1 Linear and Quadratic Functions On completion of this module you should be able to: define the terms function, domain, range, gradient, independent/dependent.

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Presentation transcript:

1 Linear and Quadratic Functions On completion of this module you should be able to: define the terms function, domain, range, gradient, independent/dependent variable use function notation recognise the relationship between functions and equations graph linear and quadratic functions calculate the function given initial values (gradient, 1 or 2 coordinates) solve problems using functions model elementary supply and demand curves using functions and solve associated problems

2 A function describes the relationship that exists between two sets of numbers. Put another way, a function is a rule applied to one set of numbers to produce a second set of numbers. Functions

3 Example: Converting Fahrenheit to Celsius This rule operates on values of F to produce values of C. The values of F are called input values and the set of possible input values is called the domain. The values of C are called output values and the set of output values produced by the domain is called the range.

4 Consider the function The x are the input values and f(x), read f of x, are the output values. The domain is the set of positive real numbers including 0 and excepting 3. (Why?) The output values produced by the domain is the range. Sometimes the symbol y is used instead of f(x). Function Notation

5 An equation is produced when a function takes on a specific output value. eg f(x) = 3x + 6 is a function. When f(x) = 0, then the equation becomes 0 = 3x + 6 which can be easily solved (to give x = −2) Function and Equations

6 This is shown graphically as follows:

7 Input and output values form coordinate pairs: (x, f(x)) or (x, y). x values measure the distance from the origin in the horizontal direction and f(x) values the distance from the origin in the vertical direction. To plot a straight line (linear function), 2 sets of coordinates (3 sets is better) must be calculated. For other functions, a selection of x values should be made and coordinates calculated. Graphing Functions

8 Graph f(x) = 2x − 4 Example: Linear Function

9

10 At the y-intercept, x = 0, so and the coordinate is (0,2). Example: Quadratic Function Graph the function:

11 At the x-intercept, f(x) = 0, so and the coordinates are (2,0) and (0.5,0).

12 Vertex: The coordinates of the vertex are: (1.25, −1.125).

x (1.25, ) (2,0) (0.5,0) (0,2)

14 All linear functions (or equations) have the following features: a slope or gradient (m) a y-intercept (b) If (x 1, y 1 ) and (x 2, y 2 ) are two points on the line then the gradient is given by: Linear Functions

15 Gradient is a measure of the steepness of the line. If m > 0, then the line rises from left to right. If m < 0, the line falls from left to right. A horizontal line has a gradient of 0; a vertical line has an undefined gradient. The y-intercept is calculated by substituting x = 0 into the equation for the line.

16 All straight line functions can be expressed in the form y = mx + b Note: The standard form equation for linear functions is Ax + By + C = 0. Equations in this form are not as useful as when expressed as y = mx + b. Equations can be derived in the following way, depending on what information is given.

17 1.Given (x 1, y 1 ) and (x 2, y 2 ) 2.Given m and (x 1, y 1 ) 3.Given m and b Deriving Straight Line Functions

18 A tractor costs $60,000 to purchase and has a useful life of 10 years. It then has a scrap value of $15,000. Find the equation for the book value of the tractor and its value after 6 years. Problem: Depreciation

19 V ? 6 15, t 60,000

20 Value (V) depends on time (t). t is called the independent variable and V the dependent variable. The independent variable is always plotted on the horizontal axis and the dependent variable on the vertical axis.

21 Given two points, the equation becomes:

22 The book value of the tractor after 6 years is $33,000. or more correctly

23 Suppose a manufacturer of shoes will place on the market 50 (thousand pairs) when the price is $35 (per pair) and 35 (thousand pairs) when the price is $30 (per pair). Find the supply equation, assuming that price p and quantity q are linearly related. Example

24

25 The supply equation is

26 For sheep maintained at high environmental temperatures, respiratory rate r (per minute) increases as wool length l (in centimetres) decreases. Suppose sheep with a wool length of 2cm have an (average) respiratory rate of 160, and those with a wool length of 4cm have a respiratory rate of 125. Assume that r and l are linearly related. (a) Find an equation that gives r in terms of l. (b) Find the respiratory rate of sheep with a wool length of 1cm. Example

27 (a) Find r in terms of l  l is independent r is dependent Coordinates will be of the form: (l, r).

28

29 When wool length is 1 cm, average respiratory rate will be per minute. (b) When l = 1

30 All quadratic functions can be written in the form where a, b and c are constants and a  0. Quadratic Functions

31 In general, the higher the price, the smaller the demand is for some item and as the price falls demand will increase. Demand curve q p Elementary Supply and Demand

32 Concerning supply, the higher the price, the larger the quantity of some item producers are willing to supply and as the price falls, supply decreases. q p Supply curve

33 Note that these descriptions of supply and demand imply that they are dependent on price (that is, price is the independent variable) but it is a business standard to plot supply and demand on the horizontal axis and price on the vertical axis.

34 The supply of radios is given as a function of price by and demand by Find the equilibrium price. Example: Equilibrium price

35 Graphically, Note the restricted domains p 70 equilibrium price 0 0

36 Algebraically, D(p) = S(p)

37 −14 is not in the domain of the functions and so is rejected. The equilibrium price is $4.

38 If an apple grower harvests the crop now, she will pick on average 50 kg per tree and will receive $0.89 per kg. For each week she waits, the yield per tree increases by 5 kg while the price decreases by $0.03 per kg. How many weeks should she wait to maximise sales revenue? Example: Maximising profit

39 Weight and Price can both be expressed as functions of time (t). W(t) = t P(t) = 0.89 − 0.03t

40 Maximum occurs at She should wait 9.83 weeks ( 10 weeks) for maximum revenue. (R = $59 per tree)