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1 MathsStart (NOTE Feb 2013: This is the old version of MathsStart. New books will be created during 2013 and 2014) Topic 3 Quadratic Functions 8 = ( 2)( 4) ( 3) (3, 1) MATHS LEARNING CENTRE Level 3, Hub Central, North Terrace Campus The Universit of Adelaide, SA, 5005 TEL FAX

2 This Topic... This topic introduces quadratic functions, their graphs and their important characteristics. Quadratic functions are widel used in mathematics and statistics. The are found in applied and theoretical mathematics, and are used to model non-linear relationships between variables in statistics. The module covers the algebra and graphing skills needed for analsing and using quadratic functions. Author of Topic 3: Paul Andrew Prerequisites You will need to solve quadratic equations in this module. This is covered in the appendi. Contents Chapter 1 Quadratic Functions and Parabolas. Chapter 2 Sketching Parabolas. Chapter 3 Transformations. Appendices A. Quadratic Equations B. s Printed: 18/02/2013

4 Height Population Quadratic Functions & Parabolas 2 The graph below shows how a population is changing (the dots are data points). In this eample, the relationship between population and time is curved, and is approimated b a quadratic function. We think of the quadratic function as a model for the population deca. The model can be used to predict how fast the population will change in the future. Years A projectile has a parabolic path. The equation of its path can be calculated from three data points, and can be used to estimate its maimum height reached and landing. Distance We use and as the independent and dependent variables when we stud the general properties of quadratic functions, and place no restriction on the domain of the functions. The important characteristics of quadratic functions are found from their graphs. These are the - and -intercepts, the verte (or turning point), the line of smmetr.

5 3 Quadratic Functions The graph of the quadratic function is shown below. The -intercepts of the parabola are (1, 0) and (3, 0), the -intercept is (0, 3) and the verte or turning point is (2, 1). You can see that the parabola is smmetric about the line = 2, in the sense that this line divides the parabola into two parts, each of which is a mirror image of the other. = (2, 1) The parabola above was drawn with a mathematical graphing package. This module shows how to draw parabolas b hand. These are not intended to be accurate representations of parabolas, but are used to guide us when solving problems. The should displa the most important features of parabolas: the intercepts, the verte and the line of smmetr (unless there is a good reason not to). Although the focus is an important feature of a parabola, it is not used in most applications and is not usuall shown on sketches. Problems 1 1. Check that 1 ( 1) 2 is a parabola b rewriting it in the general form a 2 b c. 2. Complete the following table, then use it to draw the parabolas 2 and 1 ( 1) ( 1) 2 3. What is the line of smmetr for each parabola? 4. What are the ranges of the quadratic functions 2 and 1 ( 1) 2? We need to see the graph to work out the range of a function.

6 2 Sketching Parabolas 2.1 The shape of a parabola Parabolas have two orientations: concave up and concave down. (Note that the word conve has a different meaning in mathematics, so we do not use it in this contet.) Concave up Concave down The orientation of a parabola can be found from its equation. The parabola 2 is concave up. To see this, imagine how the value of will change when we substitute ver large -values into the equation, such as = 1 000, = , = , etc. As is given larger and larger values, the value of becomes ver large and positive. So the parabola must be concave up. In comparison, the parabola 2 is concave down. As is given larger and larger values, the value of becomes ver large and negative. So the parabola must be concave down. The parabola is concave up. To see this, imagine how changes when ver large values of are substituted into the equation. When is ver large, the value of 50 2 will be much larger than the value of 230 and also much larger than 107, so the value of would be ver large and positive (check this). This would be true for all ver large values of, so the parabola must be concave up. 4

7 5 Quadratic Functions In general: We don t include the case a = 0, as b c is the equation of a straight line, not a parabola! The parabola a 2 b c is concave up if a 0 and concave down if a 0. Problems 2A Rewrite the following parabolas in the form a 2 b c, and state whether the are concave up or concave down. (a) ( 3) 2 4 (b) 3 ( 2) 2 (c) 10( 2) 2 15 (d) (e) s (t 3) 2 2t 1 (f) P (w 3) 2 (w 1) The intercepts of a parabola Before we can sketch a parabola, we need to draw scales on the - and -aes. This can be done once the - and -intercepts are known. The -intercept is where the graph meets the -ais. Find the -intercept of Put = 0, then The parabola will look something like this. The -intercept shows what scale we need to use on the -ais. The -intercept is (0, 3). Find the -intercept of (2 )(1 ). Put = 0, then (2 )(1 ) (2 0)(1 0) 2. 2

8 Sketching Parabolas 6 The -intercept is (0, 2). The -intercepts are where the graph meets the -ais. There will be either two -intercepts: The graphs cross the -ais. eactl one intercept: The graphs just touch the -ais. no intercepts: The graphs don t meet the -ais. Find (i) the -intercepts and (ii) the verte of (See Appendi: Quadratic Equations) (i) The -intercepts. Put = 0, then ( 1)( 3) 0. So either 1 0 1, or The -intercepts are (1, 0) and (3, 0). ( 1)( 3). When a product of two numbers is zero, either one number or the other must be zero. The -intercepts show what scale should be used on the - ais. 1 3 First find the line of smmetr. (ii) The line of smmetr and verte. The line of smmetr passes through the midpoint of (1, 0) and (3, 0), so it must have equation = 2. To find the verte, put = 2. ( 1)( 3) (2 1)(2 3) 1 The verte is (2, 1) Put = 2 as the verte is on the line of smmetr.

9 7 Quadratic Functions Find the -intercepts and verte of 2 2. (i) The -intercepts. Put = 0, then ( 1)( 2) 0. So either 1 0 1, or The -intercepts are ( 1, 0) and (2, 0). Solve b factorising. The midpoint of 1 and 2 is ( 1) (ii) The line of smmetr and verte. The equation of the line of smmetr is 1 2. To find the verte, put The verte is on the line of smmetr (1/2, 9/4) 9 4 The verte is ( 1 2, 9 4 ). 1 2 If ou cannot solve a quadratic equation b factorisation, then tr completing the square or the quadratic formula. These are described in the appendi. The line of smmetr can also be found from the formula below. The parabola a 2 b c has line of smmetr b 2a.

10 Sketching Parabolas 8 Find the -intercepts and verte of (i) The -intercepts. Put = 0, then Using the quadratic formula: a = 2, b = 1, c = 2 b 2 4ac ( 1) ( 2) 17 0 the equation has two solutions. b b2 4ac 2a ( 1) or (4 d.p.) The intercepts are ( , 0) and (1.2808, 0). (ii) Line of smmetr and verte. The line of smmetr is To find the verte, put b 2a The verte is on the line of smmetr The verte is ( 1 4, 17 ) (0.25, ) 2.3 Sketching a parabola A sketch of a parabola should show the intercepts, the line of smmetr and the verte. Parabolas without -intercepts will be covered in section 3. Sketch the parabola (a) Shape. The parabola is concave up as the coefficient of is greater than 0.

11 9 Quadratic Functions (b) Intercepts. Put = 0, then The -intercept is (0, 3). Put = 0, then ( 1)( 3) 0. 3 So either 1 0 1, or The -intercepts are (1, 0) and (3, 0). (c) Line of smmetr and verte. 1 3 The line of smmetr is = 2, as it passes through the midpoint of (1, 0) and (3, 0). Check: b 2a Put = 2, then The verte is (2, 1). Check using the formula for the line of smmetr (2, 1) (d) Sketch. Sketch the parabola (a) Shape. 1 ( 1) 2 1 ( 1)( 1) 1 ( 2 1) ( 1) 2. (2, 1) Epand brackets then simplif. The parabola is concave down as the coefficient of is 1. The point (4, 3) is on the parabola because the parabola is smmetric about the line = 2.

12 Sketching Parabolas 10 (b) Intercepts. Put = 0, then The -intercept is (0, 0). Put = 0, then ( 2) 0. So either 0, or The -intercepts are (0, 0) and (2, 0). (c) Line of smmetr and verte. The line of smmetr is = 1, as it passes through the midpoint of (0, 0) and (2, 0) Check: b 2a 2 2 ( 1) 1. Put = 1, then = 1 (1, 1) 2 The verte is (1, 1). (d) Sketch. 0 = 1 (1, 1) 2 Note. Sometimes the method above does not produce enough points for a sketch. If this happens, then ou should calculate more points b substitution and b using smmetr.

13 11 Quadratic Functions The parabola 2 has -intercept (0, 0), -intercept (0, 0) and line of smmetr = 0. Four more points were calculated for the sketch below: (1, 1), ( 1, 1), (2, 4), ( 2, 4). ( 2, 4) ( 1, 1) 0 (2, 4) (1, 1) Problems 2B 1. Sketch the following parabolas, showing their intercepts, line of smmetr and verte. In each case state the domain and range of the quadratic function. (a) 2 4 (b) (c) (d) 6 2 (e) f () (f) If rice plants are sown at a densit of plants per square metre in a certain location, then the ield of rice is 0.01(10 0.5) kg per square metre. What is the maimum ield of rice per acre, and what densit of plants gives this maimum?

14 3 Transformations The graph of a function is ver useful in mathematics, as it displas the important features of the function. Graphical representations can be simplified b the use of transformations such as translations, reflections, and dilations. 3.1 Translations A translation of a geometric figure is a transformation in which ever point is moved the same distance in the same direction. In the diagram below, each point on the book is shifted 4 cm to the right. 4 cm In the diagram below, each point on the book is shifted 1 cm upwards. 1 cm Question: If the parabola 2 is shifted h units to the right, what will be its new equation? ( h, ) h units (, ) (0, 0) (h, 0) 12

15 13 Quadratic Functions To answer this question, remember that the equation of a parabola is a formula which describes the relationship between the - and -coordinates of points on the parabola. If the point (, ) is on the new parabola, then the point ( h, ) must be on the original parabola, and so must satisf ( h) 2. As the points (, ) on the new parabola satisf the equation ( h) 2, this is its equation. Check. The point (0, 0) is on the original parabola 2, and the point (h, 0) is on new parabola ( h) 2. When the parabola 2 is shifted b h units to the left, we get a similar result. (, ) h units ( + h, ) ( h, 0) (0, 0) If the point (, ) is on the new parabola, then the point ( + h, ) must be on the original parabola, and so must satisf ( h) 2. Check. The point (0, 0) is on the original parabola 2, and the point ( h, 0) is on new parabola ( h) 2. If parabola 2 is shifted b h units to the right, its new equation is ( h) 2. (We interpret h 0 as being meaning shift to the left.) Question: If the parabola 2 is shifted k units upwards, what will be its new equation? (0, k) (, ) k units (, k) (0, 0) This question can be answered in a similar wa to the previous question.

16 Transformations 14 If the point (, ) is on the new parabola, then the point (, k) must be on the original parabola, and so must satisf k 2, ie. coordinate ( coordinate) 2, or 2 k. As the points (, ) on the new parabola satisf the equation 2 k, this is its equation. Check. The point (0, 0) is on the original parabola 2, and the point (0, k) is on new parabola 2 k. If the parabola 2 is shifted k units upwards, its new equation is 2 k. (We interpret k 0 as being meaning shift downwards.) If we combine the horizontal and vertical transformations we get: If the parabola 2 is shifted h units to the right and k units upwards, its new equation is ( h) 2 k. Sketch the parabola Write the equation in the form ( h) 2 k b completing the square: ( 2) ( 2) So the parabola can is obtained from 2 b shifting it b 2 units to the right and 1 unit upwards. (0, 0) = 2 (2, 1)

17 15 Quadratic Functions This eample shows how to sketch a parabola with no -intercepts. It also shows that the shape of the parabola is eactl the same as the shape of the parabola 2! Problems 3A 1. Sketch the parabola (a) , then use translations of this to sketch (b) (c) A parabola has the same shape as 2 but with verte at (2, 3). What is its equation? 3.2 Reflections A reflection of a geometric figure is a transformation that results in a mirror image of it. In the diagram below, the parabola 2 is reflected across the -ais. Each point (, ) on the original parabola is reflected to (, ) on the reflected parabola. 2 (, ) (, ) 2 The equation of the reflection of 2 is 2. Sketch the parabola Observe that parabola is the reflection of across the -ais. First sketch the parabola ( 1) 2 2 b completing the square. This parabola is concave up with verte (1, 2) and line of smmetr = 1. Net reflect our sketch across the -ais.

18 Transformations 16 = (1, 2) (1, 2) Problems 3B Sketch the parabola 2, then use translations and reflections to sketch (a) (b) Dilations A dilation of a geometric figure is a transformation that results in a figure which is similar to the original, but which ma be enlarged or reduced. In the diagram below, the second book is enlarged a factor of 2 in the -direction, and the third book is b a factor of 3. In the diagram below, the parabola 2 is enlarged b a factor of 2 in the -direction. As the -coordinate of each point is doubled, the new parabola has equation 2 2.

19 17 Quadratic Functions ( 2, 4) ( 1, 1) 0 (2, 4) (1, 1) ( 2, 8) ( 1, 2) 0 (2, 8) (1, 2) If the parabola 2 is dilated b a factor of a ( > 0) in the -direction, its new equation is a 2. (The parabola is enlarged if a 1and is reduced if a 1.) If we combine translations with dilations, we have: If the parabola 2 is dilated b a factor of a, shifted h units to the right and k units upwards, its new equation is a( h) 2 k. (The dilation has to be done first for this to be correct, but the two translations can be done in either order.)

20 A Appendi: Quadratic Equations A quadratic equation is an equation of the form a 2 b c 0. This appendi gives three methods for solving quadratic equations: factorisation, completing the square, and the quadratic formula. The first two methods are emphasised in this module to give ou more practice at algebra, but ou are more likel to use the third method in real applications. 4.1 Solving equations b factorisation When the product of two or more numbers is equal to zero, then one of the numbers must be zero. This idea can be used to solve quadratic equations. Solve the quadratic equation ( 1)( 2) 0 either or The solutions are = 1 and 2. ( 1)( 2) 0. Think of this as a product of two numbers equal to zero, where the two numbers are 1 and 2 s can be checked b substitution into the original equation. Solve the quadratic equation 10( 1) 0. 10( 2) 0 either 0 or The solutions are = 0 and 2. (alpha) and (beta) are Greek letters representing numbers. To solve an equation like , we first need to factorise into a product of two factors ( )( ), where and are some numbers. To understand how to do this, we epand the product ( )( ), giving ( )( ) 2 2 ( ). 18

21 19 Quadratic Functions You can see that we can write in the form ( )( ) if we can find two numbers and with 5and 4. We now tr to guess the numbers and. The product of and is 4, so the numbers could be 1 & 4 or 2 & 2. The sum of and is 5, so and must be 1 & 4. This shows that = ( 1)( 4). If we couldn t guess these numbers, then we would need to use another method! Write 2 6 as a product of two factors. The numbers and have product 6, and sum 1. Pairs of numbers with product 6 are 1 & 6; 2 & 3; 3 & 2; and 6 & 1. Their sum is 1, so and must be 3 & 2, and 2 6 ( 3)( 2). A common factor is a factor which is a factor of ever term. It is a good idea to factorise out the common factors first of all. Factorise ( 2 6) 10( 3)( 2) Check b multipling out the brackets. Factorising out common factor 10 Product = 6 Solve b factorising ( 2 12) 3( 4)( 3) Factorising 3( 4)( 3) 0 either or The solutions are = 3 and 4. Solving

22 Appendi: Quadratic Equations 20 Solve ( 7) 0. So either 0, or The answers are = 0 and 7. a common factor Warning If we tried to solve 2 7 b dividing both sides of the equation b, then we end up with onl one solution: = 7. The reason is that when we divide both sides b we need to assume that 0, so we end up with onl the non-zero solution. Problems 4A 1. Factorise the following quadratic epressions. (a) (b) (c) (e) 2 2 (f) b 2 b 2 (g) (h) s 2 s 12 (i) 2 12 (j) a 2 8a 7 n 2 4n (k) t 2 11t 12 (l) (m) Solve the following quadratic equations. (a) (b) (c) (d) a 2 7a 12 0 (e) b 2 8b 12 0 (f) p 2 12p Multipl out the denominators, then solve the equations below. (a) (b)

24 Appendi: Quadratic Equations 22 Solve is a perfect square. Complete the square b adding 4 to the LHS. Add 4 to the RHS as well ( 3) So either 3 2 1, or There are two numbers whose square is equal 4: +2 and 2. These are abbreviated as ±2. The solutions are = 1 and 5. Solve ( 3) So either 3 2 This equation cannot be solved using factorisation. Completing the square. or (2 d.p.), (2 d.p.). The solutions are and (2 d.p.). Solve Completing the square ( 8) 2 4

25 23 Quadratic Functions There is no solution as the square of a number cannot be negative. Problems 4B 1. Solve the following quadratic equations b completing the square, if possible. (a) (b) (c) (d) (e) a 2 3a 3 0 (f) b 2 b Multipl out the denominators, then solve the equations below. (a) (b)

27 1 Quadratic Functions B s Section ( 1) a 1, b 2, c ( 1) The line of smmetr is = 0. = 1 2 (1, 1) 2 2 (0, 0) (0, 0) (2, 0) 3. = 0 and = 1 4. { : 0}and { : 1} Section 2.1 (a) ( 3) (concave up) (b) 3 ( 2) (concave down) (c) 10( 2) (concave up) (d) (concave up) (e) s t 2 4t 10 (concave up) (f) P 2w 2 4w 10 (concave up)

28 Section 2.3 s intercept -intercepts line sm. verte domain range (a) (0, 0) (0, 0) & (4, 0) = 2 (2, 4) R { : 4} (b) (0, 6) (2, 0) & (3, 0) = 2.5 (2.5, 0.25) R { : 0.25} (c) (0, 8) (2, 0) & ( 4, 0) = 1 ( 1, 9) R { : 9} (d) (0, 6) (2, 0) & ( 3, 0) = 0.5 ( 0.5, 5.25) R { : 5.25} (e) (0, 4) (2, 0) = 2 (2, 0) R { : 0} (f) (0, 8.3) (2.05, 0) & ( 4.05, 0) = 1 ( 1, 9.3) R { : 9.3} 2. Sketch 0.01(10 0.5) concave down; -intercept (0, 0), -intercepts (0, 0) & (20, 0); line sm. = 10, verte (10, 0.5). Maimum ield occurs at the verte => densit = 10 plants per square metre & ield = 0.5 kg per square metre Section (a) (b) (c) Line sm. = 3 = 3 = 3 verte (3, 1) (3, 0) (3, 1) 2. ( 2) 2 3 Section (a) (b) concavit down down Line sm. = 1 = 1 verte (1, 7) (1, 5) Section 4.1 1(a) ( 1)( 3) (b) ( 1)( 5) (c) (a 1)(a 7) 1(e) ( 2)( 1) (f) (b 2)(b 1) (g) (n 5)(n 1) 1(h) (s 4)(s 3) (i) ( 4)( 3) (j) ( 2)( 6) 1(k) (t 12)(t 1) (l) ( 7)( 2) (m) ( 14)( 1)

29 3 Quadratic Functions 2(a) 2 & 3 (b) 2 & 3 (c) 2 & 3 2(d) 3 & 4 (e) 6 & 2 (f) 6 3(a) &4 3(b) & 2 Section 4.2 1(a) ( 2) and 1 (b) ( 2) (c) ( 2) 2 1 no solution (d) ( 3 2 ) no solution 1(e) (a 3 2 ) (f) (b 1 2 )2 3 4 a and a no solution 2(a) 1(b) and ( 2) and 2 7 Section 4.3 1(a) b 2 4ac solutions (b) b 2 4ac no solutions 1(c) b 2 4ac solutions 2(a) = 0.55 & 0.45 (b) = 0.5 (c) no solutions

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