Chapter 18 Quadratic Function 2

Transcription

1 Chapter 18 Quadratic Function Completed Square Form 1 Consider this special set of numbers - the square numbers or the set of perfect squares. 4 = = 9 = 3 = 16 = 4 = 5 = 5 = Numbers like 5, 11, 15 are not perfect squares, but can be written as a perfect square, ± a number. 5 = + 1 = + 11 = 3 + = + 15 = 4 1 = The same situation occurs in algebra. Consider the following special quadratics : , , , the are also perfect squares = ( + 1), = ( + ), = ( 4), = ( + 1), etc. An algebraic epression like is not a perfect square but we can write it as almost one. Eample 1 : = (+ ) = ( + 1) + Eample : = (+ 5) = ( + ) + 5 Eample 3 : = ( 5) = ( 4) 5 Rule :- this wa of rewriting a quadratic epression is called epressing it in completed square form. Step 1 :- Separate the first terms from the last => = ( + 10) + 7 Step :- Add on (half of the term) inside the brackets => ( ) + 7 Step 3 :- Subtract this same number outside the brackets => ( ) Step 4 :- Write the epression in completed square form => ( + 5) +. Eercise Write = in completed square form. Cop and complete :- = => = ( 6) + 10 => = ( 6 + 9) => = (...) +... (b) Cop and complete :- = => = ( 1) + 30 => = ( ) => = (...).... Write = in completed square form. 3. Write each of the following in completed square form, showing each stage of our working :- = (b) = (c) = 8 5 (d) = Do the same here. (Fractions required). = (b) = (c) = (d) = (Etension work - Hard). Chapter 18 this is page 93 Quad Function

2 Graph of Quadratic in Completed Square Form In an earlier Chapter, ou learned how to sketch quadratic functions of the form = 8 quickl b using factorisation techniques. The graph formed is called a parabola. Reminder :- Sketch the graph of = 8. step 1 :- set = 0 => 8 = 0 => ( 4)( + ) = 0 => = 4 or = step :- cuts ais at the points (4, 0) and (, 0). find the minimum turning point. take half wa between the roots 4 and. i.e. at = 1 => Now find when = 1. => = = 9 => minimum turning point at (1, 9). step 3 :- find the -intercept b replacing b 0. step 4 :- => = = 8 => -intercept is at (0, 8). sketch the graph through these 4 points. (, 0) = 8 (0, 8) (1, 9) (4, 0) If the quadratic functions has been epressed in completed square form :- = ( ) + 3 it is much easier to draw its graph. Check it reall is a quadratic function :- [( ) + 3 = = ] Eample :- Sketch the graph of = ( ) + 3. Think! The square of anthing can never be negative. The smallest value of ( ) is zero at =. => The MINIMUM value of ( ) + 3 is = 3 and this occurs at =. => The minimum turning point is (, 3). => Set = 0 to find the -intercept - (0, 7). Now sketch the parabola through these poinst. (0, 7) = (, 3) = ( ) + 3 note :- equation of ais of smmetr Rule :- An quadratic function of the form = ( a) + b will have a minimum at (a, b). = ( 4) + 1 = ( 3) = ( + 1) + (4, 1 ) (3, ) ( 1, ) Chapter 18 this is page 94 Quad Function

3 Eercise Consider the quadratic :- = ( 3) +. What must ou replace with in ( 3) + to make ( 3) = 0? (b) When is replaced with this value, what will the value of be? (c) What are the coordinates of the minimum turning point (a, b)? (d) Sketch the parabola showing this minimum turning point, write down the equation of the ais of smmetr and find the -intercept.. Consider the quadratic :- = ( ) 1. What must ou replace with in ( ) 1 to make ( ) = 0? (b) When is replaced with this value, what will the value of be? (c) What are the coordinates of the minimum turning point (a, b)? (d) Sketch the parabola showing this minimum turning point, write down the equation of the ais of smmetr and find the -intercept. a b (a, b) 5. Write down the coordinates of the minimum turning point of this parabola. (b) Assuming the parabola is of the form :- = ( a) + b, write down the equation of the parabola. 6. All of the following parabolas are of the form :- = ( a) + b. Write down the equation of each parabola and the equation of the ais of smmetr. (c) (5, 1) (6, 0) (b) (d) (, 5) ( 3, ) (4, ) 3. We can improve our sketch of :- = ( + 4) 3. (e) (f) Write down the coordinates of the minimum turning point P(a, b). (note :- 4) (b) Replace with 0 in = ( + 4) 3 to determine where the graph cuts the -ais. (c) Plot this point also and sketch the parabola. 4. Use this two-step approach to sketch the following parabolas showing both the minimum turning point and the -intercept for each. = ( ) + 5 (b) = ( 1) + 3 (c) = ( + 3) + 1 (d) = ( ) 6 (e) = ( + 1) 1 (f) = ( 5) 8 (g) = ( + ) 1 (h) = ( 3). 7. The diagram shows the basic parabola :- =. (7, 3) The parabola is then slid 1 boes right and 5 boes down and is shown dotted. Write down the equation of the dotted parabola. (b) Determine the coordinates of its -intercept. (, 1) = 1 5 Chapter 18 this is page 95 Quad Function

4 Completed Square Form () Maimum Turning Points Sometimes, quadratic functions come in completed square form a bit differentl :- = 3 ( ). Check it reall is a quadratic function :- [3 ( ) = 3 ( 4 + 4) = + 4 1] The difference this time is that the term shows that the parabola will be upside-down. Eample :- Sketch the graph of = 3 ( ). note :- equation of ais of smmetr Think! The square of anthing can never be negative. The smallest value of ( ) is zero at =. => The MAXIMUM value of 3 ( ) is 3 0 = 3 and this occurs at =. = (, 3) = 3 ( ) => The maimum turning point is (, 3). Now sketch the parabola through this point. The ais of smmetr has equation =. Rule :- An quadratic function of the form = b ( a) will have a maimum at (a, b). The equation of the ais of smmetr is = a. Eercise Consider the quadratic :- = 5 ( 3). What must ou replace with in 5 ( 3) to make ( 3) = 0? (b) When is replaced with this value, what will the value of be? (c) What are the coordinates of the maimum turning point (a, b)? (d) Sketch the parabola showing this maimum turning point and write down the equation of the ais of smmetr.. Use the above approach, along with replacing b 0 to determine the -intercept, to help sketch the following parabolas :- a b 3. Write down the coordinates of the maimum turning point of this parabola. (b) Assuming the parabola is of the form :- = b ( a), write down the equation of the parabola and the equation of the ais of smmetr. 4. Write down the equation of each parabola, the coordinates of the -intercept and the equation of the ais of smmetr. (4, 3) (b) (3, 9) (, 8) = 3 ( 1) (b) = 8 ( 3) (c) = 4 ( + ) (d) = 1 ( + 1) (e) = ( 3) (f) = 1 ( 4) (g) = 5 ( + 3) (h) = ( 5). (c) ( 1, 4) (d) (5, ) Chapter 18 this is page 96 Quad Function

5 Quadratics of the form = k You should know b now what the basic parabola = looks like. = = It has a minimum turning point at O(0, 0). An parabola, of the form = k, has a similar shape. =, = 5, = 3, =, = 1 all have a minimum (or maimum) turning point at O(0, 0). = 1 = 3 = Eample :- The parabola = k passes through the point P(, 1). Determine the value of k. = k Solution :- Since the parabola = k passes through the point P(, 1), then we should be able to replace with and with 1 in = k. (, 1) => = k => 1 = k => 1 = 4k => k = 3 => = 3 Eercise The parabola = k passes through the point (3, 18). = k (3, 18) 4. The parabola = k passes through the point (, 8). Replace = 3 and = 18 in = k to determine the value of k. (b) Write down the equation of the parabola.. = k The parabola = k passes through the (4, 80) point (4, 80). Replace = and = 8 in = k to determine the value of k. (negative) (b) Write down the equation of the parabola. (, 8) 5. The parabola = k passes through the point (5, 75). Replace = 4 and = 80 in = k to determine the value of k. (5, 75) Replace = 5 and = 75 in = k to determine the value of k. (b) Write down the equation of the parabola. 3. Determine the equation of the parabola with the origin as minimum turning point through :- (5, 50) (b) (, 16) (c) (5, 100) (d) (, 1) (e) ( 1, 7) (f) (4, 8). (b) Write down the equation of the parabola. 6. Determine the equation of the parabola with the origin as maimum turning point through :- (, 1) (b) (3, 36) (c) (1, 11) (d) ( 4, 3) (e) ( 1, 6) (f) ( 6, 1). Chapter 18 this is page 97 Quad Function

6 The Quadratic Formula We learned in Chapter 6 that when ou solve the quadratic equation = 0, ou are reall finding the values of which show where the corresponding parabola = cuts the -ais. We also found that factorising the epression was the best wa of finding the solutions (the roots). => = ( + 3)( + ) = 0 = ( 3, 0) (, 0) => In this case, the roots are at = 3 and. Quadratic Equations that do NOT Factorise - The Quadratic Formula Eample :- Solve the quadratic equation = 0. => You can tr and tr if ou like, but ou can t factorise !!! To solve quadratic equations that do not factorise (or even those which do), we have a special formula we can use, along with a calculator, to solve them. The Quadratic Formula :- Ever quadratic equation can be rearranged into the form :- a + b + c = 0. We can find the solution to this b using the following formula :- = b ± b 4ac a called the quadratic formula. The PROOF of this ma be given to ou b our class teacher. Eample :- Solve the quadratic equation = 0. Step 1 compare the two : = 0 a + b + c = 0 => a = 1, b = 5 and c = 3. Step use the formula :- = b ± b 4ac a => = 5 ± ± 5 1 => = => = 5 13 or = => = or => = ( ) or ( ) => = or Chapter 18 this is page 98 Quad Function

7 Eercise Look at the quadratic equation : = 0. Compare this with a + b + c = 0 and hence write down the values of a, b and c. (b) Cop and complete the following :- = b ± b 4ac a => = 6 ± ± => = = 6 ± 0 => = ( ) or ( ) => =... or Solve these quadratic equations : = 0 (b) = 0 (c) = 0 (d) = Another thing to be careful of! Solve : = 0. Cop down the two equations : = 0 a + b + c = 0 Cop :- a = 1, b =... and c =.... (b) Use the formula (carefull) :- = b ± b 4ac a => = 3 ± (3) 4 1 ( 5) 1 note. Solve the quadratic equation : = 0, using the method shown above. (Answer to decimal places each time). 3 ± => = etc. note 3. Solve each of these quadratic equations : = 0 (b) = 0 (c) = 0 (d) = Look at our answer to questions 3(c) and 3(d). Though ou used the quadratic formula, the whole number answers should have alerted ou to the fact the quadratic equation could have been solved much easier - b factorising!! Solve = 0 b factorising. (b) Solve = 0 b factorising. 8. Solve these quadratic equations : = 0 (b) + 6 = 0 (c) 3 5 = 0 (d) 4 = and a does not alwas have to be 1. Solve : = 0. Cop down the two equations : = 0 a + b + c = 0 Cop :- a = 3, b =... and c =.... (b) Use the formula (carefull) :- 5. This one is a little trickier. Solve 4 + = 0, Cop down the two equations :- 4 + = 0 a + b + c = 0 Cop :- a = 1, b =... and c =.... = b ± b 4ac a => = 4 ± (4 ) 4 3 ( 5) 3 => Now complete the question. note (b) Use the formula (carefull) :- note = b ± b 4ac a note => = ( 4) ± ( 4) => Now complete the question. 10. Solve these quadratic equations : = 0 (b) = 0 (c) 3 4 = 0 (d) 4 3 = Tr to solve this quadratic equation : = 0. What goes wrong? Tr to find out wh. Chapter 18 this is page 99 Quad Function

8 The Discriminant Let us look a bit more carefull at the quadratic formula :- = b ± b 4ac. a In particular, we are going to consider just the b 4ac part. This is referred to as the discriminant. The quadratic formula requires that ou find the square root of the discriminant. There are 3 possibilities when ou attempt to find the square root of a number :- If the b 4ac part is positive, ou can find it and the ± b 4ac means ou get answers. If the b 4ac part is zero, then the ± b 4ac = ± 0 and this means ou onl get 1 answer. If the b 4ac part is negative, ou can t find the square root of a negative number => 0 answers. The solutions we get when solving a quadratic equation are referred to as the roots of the equation. The discriminant ver quickl tells us how man roots there are, and it also tells us how man times the quadratic graph (the parabola) cuts the -ais. This helps us quickl see what the parabola looks like. If b 4ac > 0, the quadratic has real roots and the parabola cuts the -ais at points. Eample :- = => b 4ac = = 1 => There are real roots. = If b 4ac = 0, the quadratic has equal roots and the parabola cuts the -ais at onl 1 point. Eample :- = => b 4ac = = 0 => There is onl 1 real root. = If b 4ac < 0, the quadratic has 0 real roots. The parabola does not cross the -ais at all. Eample :- = 4 5 b 4ac = = 4 => There are 0 real roots. = 4 5 ( 3, 0) (, 0) ( 3, 0) Eercise Find the discriminant for each equation : = 0 (b) = 0 (c) = 0 (d) = 0.. Find the discriminant for each of these and use it to determine the nature of the roots = 0 (b) = 0 (c) + 3 = 0 (d) = 0 (e) + 5 = 0 (f) = 0 (g) + 3 = 0 (h) = 0 (i) = 0 (j) = 0 (k) = (l) 3 = Assume p = 0 has 1 root. Find p. Cop and complete :- If there is onl 1 root, then b 4ac = 0 => 6 4 p = 0 => p = a + 4 = 0 has equal roots. Find a b + 5 = 0 has 1 root. Find values for b. 6. p + 8 = 0 has real roots. Set up an inequalit in p, and solve for p. 7. m m= 0 has equal roots. Find m t = 0 has no real roots. Solve for t. Chapter 18 this is page 100 Quad Function

9 1. Epress in completed square form :- Remember Remember...? = (b) = Consider the quadratic :- = ( 5) + 3. What must ou replace with in ( 5) + 3 to make ( 5) = 0? (b) When is replaced with this value, what will the value of be? (c) Find the minimum turning point at (a, b). (d) Sketch the parabola showing this minimum turning point. 3. B finding both the minimum turning point and the -intercept, make neat sketches of the following parabolas :- = ( 1) + 4 (b) = ( + ) + 5 a b Topic in a Nutshell 8. All of the following parabolas are of the form :- = ( a) + b or = b ( a). Write down the equation of each parabola and the equation of the ais of smmetr each time. (c) (3, 6) (7, 5) (b) (d) ( 4, 3) (5, 0) (c) = ( 4) (d) = ( 3). 4. Use the above approach, along with replacing b zero to determine the -intercept, to help sketch the following parabolas :- (All have a maimum turning point). = 6 ( ) (b) = 4 ( 1) (c) = 5 ( + 3) (d) = 1 ( + ) 5. = k The parabola = k passes through the (3, 45) point (3, 45). Replace = 3 and = 45 in = k to find k. (b) Write down the equation of the parabola. 6. Determine the equation of the parabola with the origin as minimum turning point through :- (5, 50) (b) (, 16) (c) (5, 100). 7. Determine the equation of the parabola with the origin as maimum turning point through :- (4, 80) (b) (1, 10) (c) (, 36). a b (e) 9. Use the discriminant to determine the nature of roots of these quadratic equations :- (f) = 0 (b) = Solve the following quadratic equations b using the quadratic formula : = 0 (b) = 0 (c) = 0 (d) = Shown is the parabola :- (, 5) = Use the quadratic formula to determine the coordinates of the two points A and B. (7, 3) 1. It is known that the quadratic equation n = 0 has onl 1 root. Find n. (b) If 6 + 3d = 0 has no real roots, find the range of values for d. A B Chapter 18 this is page 101 Quad Function

10 Chapter 18 Answers Chapter 18 - Quad Function Eercise a = ( 3) + 1 b = ( + 6) 6. = ( + 1) a = ( ) + 3 b = ( + 10) 10 c = ( 4) 1 d = ( + 7) a = ( + 3 / ) + 7 / 4 b = ( + 5 / ) + 11 / 4 c = ( 1 / ) 9 / 4 d = ( + 1) + 4 Eercise a 3 b c (3, ) d sketch of V parabola turning at (3, ), ais of smmetr = 3.. a b 1 c (, 1) d sketch of parabola turning at (, 1), ais of smmetr =. 3. a ( 4, 3) b (0, 13) c sketch of parabola turning at ( 4, 3), crossing -ais at (0, 13) 4. a sketch of parabola turning at (, 5), crossing -ais at (0, 9) b sketch of parabola turning at (1, 3), crossing -ais at (0, 4) c sketch of parabola turning at ( 3, 1), crossing -ais at (0, 10) d sketch of parabola turning at (, 6), crossing -ais at (0, ) e sketch of parabola turning at ( 1, 1), crossing -ais at (0, 0) f sketch of parabola turning at (5, 8), crossing -ais at (0, 17) g sketch of parabola turning at (, 1), crossing -ais at (0, 3) h sketch of parabola turning at (3, 0), crossing -ais at (0, 9) 5. a (4, ) b = ( 4) + 6. a = ( 5) + 1, = 5 b = ( ) + 5, = c = ( 6), = 6 d = ( + 3) +, = 3 e = ( 7) 3, = 7 f = ( + ) 1, = 7. a = ( 1) 5 b (0, 139) Eercise a 3 b 5 c (3,5) d sketch of upside down parabola turning at (3,5), ais of smmetr = 3.. a sketch of upside down parabola turning at (1,3), ais of smmetr = 1. intercept at (0,) b sketch of upside down parabola turning at (3,8), ais of smmetr = 3. intercept at (0, 1) c sketch of upside down parabola turning at (,4), ais of smmetr =. intercept at (0,0) d sketch of upside down parabola turning at ( 1,1), ais of smmetr = 1. intercept at (0,0) e sketch of upside down parabola turning at (3, ), ais of smmetr = 3. intercept at (0, 11). f sketch of upside down parabola turning at (4, 1), ais of smmetr = 4. intercept at (0, 17) g sketch of upside down parabola turning at ( 3, 5), ais of smmetr = 3. intercept at (0, 14) h sketch of upside down parabola turning at (5,0), ais of smmetr = 5. intercept at (0, 5) 3. a (3,9) b = 9 ( 3) ais of smmetr = 3 4. a = 3 ( 4), = 4, -intercept (0, 13) b = 8 ( ), =, -intercept (0, 4) c = 4 ( + 1), = 1, -intercept (0, 3) d = ( 5), = 5, -intercept (0, 7) Eercise a k = b =. a k = 5 b = 5 3. a = b = 4 c = 4 d = 3 e = 7 f = a k = b = 5. a k = 3 b = 3 6. a = 3 b = 4 c = 11 d = e = 6 f = 1 /3 Eercise a a = 1 b = 6 c = 4 b = 5 4 = = 0 6 = a = 0 63 = 6 37 b = 0 76 = 9 4 c = = 4 d = = 5 4. a = = 4 b = = 5 5. a a = 1 b = 4 c = b = 3 41 = a = 5 45 = 0 55 b = 7 3 = 0 68 c = 6 85 = 0 15 d = 3 = 5 7. a a = 1 b = 3 c = 5 b = 1 19 = a = 1 16 = 5 16 b = 0 3 = 6 3 c = 4 19 = 1 19 d = 56 = a a = 3 b = 4 c = 5 b = 0 79 = a = 1 = 1 5 b = 0 1 = 3 1 c = 35 = 0 85 d = 1 18 = Cannot get square root of a negative. Parabola does not cut -ais. Chapter 18 this is page 10 Quad Function

11 Eercise a 37 b 0 c 7 d 1. a 1, real roots b 15, no real roots c 17, real roots d 0, equal roots (1 root) e 16, no real roots f 1, real roots g 5, real roots h 15 no real roots i 0, equal roots j 3, no real roots k 1, real roots l 61, real roots 3. p = 9 4. a = 5. b = 10 or p > 0 => p > 8 7. m = 3 or t < 0 => t < 9 Remember Remember 1. a = ( + 3) + 3 b = ( + 5) 8 5. a 5 b 3 c (5, 3) d sketch of parabola turning at (5, 3), ais of smmetr = a sketch of parabola turning at (1, 4), crossing -ais at (0, 5) b sketch of parabola turning at (, 5), crossing -ais at (0, 9) c sketch of parabola turning at (4, ), crossing -ais at (0, 14) d sketch of parabola turning at (3, 0), crossing -ais at (0, 9) 4. a sketch of upside down parabola turning at (, 6), ais of smmetr =. intercept at (0, ) b sketch of upside down parabola turning at (1, 4), ais of smmetr = 1. intercept at (0, 3) c sketch of upside down parabola turning at ( 3, 5), ais of smmetr = 3. intercept at (0, 4) d sketch of upside down parabola turning at (, 1), ais of smmetr =. intercept at (0, 5) 5. a k = 5 b = 5 6. a = b = 4 c = 4 7. a = 5 b = 10 c = 9 8. a = ( 3) + 6, = 3 b = ( + 4) + 3, = 4 c = 5 ( 7), = 7 d = ( 5), = 5 e = 5 ( + ), = f = 3 ( 7), = 7 9. a 0, equal roots (1 root) b 3, no real roots 10. a = 0 7 = 4 3 b = 5 4 = 0 76 c = 0 7 = 78 d = 1 68 = A(0 7,0) B(4 3,0) 1. a 64 8n = 0 => n = 8 b 36 1d < 0 => d > 3 Chapter 18 this is page 103 Quad Function