Grade 10 Academic Math Chapter 3 – Analyzing and Applying Quadratic Models

Grade 10 Academic Math Chapter 3 – Analyzing and Applying

Quadratic Models

Solving Word Problems Given the Quadratic Equation

Learning GoalBy the end of the lesson, students will

be able to…apply knowledge of quadratics to

solving word problems

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

Mathematical Process Expectations

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

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)

Mental Health Break

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

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)

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)

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

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

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



• 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 = 0....... Isolate mini-zero

equations• t = 0 and t = 5 ....... These are the two zeros


• h = 25t – 5t²• Since the zeros are 0 and 5, the football hits

the ground at t = 5 or 5 seconds


• h = 25t – 5t²• (b) What are the coordinates of the vertex• Zeros are 0 and 5• tv = tzero 1 + tzero 2

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2• tv = (0 + 5) --------................ tv = 2.5 2


• 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 – 31.25• h = 31.25m• (d) The maximum height of 31.25m occurs at time

of 2.5 seconds

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

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

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



P = -2b² + 14b – 20(a) When does Wheely Fast break even (P = 0)0 = -2b² + 14b – 20 ....... set P = 00 = -2(b² + 7b – 10) ....... factor out -2, GCF0 = -2(b – 5)(b – 2) ....... factor using butterflyb – 5 = 0 and b – 2 = 0 .... Identify zero equationsb = 5 and b = 2 ....... The zeros are 5 and 2 so this

happens when they produce either 2000 or 5000 skateboards



• P = -2b² + 14b – 20• (b) How many skateboards does Wheely Fast need to

produce to maximize profit• Zeros are 2 and 5• bv = bzero 1 + bzero 2

---------------------------

2• bv = (2 + 5) --------................ bv = 3.5 (which is 3500) 2


• 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 = -24.5 + 49 – 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

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?



• 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 = 0....... Isolate mini-zero

equations• w = 0 and w = 18 ....... These are the two zeros


• A = -2w² + 36w• What are the coordinates of the vertex• Zeros are 0 and 18• wv = wzero 1 + wzero 2

---------------------------

2• wv = (0 + 18) --------................ wv = 9 2


• 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) + 324• A = 162• The maximum height of area of 162m² occurs

when w is 9 metres


Maximum area (vertex)

ZeroZero

Homework (Given the Equation)

• Handout 1 – Quadratic Word Problems - #3 & 4

• Handout 2 – Application Problems - #1, 5 & 6

Documents

Grade 10 Academic Math Chapter 3 – Analyzing and Applying Quadratic Models