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Grade 10 Academic Math Chapter 3 – Analyzing and Applying Quadratic Models Solving Word Problems Given the Quadratic Equation

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Learning Goal By the end of the lesson, students will be able to… apply knowledge of quadratics to solving word problems

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Curriculum Expectations By the end of the lesson, students will… Solve problems arising from realistic situations represented by a graph or an equation of a quadratic relationship

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Mathematical Process Expectations Connecting – make connections among mathematical concepts and procedures; and relate mathematical ideas to situations or phenomena drawn from other contexts

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Agenda Solving quadratic word problems generally Movement problems (given the equation) Engineering problems (given the equation) Revenue problems (given the equation) Area problems (given the equation)

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Mental Health Break

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Solving Quadratic Word Problems Generally For quadratic word problems, you will either be given, or need to set-up a quadratic equation in either standard or factored form If the problem asks when something happens, you must find the x value of the vertex, or sometimes the zeros (ex. When ball hits ground) If the problem asks for the maximum, minimum or optimal value, you must find the y value of the vertex

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Solving Quadratic Word Problems Generally Problems usually fall into one of the following categories – Movement (ex. Ball thrown) – Engineering (ex. Bridge) – Revenue (ex. Price change or maximum revenue) – Area (ex. Maximum area or how much border can you add) – Integers (ex. Squares of consecutive odd integers) – Triangles (ex. Find side length of right angle triangle)

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Solving Quadratic Word Problems Given the Equation Today, we’re going to look at some problems where you are given the equation – Movement (ex. Ball thrown) – Revenue (ex. Break-even and maximum revenue) – Area (ex. Maximum area)

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Quadratic Word Problems Movement Given the Equation Factor (GCF) if possible Set y = 0 and determine the zeros Answer questions about time (x-axis) or height (y-axis) Typically have to determine the vertex Remember, average of zeros gives x of vertex Then, substitute x into equation and solve for y of vertex

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Quadratic Word Problems Movement Given the Equation If asked for the x value for a given y value other than zero, substitute it in for y but then move it to the other side so that there is a zero where the y was again You then should end up with a trinomial that you can factor and solve for x again These won’t be the same x’s as the zeros

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Quadratic Word Problems Movement Given the Equation, ex. p.268, #11 A football is kicked straight up into the air. Its height above the ground is approximated by the relation h = 25t – 5t², where h is the height in metres and t is the time in seconds

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Quadratic Word Problems Movement Given the Equation, ex. p.268, #11

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h = 25t – 5t² (a) What are the zeros of the relation and when does the football hit the ground? 0 = 25t – 5t² Set h = 0 0 = 5t(5 – t) Factor our the GCF 5t = 0 and 5 – t = Isolate mini-zero equations t = 0 and t = These are the two zeros

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Quadratic Word Problems Movement Given the Equation, ex. p.268, #11 h = 25t – 5t² Since the zeros are 0 and 5, the football hits the ground at t = 5 or 5 seconds

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Quadratic Word Problems Movement Given the Equation, ex. p.268, #11 h = 25t – 5t² (b) What are the coordinates of the vertex Zeros are 0 and 5 t v = t zero 1 + t zero t v = (0 + 5) t v = 2.5 2

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Quadratic Word Problems Movement Given the Equation, ex. p.268, #11 h = 25t – 5t² (b) What are the coordinates of the vertex t of vertex is 2.5. Substitute this value in for t and solve for h h = 25(2.5) – 5(2.5) ² h = 62.5 – h = 31.25m (d) The maximum height of 31.25m occurs at time of 2.5 seconds

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Quadratic Word Problems Movement Given the Equation, ex. p.268, #11(c) (Graph)

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Quadratic Word Problems Revenue Given the Equation Equation given is usually either a Revenue Equation or a Profit Equation Revenue basically means money from sales and is calculated by taking the price of a item and multiplying it by the number of items sold R = price x # sold Profit is the revenue of a company after expenses are subtracted Breakeven takes place when profit (not revenue) is zero Maximum profit or revenue takes place at the vertex of the graph

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Quadratic Word Problems Revenue Given the Equation, ex. p.332, #15 The Wheely Fast Co. makes custom skateboards for professional riders. They model their profit with the relation P = -2b² + 14b – 20, where b is the number of skateboards they produce (in thousands), and P is the company’s profit in hundreds of thousands of dollars

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Quadratic Word Problems Revenue Given the Equation, ex. p.332, #15

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P = -2b² + 14b – 20 (a)When does Wheely Fast break even (P = 0) 0 = -2b² + 14b – set P = 0 0 = -2(b² + 7b – 10) factor out -2, GCF 0 = -2(b – 5)(b – 2) factor using butterfly b – 5 = 0 and b – 2 = Identify zero equations b = 5 and b = The zeros are 5 and 2 so this happens when they produce either 2000 or 5000 skateboards

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Quadratic Word Problems Revenue Given the Equation, ex. p.332, #15

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P = -2b² + 14b – 20 (b) How many skateboards does Wheely Fast need to produce to maximize profit Zeros are 2 and 5 b v = b zero 1 + b zero b v = (2 + 5) b v = 3.5 (which is 3500) 2

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Quadratic Word Problems Revenue Given the Equation, ex. p.332, #15 P = -2b² + 14b – 20 Substitute b = 3.5 into equation P = -2(3.5)² + 14(3.5) – 20 P = -2(12.25) + 49 – 20 P = – 20 P = 4.5 (so 4.5 x 100,000 = $450,000) So, the Wheely Fast maximizes profits of $450,000 by producing 3,500 skateboards

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Quadratic Word Problems Area Given the Equation, ex. p.308, #9 A rectangular enclosure has an area in square metres given by A = -2w² + 36w, where w is the width of the rectangle in metres. What is the maximum area of the enclosure?

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Quadratic Word Problems Area Given the Equation, ex. p.308, #9

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A = -2w² + 36w First, find the zeros 0 = -2w² + 36w Set A = 0 0 = -2w(w - 18) Factor our the GCF -2w = 0 and w – 18 = Isolate mini-zero equations w = 0 and w = These are the two zeros

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Quadratic Word Problems Area Given the Equation, ex. p.308, #9 A = -2w² + 36w What are the coordinates of the vertex Zeros are 0 and 18 w v = w zero 1 + w zero w v = (0 + 18) w v = 9 2

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Quadratic Word Problems Area Given the Equation, ex. p.308, #9 Now we need to find the A of the vertex. The w of the vertex is 9. Substitute this value in for w and solve for A A = -2w² + 36w A = -2(9)²+ 36(9) A = -2(81) A = 162 The maximum height of area of 162m² occurs when w is 9 metres

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Quadratic Word Problems Area Given the Equation, ex. p.308, #9 Maximum area (vertex) Zero

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Homework (Given the Equation) Handout 1 – Quadratic Word Problems - #3 & 4 Handout 2 – Application Problems - #1, 5 & 6

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