(chapter 2) Linear Function

QQM1023 Managerial Mathematics Y = mx + C

3.1 INTRODUCTION •

A function f is a linear function if and only if f ( x) can be written in the form f ( x) = mx + c , where m and c are constant and m ≠ 0 .

•

Let y = f ( x) , then y = mx + c , is an equation of a straight line with slope

m and y-intercept c.

y = mx + c slope

•

y-intercept

Thus, the graph of a linear function is a straight line.

y

y=mx+c c

x

Example 1: Determine whether the functions given is a linear function or not. If it is, then identify the slope (m) and the y-intercept (c). a) y = -3x + 4 b) 2y = 4x3 – 6 c) y = x½ d) 4y = 24x + 16 e) y = (x2 – 81) (x-9)

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3.2 Slope of a line (m) • •

The slope determine whether the line is skewed to the left (m is negative) or to the right (m is positive) To calculate the slope of a line connected by two points namely (x1,y1) and (x2,y2) :

Slope (m) = y2 - y1 x2 - x1 Slope(m) = - y-intercept x-intercept •

Types of slope :

a)

Positive slope -

c)

Skewed to the right eg : y = 2x

Horizontal line -m=0 - eg : y = 3

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b) Negative slope - skewed to the left - eg : y = -2x

d) Vertical Line - m = undefine - eg : x = 3

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Example 2: Find the slope of the straight line connected by these two points a) A( 5,4) andB(-1,2)

b) P(11,-3) andQ(-3,7)

c) R(k, 2k2) and S(3, 6k)

d) T(-5, 4) andU(3,4)

e) L(3.5, 0) and M(3.5, 4.5)

f) Y(0, -9) and Z(1.5, 0)

3.3 Equation of a straight line There are 4 ways to form an equation of straight line: a) Given : Slope(m) and y-intercept (c):

y = mx + c Example 3: If the slope of a straight line L is 2 and the y-intercept is 5, Form the equation that define the straight line. Equation:

m = 2 and c =5

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QQM1023 Managerial Mathematics b) Given : Slope (m) and a point (x1,y1) on the line :

y – y1 = m (x – x1) Example 4: Find the equation of the straight line with the slope m and pass through the point P : i) m = 4, P(-3, 2) Equation :

Exercise: ii) m = -5, P(-2,1)

iii)

m = 1/3, P(4,-2)

c) Given : 2 points on the line : (x1,y1) dan (x2, y2) - Start by calculating the slope of the line using the formula:

Slope (m) = y2 – y1 x2 – x1 - Then pick a point, and replace the x coordinate and y coordinate as well as the slope(m) in the formula :

y – y1 = m (x – x1) Chapter 3 : Linear Function

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Example 5: Find the equation that pass through the given points. A( 4, 3) dan B(-2, 0) x1 y1

x2 y2

Slope = m

=

Equation :

Exercise: Find the equation that pass through the given points. , a. P(4,8) and Q(0,-3)

b. R(1,0) and S(5,8)

d) Given : x-intercept and y-intercept :

(a,0) and (0,b)

x + y=1 a b x-intercept

y-intercept

Example 6: Find the straight line equation that intercept the x-axis at (3,0) and yaxis at (0,4) Equation :

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QQM1023 Managerial Mathematics Exercise: Find the straight line equation that has i) x-intercept = 1 and y-intercept = -3 2

3.4 Sketching the graph of linear function The graph of a linear function is in a form of a straight line. Thus, we need only determine two different points on graph in order to sketch it. To make it easier, we could find the y-intercept and the x-intercept. Steps: 1. Find the y-intercept (0,c) : replace x=0 to the equation to find the y coordinate (c). 2. Find the x-intercept (b,0) : replace y=0 to the equation to find the x coordinate(b). 3. Draw the x and y-axis. 4. Tick the points (x and y-intercept) and draw a line that connects the two point. 5. Label the graph. Example: Sketch the graph of 2 x + 3 y + 6 = 0

y

x

(-3,0)

(0,-2)

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Exercise: Sketch the graph for

1. 2. 3. 4.

y = 4x − 6 y =1 x = −5 x + 2y −3 = 0

3.5 Parallel Lines and Perpendicular Lines

Suppose that we have two straight lines : L1 and L2 y = m1x + a L1 : L2 : y = m2x + b , therefore a) Two straight lines L1 and L2 are parallel if L1 and L2 have the same slope, m1 = m2 .

m1 = m2

y L1 with m1 slope L2 with m2 slope

x

b) Two straight lines L1 and L2 are intersect if L1 and L2 have different slope. And, the two lines with slopes m1 and m2 are perpendicular to each other if and only if

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m1 x m2 = -1

Moreover, a horizontal line and a vertical line are perpendicular to each other.

y L1 with m1 slope L2 with m2 slope 90o

x

Example 7: Determine whether the two lines given are parallel or not. a) y = 2x + 4 and y = -2x + 4 b) 2y = 16x + 8 and y = 8x c) 4y = -3x + 24 and 3y = -4x + 2

Example 8: Determine whether the two lines given are perpendicular or not. a) y = 2x + 4 and y = - x + 8 2 b) 2y = -8x + 3 and 4y = 8x + 9 c) 3y = 9x + 1 and y = -x 3

Example 9: Find the straight line equation that pass through (1,2) and paralle to the y =2x + 3 line.

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Example 10: Find the straight line that pass through (-3, 4) and perpendicualr to the y = 3x – 5 line.

Example 11: Find the straight line equation that pass through (4, -5) and perpendicular to the y = - x + 7 line 4

3.6 Intersection point of two lines • •

The intersection point is the point where two or more lines met. In determining the intersection point of straight line, there are 3 possibilities:

a) 2 or more straight line intersect (either perpendicularly or not) at ONE point y

L2 intersection L1

X Eg : the y = 2x line and y = -2x line REMEMBER: two straight lines with different slope will intersect at ONE point!!

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QQM1023 Managerial Mathematics b) 2 or more straight line DO NOT intersect to each other – Parallel If two lines have the same slope(m1 = m2) – then the two lines are parallel. Therefore, this two lines will not intersect each other. y L1

L2

x Example : y = 2x + 4 line and y = 2x line will not intersect each other

c) 2 or more lines are identical - therefore each point on the lines are the intersect points. x

L1 and L2

y Example : y = 2x + 4 line and 3y = 6x + 12 line

Determining the intersect point of two straight lines We can find the intersection point between two straight lines by solving both equation simultaneously. There are two ways of doing so; - Substitution Method - Elimination Method Otherwise we can also use the methods we have learnt in chapter one (using Inverse matrix or Cramer’s Rule)

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Example 12: Substitution Method Find the intercept points of the given lines : y = 2x + 4 and 2y = x + 5

Example 13: Elimination Method: Find the intercept points of the given lines: 3y = 4x + 7 dan 9y = 6x – 15

* p/s: Otherwise, we also can solve this system of equations using method learnt in

Chapter 1. We can use either inverse method or Cramer’s Rule. Let’s try to apply the method and we should obtain the same answer.

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3.7 Applications of Linear Function • •

In this chapter we will be introduce to two function that takes the linear form: the demand and supply function. In economy, normally the y-axis will take the value of the price(p) and the x-axis will take the value of the quantity(q) or volume of production.

a) Demand function: • Describe the customers’ behaviors towards the demand for a commodity. • Whenever the price (p) for a commodity is high, the demanded units or quantity (q) will be less. • This shows that the price and the quantity demanded have a negative relation – demand function has a negative slope. p

b

(a,b)

q a At the price of $ b per unit, the demanded quantity is at a unit b) Supply function : • For supply function, whenever the price (p) for a commodity is high, the quantity of supply (q) will increase too. • This shows that the price and the quantity have a positive relation – supply function has a positive slope. p

(c,d) d

q c At the price of $ d per unit, the quantity of supply at c unit ATTENTION: For this course, the demand and supply function will only be considered in the first quarter of the plane.

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QQM1023 Managerial Mathematics There are also situations where the demand and supply function is in a form of horizontal line (fixed price) or vertical line (fixed quantity). c) Case 1 : Fixed price • The price for a commodity is fixed no matter how many units is demanded/supplied. • Example 1 :

• Example 2 :

p

a

q b

c

d) Case 2 : Fixed Quantity • The price for a commodity is varied for the same amount or quantity demanded/supplied.

p 200 140 a

q

• Example 1: • Example 2:

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Example 14: When the price of a watch is set to RM100, none of it can be sold. However, if it is free, 50 of its were demanded. Determine the demand function?

Example 15: Every year, 50 generators are bought by TNB no matter how much it costs. Determine the demand function.

Example 16: When the price of a camera is set to RM25, there are no camera in the stock (to be sold). If the price is set to RM40, 20 units of camera can be sold. Find the supply function.

Example 17: Based on the agreement between Telekom Malaysia and FTM, RM500 need to be paid per month by FTM no matter how many calls are made and the distance of the calls. Find the supply function.

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e) Equilibrium : • When the demand and supply of a product are drawn on the same coordinate plane, the point (m, n) where both functions intersect is call the point of equilibrium. • The price ( n ), called the price of equilibrium, is the price at which consumers will purchase a number of products (m) that producers is willing to supply (and sells at the same price). In short, m is called the quantity of equilibrium. • Attention : in this course we only consider the equilibrium which lies in the first quarter of the plane. p

Equilibrium (m,n)

Supply

n Demand

q m

Example 18: Find the equilibrium point for the given demand and supply function y = 10 – 2x y = 3x + 1 2

Example 19: Find the equilibrium point for the given demand and supply function y = 5 – 3x y = 4x + 12

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3.8 Total Cost, Revenue, Profit / Loss and Break-Event Point(BEP) a. Total Cost: • Total icost is the amount invested in order to come out with products or services. • There are two types of cost: - Fixed Cost : The sum of all costs that are independent of the level of production. Eg: rent, insurance. This cost must be paid whether or not output is produced. - Variable Cost : The sum of all costs that are dependent on the level of output. Eg: labors, materials • Total Cost - the sum of variable cost and fixed cost

Total Cost = Fixed Cost + Variable Cost

b. Total Revenue : • Total revenue is the money that the manufacturer receives (gross) for selling the outputs. • Therefore, the total revenue:

Total Revenue = Price per unit x Quantity sold c. Profit / Loss : • Profit is the nett income obtained by selling the outputs. • Profit is a measure of difference between Total Cost and Total Revenue. • Profit is obtain if the Total Revenue is greater than the Total Cost. Meanwhile Loss happens when the Total Cost is greater than Total Revenue.

Profit / Loss = Total Revenue – Total Cost

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QQM1023 Managerial Mathematics d. Break-Event point (BEP) : • Break-event point is the point where the Total Cost and Total Revenue are intersected. In this case, no profit or loss is obtain. • Therefore, at the BEP ,

Profit / Loss = 0 Total Cost = Total Revenue

Example 19: Assume that the total cost to produce 10 units of a product is RM40, and total cost to produce 20 units is RM70. If the total cost (TC) have a linear relation with the output (q), get the a. Function that define TC as a linear function of q

b. Total Cost to produce 35 unit of the products.

Example 20: A company sells a product at the price RM45 per unit. The varaible cost for each unit of the product is RM33 and the fixed cost is RM450,000. a) Find the break-event quantity?

b) How many unit of the product should be sells in order to gain profit at least RM150,000?

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