials by systematically looking for values of x that will make the polynomial equal to 0. For example: Factorize the cubic polynomial x 3 – 3 x 2 – 6 x + 8. Let f ( x ) = x 3 – 3 x 2 – 6 x + 8. f ( x ) has a constant term of 8. So possible factors of f ( x ) are: ( x ± 1),( x ± 2),( x ± 4)( x ± 8)or f (1) = 1 – 3 – 6 + 8 = 0  ( x – 1) is a factor of f ( x ). f (–1) = –1 – 3 + 6 + 8 ≠ 0  ( x + 1) is not a factor of f ( x ). f (2) = 8 – 12 – 12 + 8 ≠ 0  ( x – 2) is not a factor of f ( x ). f (–2) = – 8 – 12 + 12 + 8 = 0  ( x + 2) is a factor of f ( x ). f (4) = 64 – 48 – 24 + 8 = 0  ( x – 4) is a factor of f ( x ). We have found three factors and so we can stop. x 3 – 3 x 2 – 6 x + 8 = ( x – 1)( x + 2)( x – 4) The given polynomial can therefore be fully factorized as:

55. The Factor Theorem We don’t know any other factors but we do know that the expression x + 1 must be multiplied by a quadratic expression to give x 3 + 1. We can therefore write x 3 + 1 = ( x + 1)( ax 2 + bx + c ) We can see immediately that a = 1 and c = 1 so x 3 + 1 = ( x + 1)( x 2 + bx + 1) = x 3 + bx 2 + x + x 2 + bx + 1 = x 3 + ( b + 1) x 2 + ( b + 1) x + 1 Equating coefficients of x 2 gives b + 1 = 0 b = –1 So x 3 + 1 can be fully factorized as x 3 + 1 = ( x + 1)( x 2 – x + 1) f (1) = 1 + 1 ≠ 0  ( x – 1) is not a factor of f ( x ). f (–1) = (–1) 3 + 1 = 0  ( x + 1) is a factor of f ( x ). Factorize f ( x ) = x 3 + 1 f ( x ) has a constant term of 1 so the only possible factors of f ( x ) are ( x – 1) or ( x + 1).

56. Dividing polynomials Suppose we want to divide one polynomial f ( x ) by another polynomial of lower order g ( x ). g ( x ) will divide exactly into f ( x ). In this case, g ( x ) is a factor of f ( x ) and the remainder is 0. There are two possibilities. Either: g ( x ) will leave a remainder when divided into f ( x ). We can use either of two methods to divide one polynomial by another. These are by: using long division, or writing an identity and equating coefficients.

57. Divisor Dividend ) Quotient + Remainder Dividend ÷ Divisor = Quotient + Remainder Dividend = Divisor x Quotient + Remainder Cubic = Linear x Quadratic + Remainder For polynomials

58. Dividing polynomials by long division Using long division The method of long division used for numbers can be applied to the division of polynomial functions. Let’s start by looking at the method for numbers. For example, we can divide 5482 by 15 as follows: This tells us that 15 divides into 5482 365 times, leaving a remainder of 7. We can write or 5482 = 15 × 365 + 7 The dividend 5482 ÷ 15 = 365 remainder 7 The divisor = The quotient × The remainder + 515482 3 –45 98 6 90– 82 5 75– 7

59. Dividing polynomials by long division We can use the same method to divide polynomials. For example: What is f ( x ) = x 3 – x 2 – 7 x + 3 divided by x – 3? x 3 – x 2 – 7 x + 3 x – 3 x 3 – 3 x 2 2x22x2 – 7 x + 2 x 2 x 2 – 6 x – x + 3 – 1 – x + 3 0 This tells us that x 3 – x 2 – 7 x + 3 divided by x – 3 is x 2 + 2 x – 1. The remainder is 0 and so x – 3 is a factor of f ( x ). We can write x 3 – x 2 – 7 x + 3 =( x – 3)( x 2 + 2 x – 1) x2x2 or

60. Dividing polynomials by long division Here is another example: What is f ( x ) = 2 x 3 – 3 x 2 + 1 divided by x – 2? 2 x 3 – 3 x 2 + 0 x + 1 x – 2 2 x 3 – 4 x 2 x2x2 + 0 x + x x 2 – 2 x 2x2x + 1 + 2 2 x – 4 5 This tells us that 2 x 3 – 3 x 2 + 1 divided by x – 2 is 2 x 2 + x + 2 remainder 5. There is a remainder and so x – 2 is not a factor of f ( x ). We can write 2 x 3 – 3 x 2 + 1 =( x – 2)(2 x 2 + x + 2) + 5 2x22x2 or

61. Objective: Sketching a cubic curve - steps Slope:+ve-ve Intercept:where does it cross y axis (x = 0) Roots:where it crosses x axis (y = 0) Factor Theorem Does quadratic factorise Quadratic Formula: Discriminant / Repeat roots Quadrant:Sub for x close to y check slope.

62. y = (2x + 3)(x + 4)(5 – 2x)

63. y = (x + 1)(x 2 –x + 1) Slope:+ve Intercept:y = 1 Roots:x = -1 Quadrant: na

64. y = (2x + 3)(x - 1) 2