For every input number the output involves squaring a number.

Transcription

1 Quadratic Functions The function For every input number the output involves squaring a number. eg. y = x, y = x + 3x + 1, y = 3(x 5), y = (x ) 1 The shape parabola (can open up or down) axis of symmetry The parts Vertex: the maximum or minimum point on the curve Axis of symmetry: the line of symmetry The equation Vertex Standard form: y = ax + bx + c, x 0 (also called Expanded form) a : positive: parabola opens up has a minimum negative: parabola opens down has a maximum closer to 0: wider further from 0: narrower b : moves the vertex left or right. c : is the y-intercept eg. y = x + 3x + 5 Graph the parabola and determine the following: y 5 y-intercept: 5 x-intercepts: 1,.5 vertex: (0.75, 6.15) i.e. let x = 0 in the equation. y = y = 5 x axis of symmetry: x = 0.75 From Graphing Calculator domain: x R range: y 6.15 maximum value of 6.15 when the x is 0.75

2 Calculator Skills Vertex: Enter quadratic function in and determine window settings that show a vertex. maximum if a < 0, minimum if a > 0. Note: The vertex is (3, ) not ( , ). The calculator does its best estimate of the vertex. The value for the x-coordinate may need rounding. y-intercept: x-intercept(s): Enter quadratic function and the function y = 0 in and determine window settings that show the x-intercepts. intersection

3 The equation Factored form: y = a(x r)(x s) a : same characteristics as the a in the standard form. r, s : are the x-intercepts y eg. y = (x 3)(x + 1) Graph the parabola and determine the following: y-intercept: 6 i.e. let x = 0 in the equation x x-intercepts: 3, 1 from the factors above From Graphing Calculator vertex: (1, 8) axis of symmetry: x = 1 domain: x R range: y 8 6 eg. From the graph of the quadratic function, determine the equation. Solution: Notice the x-intercepts are and 6. y 3 These are r and s in the factored form: y = a(x )(x 6) 6 x Now use another point on the graph to find a. Using (7, 4) in the equation above: 4 = a(7 )(7 6) 4 = 5a 4 a = 4 5 Thus the equation of the parabola is y = 4 (x )(x 6) 5

4 x = 3 axis of symmetry The equation Vertex form: y = a(x h) + k a: same characteristics as the a in the standard form. h: the x-coordinate of the vertex. Changes in h move the function left or right. k: the y-coordinate of the vertex Changes in k move the function up or down. x = h: the equation of the axis of symmetry y 1 f x x 3 1 domain: x R range: y 1 Vertex (3, 1) x Domain: {x x R} Range: {y y k, y R} if a > 0, {y y k, y R} if a < 0 Minimum if a > 0 or Maximum if a < 0 Algebraic Methods: y-intercept: substitute x = 0 into the function. x-intercept: substitute y = 0 into the function. see chapter on Quadratic equations vertex: ( b, f ( b )) a a a, b are from the standard form of a parabola f(x) = ax + bx + c Basic Problem Determine the equation of the quadratic function given the vertex and a point on the function. eg. a parabola has a vertex of (3, 47) and goes through the point (7, ). Determine the equation of the quadratic function. Start with y = a(x h) + k, replace h and k, and determine the value of a. y = a(x 3) + 47 = a(7 3) + 47 a = Thus the equation is y = (x 3) + 47

5 Changing Forms Eg. Change y = x + 0x + 3 to vertex form. Solution: Find the vertex. i.e. ( 5, 18) from calculator or algebraic method. Use the same value for a. y = (x + 5) 18 Eg. Change y = x + 0x + 3 to factored form. Solution: Factor the trinomial part using factoring methods. In this situation there is a common factor of. y = [x + 10x + 16] Then use product sum factoring method. y = (x + )(x + 8) Eg. Change y = (x + 5) 18 to standard form. Solution: Expand the squared binomial. y = (x + 5)(x + 5) 18 y = [x + 5x + 5x + 5] 18 y = (x + 10x + 5) 18 Multiply the trinomial by a and combine constants. y = x + 0x y = x + 0x + 3 Eg. Change y = (x + )(x + 8) to standard form. Solution: Expand the binomial factors by FOIL. y = [x + 8x + x + 16] y = [x + 10x + 16] Multiply the trinomial by a. y = x + 0x + 3

6 Word Problems 1. Projectile Problem - find maximum height or time at certain height(s) Example: A frog sitting on a rock jumps into a pond. The height, h, in centimetres, of the frog above the surface of the water as a function of time, t, in seconds, since it jumped can be modelled by the function h(t) = 490t + 150t + 5. Where appropriate, answer the following questions to the nearest tenth. a) Graph the function b) What is the y-intercept? What does it represent in this situation? The y-intercept is 5. It represents the initial height of the frog, 5 cm. c) What maximum height does the frog reach? When does it reach that height? The frog reaches at maximum height of 36.5 cm after 0. seconds. d) When does the frog hit the surface of the water? The frog hits the water after 0.4 seconds. e) What are the domain and range in this situation? Domain: {t 0 t 0.4, t R} Range: {h 0 h 36.5, h R} f) How high is the frog 0.5 s after it jumps? The frog is 31.9 cm high after 0.5 seconds. g) When was the frog 8 cm high? The frog was 8 cm high at 0.0 s and at 0.8 s.

7 . Maximum Revenue Example: The student council at a high school is planning a fundraising event with a professional photographer taking portraits of individuals or groups. The student council gets to charge and keep a session fee for each individual or group photo session. Last year, they charged a $10 session fee and 400 sessions were booked. In considering what price they should charge this year student council members estimate that for every $1 increase in the price, they expect to have 0 fewer sessions booked. a) Write a function to model this situation. Revenue = Cost per item Number Sold = $ sessions Let x be the number of changes = (10 + x)(400 0x) b) What is the maximum revenue they can expect based on these estimates. What session fee will give that maximum? y x x Vertex: (5, 4500) (10 + 5) (400 0(5)) 5 changes $4500 is maximum revenue $15 charge per ticket 300 sessions Example: A company sells canoes for $500. At this price, it sells 60 canoes per season. If the price increases by $50 each, then the number of canoes sold will drop by 4. a. Find an equation for the Revenue. Let x be the number of changes R = (selling price) (number of canoes sold) = ( x)(60 4x) 1 b. What is the maximum revenue? Expanding the above gives: R = 00x x Vertex = b = 1000 =.5 a ( 00) Letting x =.5 in the Revenue equation gives 8750 Therefore maximum revenue would be $8750 c. What is the selling price of the canoe which gives the maximum revenue? From 1 above, with.5 changes to the price ( (.5)) = 65 Therefore the selling price should be $65.

8 3. Maximum Area Example: Given 40 m of fence, find the dimensions which give the maximum area if the pen must be a rectangle. w w P w 40 w 0 w w 0 A w 0 substitution 0 expanding b 0 10, A a 1 Vertex (10, 100) find the width: w Therefore the maximum area of 100m is found when both sides are 10m. 4. Minimizing Cost Example: Assume that you are an advisor to business owners who want to analyse their production costs. The production costs, C, to produce n thousand units of their product can be modeled by the function C(n) = 0.3n 48.6n Determine the number of units that should be produced to minimize their costs. They should produce units to have a production cost of $ Find two numbers with a maximum(or minimum) product Example: Find two numbers whose difference is 150 and whose product is a minimum. Let the numbers be x and y. Thus y x = 150 or rearranged y = x. Product = xy = x(150 + x) = x + 150x Since the x value for the vertex would be x = b = 150 = 75, and the y value a (1) would be x = ( 75) = 75. Therefore the two numbers are 75 and 75.