1. 1 Topic 1 Topic 1 : Elementary functions Reading: Jacques Section 1.1 - Graphs of linear equations Section 2.1 – Quadratic functions Section 2.2 – Revenue, Cost and Profit

2. 2 Linear Functions The function f is a rule that assigns an incoming number x, a uniquely defined outgoing number y. y = f(x) The Variable x takes on different values…... The function f maps out how different values of x affect the outgoing number y. A Constant remains fixed when we study a relationship between the incoming and outgoing variables

3. 3 Simplest Linear Relationship: y = a+bx  independent dependent   variable variableintercept This represents a straight line on a graph i.e. a linear function has a constant slope b = slope of the line = change in the dependent variable y, given a change in the independent variable x. Slope of a line=  y /  x = (y2-y1) / (x2-x1)

4. 4 Example: Student grades Example: y = a + bx y : is the final grade, x : is number of hours studied, a%: guaranteed Consider the function: y = 5+ 0x What does this tell us? Assume different values of x ………

5. 5 Example Continued: What grade if you study 0 hours? 5 hours? y=5+0x Output = constant slope Input yabX 5500 5501 5502 5503 5504 5505

6. 6 Example Continued…. y=5+15x If x = 4, what grade will you get? Y = 5 + (4 * 15) = 65

7. 7 Demand functions: The relationship between price and quantity If p =5, how much will be demanded? D = 10 - (2 * 5) = 0

8. 8 Inverse Functions: Definition If y = f(x) then x = g(y) f and g are inverse functions Example Let y = 5 + 15x If y is 80, how many hours per week did they study?

9. 9 Example continued….. If y is 80, how many hours per week did they study? Express x as a function of y: 15x = y – 5.... So the Inverse Function is: x = (y-5)/15 Solving for value of y = 80 x = (80-5 / 15) x = 5 hours per week

10. 10 An inverse demand function If D = a – bP then the inverse demand curve is given by P = (a/b) – (1/b)D E.g. to find the inverse demand curve of the function D= 10 -2P …… First, re-write P as a function of D 2P = 10 – D Then, simplify So P= 5 – 0.5D is the inverse function

11. More Variables: Student grades again: y = a + bx + cz y : is the final grade, x : is number of hours studied, z: number of questions completed a%: guaranteed Example: If y = 5+ 15x + 3z, and a student studies 4 hours per week and completes 5 questions per week, what is the final grade? Answer: y = 5 + 15x + 3z y = 5 + (15*4) + (3*5) y = 5+60+15 = 80

12. 12 Another example: Guinness Demand. The demand for a pint of Guinness in the Student bar on a Friday evening is a linear function of price. When the price per pint is € 2, the demand ‘is € 6 pints. When the price is € 3, the demand is only 4 pints. Find the function D = a + bP 6 = a + 2b => a = 6-2b 4 = a + 3b => a = 4-3b 6-2b = 4-3b Solving we find that b = -2 If b = -2, then a = 6-(-4) = 10 The function is D = 10 – 2P What does this tell us?? Note, the inverse Function is P = 5- 0.5D

13. 13 A Tax Example…. let €4000 be set as the target income. All income above the target is taxed at 40%. For every €1 below the target, the worker gets a negative income tax (subsidy) of 40%. Write out the linear function between take- home pay and earnings. Answer: THP = E – 0.4 (E – 4000) if E>4000 THP = E + 0.4 (4000-E) if E<4000 In both cases, THP = 1600 +0.6E So i) If E = 4000 => THP = 1600+2400=4000 ii) If E = 5000 => THP = 1600+3000=4600 iii) If E = 3000 => THP = 1600+1800=3400

14. 14 Tax example continued…. THP = 1600 +0.6E If the hourly wage rate is equal to €3 per hour, rewrite take home pay in terms of number of hours worked? Total Earnings E = (no. hours worked X hourly wage) THP = 1600 + 0.6(3H) = 1600 + 1.8H Now add a (tax free) family allowance of €100 per child to the function THP = 1600 +0.6E THP = 1600 + 0.6E + 100Z (where z is number of children) Now assume that all earners are given a €100 supplement that is not taxable, THP = 1600 + 0.6E + 100Z + 100 = 1700 + 0.6E + 100Z

15. 15 Topic 1 continued: Non- linear Equations Jacques Text Book: Sections 2.1 and 2.2

16. 16 Quadratic Functions Represent Non-Linear Relationships y = ax 2 +bx+c where a  0, c=Intercept a, b and c are constants So the graph is U-Shaped if a>0, And ‘Hill-Shaped’ if a<0 And a Linear Function if a=0

17. 17 Solving Quadratic Equations: 1) Graphical Approach: To find Value(s), if any, of x when y=0, plot the function and see where it cuts the x-axis If the curve cuts the x-axis in 2 places: there are always TWO values of x that yield the same value of y when y=0 If it cuts x-axis only once: when y=0 there is a unique value of x If it never cuts the x-axis: when y=0 there is no solution for x

18. 18 e.g. y = -x 2 +4x+5 Since a ‘Hill Shaped Graph’

19. 19 The graph y=0, then x= +5 OR x = -1

20. 20 Special Case: a=1, b=0 and c=0 So y = ax 2 +bx+c => y = x2 Min. Point: (0,0) Intercept = 0

21. 21 Practice examples Plot the graphs for the following functions and note (i) the intercept value (ii) the value(s), if any, where the quadratic function cuts the x-axis y = x 2 -4x+4 y = 3x 2 -5x+6

22. 22 Solving Quadratic Equations: 2) Algebraic Approach: find the value(s), if any, of x when y=0 by applying a simple formula…

23. 23 Example e.g.y = -x 2 +4x+5 hence, a = -1; b=4; c=5 Hence, x = +5 or x = -1 when y=0 Function cuts x-axis at +5 and –1

24. 24 Example 2 y = x 2 -4x+4 hence, a = 1; b= - 4; c=4 If y = 0 x = 2 when y = 0 Function only cuts x-axis at one point, where x=2

25. 25 Example 3 y = 3x 2 -5x+6 hence, a = 3; b= - 5; c=6 If y = 0 when y = 0 there is no solution The quadratic function does not intersect the x-axis

26. 26 Understanding Quadratic Functions  intercept where x=0 is c  a>0 then graph is U-shaped  a<0 then graph is inverse-U  a = 0 then graph is linear b 2 – 4ac > 0 : cuts x-axis twice b 2 – 4ac = 0 : cuts x-axis once b 2 – 4ac < 0 : no solution

27. 27 Essential equations for Economic Examples: Total Costs = TC = FC + VC Total Revenue = TR = P * Q  = Profit = TR – TC Break even:  = 0, or TR = TC Marginal Revenue = MR = change in total revenue from a unit increase in output Q Marginal Cost = MC = change in total cost from a unit increase in output Q Profit Maximisation: MR = MC

28. 28 An Applied Problem A firm has MC = 3Q 2 - 32Q+96 And MR = 236 – 16Q What is the profit Maximising Output? Solution Maximise profit where MR = MC 3Q 2 – 32Q + 96 = 236 – 16Q 3Q 2 – 32Q+16Q +96 – 236 = 0 3Q 2 – 16Q –140 = 0 Solve the quadratic using the formula where a = 3; b = -16 and c = -140 Solution: Q = +10 or Q = -4.67 Profit maximising output is +10 (negative Q inadmissable)

29. 29 Graphically

30. 30 Another Example…. If fixed costs are 10 and variable costs per unit are 2, then given the inverse demand function P = 14 – 2Q: 1.Obtain an expression for the profit function in terms of Q 2.Determine the values of Q for which the firm breaks even. 3.Sketch the graph of the profit function against Q

31. 31 Solution: 1. Profit = TR – TC = P.Q – (FC + VC)  = (14 - 2Q)Q – (2Q + 10)  = -2Q 2 + 12Q – 10 2.Breakeven: where Profit = 0 Apply formula to solve quadratic where  = 0 so solve -2Q 2 + 12Q – 10 = 0 with Solution: at Q = 1 or Q = 5 the firm breaks even

32. 32 3. Graphing Profit Function STEP 1: coefficient on the squared term determines the shape of the curve STEP 2: constant term determines where the graph crosses the vertical axis STEP 3: Solution where  = 0 is where the graph crosses the horizontal axis

33. 33

34. 34 Questions Covered on Topic 1: Elementary Functions Linear Functions and Tax…… Finding linear Demand functions Plotting various types of functions Solving Quadratic Equations Solving Simultaneous Linear (more in next lecture) Solving quadratic functions