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© Boardworks Ltd 2005 1 of 50 © Boardworks Ltd 2005 1 of 50 AS-Level Maths: Core 1 for Edexcel C1.2 Algebra and functions 2 This icon indicates the slide.

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Presentation on theme: "© Boardworks Ltd 2005 1 of 50 © Boardworks Ltd 2005 1 of 50 AS-Level Maths: Core 1 for Edexcel C1.2 Algebra and functions 2 This icon indicates the slide."— Presentation transcript:

1 © Boardworks Ltd 2005 1 of 50 © Boardworks Ltd 2005 1 of 50 AS-Level Maths: Core 1 for Edexcel C1.2 Algebra and functions 2 This icon indicates the slide contains activities created in Flash. These activities are not editable. For more detailed instructions, see the Getting Started presentation.

2 © Boardworks Ltd 2005 2 of 50 Contents © Boardworks Ltd 2005 2 of 50 Quadratic expressions Factorizing quadratics Completing the square Solving quadratic equations The discriminant Graphs of quadratic functions Examination-style questions

3 © Boardworks Ltd 2005 3 of 50 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:

4 © Boardworks Ltd 2005 4 of 50 Contents © Boardworks Ltd 2005 4 of 50 Factorizing quadratics Quadratic expressions Factorizing quadratics Completing the square Solving quadratic equations The discriminant Graphs of quadratic functions Examination-style questions

5 © Boardworks Ltd 2005 5 of 50 Factorizing quadratic expressions Factorizing an expression is the inverse of expanding it. Expanding or multiplying out Factorizing When we expand an expression we multiply out the brackets. ( x + 1)( x + 2) x 2 + 3 x + 2 When we factorize an expression we write it with brackets.

6 © Boardworks Ltd 2005 6 of 50 Factorizing quadratic expressions Quadratic expressions of the form ax 2 + bx can always be factorized by taking out the common factor x. For example: No constant term 3 x 2 – 5 x = x (3 x – 5) When a quadratic has no term in x and the other two terms can be written as the difference between two squares, we can use the identity The difference between two squares a 2 – b 2 = ( a + b )( a – b ) to factorize it. For example: 9 x 2 – 49 =(3 x + 7)(3 x – 7)

7 © Boardworks Ltd 2005 7 of 50 Factorizing quadratic expressions Quadratic expressions of the form x 2 + bx + c can be factorized if they can be written using brackets as ( x + d )( x + e ) where d and e are integers. If we expand ( x + d )( x + e ), we have ( x + d )( x + e ) = x 2 + dx + ex + de = x 2 + ( d + e ) x + de Quadratic expressions with a = 1

8 © Boardworks Ltd 2005 8 of 50 Factorizing quadratic expressions Quadratic expressions of the general form ax 2 + bx + c can be factorized if they can be written using brackets as ( dx + e )( fx + g ) where d, e, f and g are integers. If we expand ( dx + e )( fx + g ), we have ( dx + e )( fx + g )= dfx 2 + dgx + efx + eg = dfx 2 + ( dg + ef ) x + eg The general form

9 © Boardworks Ltd 2005 9 of 50 Contents © Boardworks Ltd 2005 9 of 50 Completing the square Quadratic expressions Factorizing quadratics Completing the square Solving quadratic equations The discriminant Graphs of quadratic functions Examination-style questions

10 © Boardworks Ltd 2005 10 of 50 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 How could the quadratic expression x 2 + 8 x be made into a perfect square? We could add 16 to it. In general: x 2 + 2 ax + a 2 = ( x + a ) 2 or x 2 – 2 ax + a 2 = ( x – a ) 2

11 © Boardworks Ltd 2005 11 of 50 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:

12 © Boardworks Ltd 2005 12 of 50 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

13 © Boardworks Ltd 2005 13 of 50 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:

14 © Boardworks Ltd 2005 14 of 50 Completing the square Complete the square for x 2 + 12 x – 5. Compare this expression to ( x + 6) 2 = x 2 + 12 x + 36 = ( x + 6) 2 – 41 x 2 + 12 x – 5 = x 2 + 12 x + 36 – 36 – 5 Complete the square for x 2 – 5 x + 7. Compare this expression to

15 © Boardworks Ltd 2005 15 of 50 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:

16 © Boardworks Ltd 2005 16 of 50 Completing the square Complete the square for 5 + 6 x – 3 x 2. By completing the square, x 2 – 2 x = ( x – 1) 2 – 1 so 5 + 6 x – 3 x 2 = 5 – 3(( x – 1) 2 – 1) = 5 – 3( x – 1) 2 + 3 = 8 – 3( x – 1) 2 Take out the coefficient of x 2 as a factor from the terms in x : 5 + 6 x – 3 x 2 = 5 – 3(–2 x + x 2 ) = 5 – 3( x 2 – 2 x )

17 © Boardworks Ltd 2005 17 of 50 Complete the square

18 © Boardworks Ltd 2005 18 of 50 Contents © Boardworks Ltd 2005 18 of 50 Solving quadratic equations Quadratic expressions Factorizing quadratics Completing the square Solving quadratic equations The discriminant Graphs of quadratic functions Examination-style questions

19 © Boardworks Ltd 2005 19 of 50 Quadratic equations Quadratic equations can be solved by: completing the square, or factorization using the quadratic formula. ax 2 + bx + c = 0 (where a ≠ 0) The general form of a quadratic equation in x is: The solutions to a quadratic equation are called the roots of the equation. A quadratic equation may have: one repeated root, or two real distinct roots no real roots.

20 © Boardworks Ltd 2005 20 of 50 The roots of a quadratic equation If we sketch the graph of a quadratic function y = ax 2 + bx + c the roots of the equation coincide with the points where the function cuts the x -axis. As can be seen here, this can happen twice, once or not at all.

21 © Boardworks Ltd 2005 21 of 50 Solving quadratic equations by factorization Start by rearranging the equation so that the terms are on the left-hand side: Factorizing the left-hand side gives us Solve the equation 5 x 2 = 3 x 5 x 2 – 3 x = 0 x (5 x – 3) = 0 or5 x – 3 = 0 5 x = 3 x = 0So Don’t divide through by x !

22 © Boardworks Ltd 2005 22 of 50 Solving quadratic equations by factorization Start by rearranging the equation so that the terms are on the left-hand side. We need to find two integers that add together to make –5 and multiply together to make 4. Factorizing the left-hand side gives us Solve the equation x 2 – 5 x = –4 by factorization. x 2 – 5 x + 4 = 0 ( x – 1)( x – 4) = 0 x – 1 = 0or x – 4 = 0 x = 4 Because 4 is positive and –5 is negative, both the integers must be negative. These are –1 and –4. x = 1

23 © Boardworks Ltd 2005 23 of 50 Solving quadratics by completing the square Quadratic equations that cannot be solved by factorization 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 –

24 © Boardworks Ltd 2005 24 of 50 Solving quadratics by completing the square Solve the equation 2 x 2 – 4 x + 1 = 0 by completing the square. Write the answer to 3 significant figures. = 2(( x – 1) 2 – 1) + 1 = 2( x – 1) 2 – 2 + 1 = 2( x – 1) 2 – 1 Start by completing the square for 2 x 2 – 4 x + 1: 2 x 2 – 4 x + 1 = 2( x 2 – 2 x ) + 1

25 © Boardworks Ltd 2005 25 of 50 Solving quadratics by completing the square 2( x – 1) 2 = 1 2( x – 1) 2 – 1 = 0 x = 1.71 x = 0.293 (to 3 s.f.) Now solving the equation 2 x 2 – 4 x + 1 = 0: or

26 © Boardworks Ltd 2005 26 of 50 Using the quadratic equation formula Any quadratic equation of the form can be solved by substituting the values of a, b and c into the formula ax 2 + bx + c = 0 This formula can be derived by completing the square on the general form of the quadratic equation.

27 © Boardworks Ltd 2005 27 of 50 Using the quadratic formula Use the quadratic formula to solve 2 x 2 + 5 x – 1 = 0. 2 x 2 + 5 x – 1 = 0 x = 0.186 x = –2.69 (to 3 s.f.) or

28 © Boardworks Ltd 2005 28 of 50 Using the quadratic formula Use the quadratic formula to solve 9 x 2 – 12 x + 4 = 0. 9 x 2 – 12 x + 4 = 0 There is one repeated root:

29 © Boardworks Ltd 2005 29 of 50 Equations that reduce to a quadratic form Some equations, although not quadratic, can be written in quadratic form by using a substitution. For example: Solve the equation t 4 – 5 t 2 + 6 = 0. This is an example of a quartic equation in t. Let’s substitute x for t 2 : x 2 – 5 x + 6 = 0 This gives us a quadratic equation that can be solved by factorization: ( x – 2)( x – 3) = 0 x = 2 So t 2 = 2 t = or x = 3 t 2 = 3 t = or

30 © Boardworks Ltd 2005 30 of 50 Contents © Boardworks Ltd 2005 30 of 50 The discriminant Quadratic expressions Factorizing quadratics Completing the square Solving quadratic equations The discriminant Graphs of quadratic functions Examination-style questions

31 © Boardworks Ltd 2005 31 of 50 The discriminant By solving quadratic equations using the formula we can see that we can use the expression under the square root sign, b 2 – 4 ac, to decide how many roots there are. When b 2 – 4 ac > 0, there are two real distinct roots. When b 2 – 4 ac = 0, there is one repeated root: When b 2 – 4 ac < 0, there are no real roots. Also, when b 2 – 4 ac is a perfect square, the roots of the equation will be rational and the quadratic will factorize. b 2 – 4 ac is called the discriminant of ax 2 + bx + c

32 © Boardworks Ltd 2005 32 of 50 The discriminant We can demonstrate each of these possibilities graphically.

33 © Boardworks Ltd 2005 33 of 50 Contents © Boardworks Ltd 2005 33 of 50 Graphs of quadratic functions Quadratic expressions Factorizing quadratics Completing the square Solving quadratic equations The discriminant Graphs of quadratic functions Examination-style questions

34 © Boardworks Ltd 2005 34 of 50 Plotting graphs of quadratic functions Plot the graph of y = x 2 – 4 x + 2 for –1 < x < 5. We can plot the graph of a quadratic function using a table of values. For example: x x2x2 – 4 x + 2 y = x 2 – 4 x + 2 –1012345 101491625 + 4+ 0– 4– 8– 12– 16– 20 + 2 72–1–2–127 y = ax 2 + bx + c (where a ≠ 0) A quadratic function in x can be written in the form:

35 © Boardworks Ltd 2005 35 of 50 x 1 01–13 4 5 2 3 4 5 6 y 2 Plotting graphs of quadratic functions x y = x 2 – 4 x + 2 –1012345 72 –2–127 The points given in the table are plotted … … and the points are then joined together with a smooth curve. The shape of this curve is called a parabola. It is characteristic of a quadratic function.

36 © Boardworks Ltd 2005 36 of 50 Parabolas When the coefficient of x 2 is positive the vertex is a minimum point and the graph is  -shaped. When the coefficient of x 2 is negative the vertex is a maximum point and the graph is  -shaped. Parabolas have a vertical axis of symmetry … …and a turning point called the vertex.

37 © Boardworks Ltd 2005 37 of 50 Exploring graphs of the form y = ax 2 + bx + c

38 © Boardworks Ltd 2005 38 of 50 Sketching graphs of quadratic functions When a quadratic function factorizes we can use its factorized form to find where it crosses the x -axis. For example: Sketch the graph of the function y = x 2 – 2 x – 3. The function crosses the x -axis when y = 0. x 2 – 2 x – 3 = 0 ( x + 1)( x – 3) = 0 x + 1 = 0or x – 3 = 0 x = 3 The function crosses the x -axis at the points (–1, 0) and (3, 0). x = –1

39 © Boardworks Ltd 2005 39 of 50 Sketching graphs of quadratic functions By putting x = 0 in y = 2 x 2 – 5 x – 3 we can also find where the function crosses the y -axis. y = 2(0) 2 – 5(0) – 3 y = – 3 So the function crosses the y -axis at the point (0, –3). The quadratic function y = ax 2 + bx + c will cross the y -axis at the point (0, c ). We now know that the function y = x 2 – 2 x – 3 passes through the points (–1, 0), (3, 0) and (0, –3) and so we can place these points on our sketch. In general:

40 © Boardworks Ltd 2005 40 of 50 Sketching graphs of quadratic functions 0 y x (–1, 0) (3, 0) (0, –3) (1, –4) We can also use the fact that a parabola is symmetrical to find the coordinates of the vertex. The x coordinate of the vertex is half-way between –1 and 3. When x = 1, y = (1) 2 – 2(1) – 3 y = –4 So the coordinates of the vertex are (1, –4). We can now sketch the graph.

41 © Boardworks Ltd 2005 41 of 50 Sketching graphs of quadratic functions When a quadratic function is written in the form y = a ( x – α )( x – β ), it will cut the x -axis at the points ( α, 0) and ( β, 0). α and β are the roots of the quadratic function. When a quadratic function is written in the form y = a ( x – α )( x – β ), it will cut the x -axis at the points ( α, 0) and ( β, 0). α and β are the roots of the quadratic function. For example, write the quadratic function y = 3 x 2 + 4 x – 4 in the form y = a ( x – α )( x – β ) and hence find the roots of the function. This function can be factorized as follows, y = (3 x – 2)( x + 2) It can be written in the form y = a ( x – p )( x – q ) as Therefore, the roots are In general:

42 © Boardworks Ltd 2005 42 of 50 Exploring graphs of the form y = a ( x – α )( x – β )

43 © Boardworks Ltd 2005 43 of 50 Sketching graphs by completing the square When a function does not factorize we can write it in completed square form to find the coordinates of the vertex. For example: 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.

44 © Boardworks Ltd 2005 44 of 50 Sketching graphs by completing the square 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 (–1, 0). Using symmetry we can now sketch the graph. y x 0 (–1, 0) x = –2 y = x 2 + 4 x – 1 (–2, –5)

45 © Boardworks Ltd 2005 45 of 50 Sketching graphs by completing the square In general, when the quadratic function y = ax 2 + bx + c is written in completed square form as a ( x + p ) 2 + q 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.

46 © Boardworks Ltd 2005 46 of 50 Exploring graphs of the form y = a ( x + p ) 2 + q

47 © Boardworks Ltd 2005 47 of 50 Contents © Boardworks Ltd 2005 47 of 50 Examination-style questions Quadratic expressions Factorizing quadratics Completing the square Solving quadratic equations The discriminant Graphs of quadratic functions Examination-style questions

48 © Boardworks Ltd 2005 48 of 50 Examination-style question a) Write 2 x 2 – 8 x + 7 in the form a ( x + b ) 2 + c. b) Write down the minimum value of f ( x ) = 2 x 2 – 8 x + 7 and state the minimum value of x where this occurs. c) Solve the equation 2 x 2 – 8 x + 7 = 0 leaving your answer in surd form. d) Sketch the graph of y = 2 x 2 – 8 x + 7. a)2 x 2 – 8 x + 7 = 2( x 2 – 4 x ) + 7 = 2(( x – 2) 2 – 4) + 7 = 2( x – 2) 2 – 8 + 7 = 2( x – 2) 2 – 1

49 © Boardworks Ltd 2005 49 of 50 Examination-style question b) From this we can see that the minimum value of f ( x ) is –1. f ( x ) can be written as f ( x ) = 2( x – 2) 2 – 1 This occurs when x = 2. c)2 x 2 – 8 x + 7 = 0 2( x – 2) 2 – 1 = 0 2( x – 2) 2 = 1 ( x – 2) 2 = x – 2 = x = 2 or x = 2

50 © Boardworks Ltd 2005 50 of 50 Examination-style question d)When y = 0, x = 2 – or x = 2 + When x = 0, y = 7 So the graph cuts the coordinate axes at (2 +, 0), (2 –, 0) and (0, 7). The parabola has a minimum at the point (2, –1). y x 7 –1 2 +2 –


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