(chapter 3) quadratic function

QQM1023 Managerial Mathematics

Chapter 4: Quadratic Function 90

4.1 : INTRODUCTION TO QUADRATIC FUNTION • Quadratic function was described as polynomial function of degree 2. • A function f is a quadratic function if and only if f(x) can be written in

the form of: where a ≠ 0 and a, b and c are constant

• The graph of the quactratic function is called parabola. • If the value of a for a quadratic function is positive, therefore the graph

(parabola) will open upward (concave up U) – minimum • Meanwhile, if the value for a is negative, therefore the graph

(parabola) will open downward (concave down - ∩) - maximum.

ƒ(x) = ax2 + bx + c

y x-intercept c y-intercept x vertex(min) eg: y = x2 + 7x + 10 - a is positive (a = 1) - concave up - minimum vertex point

y vertex (max) y-intercept c x x-intercept eg: y =-3 x2 + 6x + 9 - a is negative (a = 1) - concave down - maximum vertex point



• The lowest (minimum) or the highest point (maximum) of a quadratic function is called the “vertex”.

Determine whether each given function below is a quadratic function or not. If it is, then state the value of a, b and c and the shape of the graph (parabola) – concave up/down? a) g(x) = 5x2 b) f(x) = 7x-2 c) y = 2x3 + 4x2 – 2x + 5 d) f(v) = -10v2 – 6

4.1.1: Vertex point

• If the value of a is greater than 0 (positive), then the quadratic function

will have a minimum vertex point. • Meanwhile, if the value of a is less than 0 (negative) then the quadratic

function will have a maximum vertex. y = ax2 +bx +c ; a > 0 y = -ax2 + bx + c ; a <0 Maximum vertex Minimum vertex Formula used to obtain the coordinate for the vertex point (x,y) of the function y = ƒ(x) = ax2 + bx + c

x-coordinate = - b 2a

y-coordinate = 4ac – b2 4a or simply finds,

−

abf

2

Example 1:



Find the vertex point for the given functions, and then determine whether it is a maximum or minimum point: a) f(x) = x2 + x -12 b) f(x) = x2 + x

c) f(x) = -3x2 + 2x + 8 d) f(x) = 2x2 + 5x – 3

e) f(x) = x2 + 4x + 6 4.1.2: y-intercept (0, c) • The y-intercept (0,c) is the point where the parabola pass through

the y-axis (or when x = 0 ). • To find the value of the y-intercept, simply replace x = 0 into the

function. y/f(x) y/f(x) x y-intercept y-intercept x Given the function f(x) = ax2+bx+c, then the y-intercept is c.

Example 2:



x = aacbb

242 −±−

Find the y-intercept for the following quadratic functions: a) f(x) = x2 + x -12 b) f(x) = x2 + x

c) f(x) = -3x2 + 2x + 8 d) f(x) = 2x2 + 5x – 3

e) f(x) = x2 + 4x + 6

4.1.3 : x-intercept • x-intercept (root/s) is the point where the parabola pass through x-axis

(or when y=0). • when y = 0, : ax2 + bx + c = 0 • There are 3 conditions for the root of a given quadratic function:

• The value/s of the x-intercept/s can be gain in 2 ways: - quadratic formula - factorization

a) Quadratic Formula :

Given the quadratic equation: ax2 + bx + c = 0, The value for x can be determine using the formula;

Attention !!! : If b2 – 4ac < 0 ; therefore they do not intercept x-axis.

Example 3:

y x • The parabola pass through

TWO points at the x-axis. • b2-4ac > 0

y x • The parabola touch the x-

axis at ONE point. • b2-4ac = 0

y x • The parabola DO NOT

pass/touch the x-axis. • b2-4ac < 0



Solve the following quadratic equation (to find the value/s of x) : a) 0 = x2 + x -12

Solution:

i. Determine the value of a, b and c: a = 1, b = 1 and c = -12 ii. Replace the value into the formula:

x = )1(2

)12)(1(411 2 −−±−

x = 2

491±−

x = 2

71+− or x =2

71−−

x = 3 x = -4 therefore the function intercept the x-axis at (3,0) and (-4,0)

b) 0 = x2 + x

c) 0 = -3x2 + 2x + 8

d) 0 = 2x2 + 5x – 3

e) 0 = x2 + 4x + 6

Example 4:

x = aacbb

242 −±−



b) Solving a quadratic equation using factorization : Solve the following quadratic equation (to find the value/s of x ): a) 0 = x2 + x -12

x x x2

(x+4) (x-3) 4 -3 -12 4x -3x x

therefore x2 + x -12 = 0

(x+4)(x-3) = 0 x + 4 = 0 or x – 3 = 0 thus, x = -4 or x = 3

b) 0 = x2 + x

c) 0 = -3x2 + 2x + 8

d) 0 = 2x2 + 5x – 3

+

x

x x x

Example 5:



4.2 : SKETCHING THE GRAPH OF QUADRATIC FUNCTION

• The graph of a quadratic function is in the form of parabola • Steps to sketch the quadratic function graph;

y = ƒ(x) = ax2 + bx + c: 1. Determine shape of the graph (concavity) :

Look at the value of a: a : positive concave up (U) a : negative concave down (∩)

2. Find the vertex point (x,y) using the formula :

3. Find the y-intercept : replace x = 0 into the function y = (a x 02) + (b x 0) + c y = c

4. Find the x-intercept : replace y=0 into the function and

find the value of x using the quadratic formula or the factorization method:

5. Draw the axis and tick all of the points (vertex, y-intercept, x-intercept/s)

6. Draw a parabola that connects all of the points and label the graph.

x = - b y = 4ac – b2 2a 4a

x = aacbb

242 −±−



Sketch the graph for each of the following quadratic functions: a) f(x) = x2 + x -12 b) f(x) = x2 + x

c) f(x) = -3x2 + 2x + 8 d) f(x) = 2x2 + 5x – 3

e) f(x) = x2 + 4x + 6

4.3 FORMING A QUADRATIC EQUATION • To form a quadratic equation, we need to know at least 3 points that

reside on the function/parabola. • Steps:

- Substitute all three coordinates of x and y into the general form of the quadratic equation; y = ax2 + bx + c.

- Therefore, we will have 3 equations in the mean of a, b and c. - Solve this three equations simultaneously (using either the

substitution, elimination, or inverse matrix, Cramer’s rule method) to find the value of a, b dan c that satisfy the three equations.

- Finally, rewrite the equation by replacing the value of a, b and c.

Example 6:



Form a quadratic equation that passes through the points (1,8), (3,20) and (-2,5) i. Substitude all three coordinates into the gerenal form of

quadratic equation y = ax2 + bx + c. ii. Solve all three equations simultaneously to find the value of a, b

and c: iii. Rewrite the equation y = ax2 + bx + c by replacing the value of a,

b and c into the equation. Form a quadratic equation that passes through the points (0,12) , (-6,0) and (2,0).

Example 7:

Example 8:



4.4 APPLICATIONS - DEMAND AND SUPPLY FUNCTION, EQUILIBRIUM Many situations in economics can be described by using quadratic functions. a) Demand and Supply Function :

• The function that relates price per unit and demanded quantity is

called a demand function. Meanwhile the function that relates

price per unit and supplied quantity is called supply function.

• For quadratic function :

i. If a is positive (a>0), : - the function has a minimum point/vertex (U) - supply function

ii. If a is negative (a<0), : - the function has a maximum point/vertex (∩), - demand function.

Price per unit (p) Supply Function

Quantity Supplied(q)

Price per unit (p) Demand Function Quantity Demanded(q)



ATTENTION : In this course, we only consider the price per unit and the quantity in the first quarter of the plane!! The Supply function y = f(q) for a product is in a form quadratic function. Three points that reside on the functions are (1,11), (0,6) and (2,18). Form the supply function of the product.

A market research done by manufacturers of a product comes out with a supply function in a form of quadratic. The manufacturers were asked on the amount (quantity) of products they will produced at a certain price per unit. The result of the research found that, at the price of RM6, RM30 and RM48, the manufacturers will produce 4, 8 and 10 units of the product. a) Form a quadratic supply function based on the informations given.

Example 9:

Example 10:



• The equilibrium point is the point where the supply meets demand.

b) If 20 units of the product were produced, how much is the price per unit?

c) If the price per unit of the product is set to RM70, how many of the

product should be produce?

b) Equilibrium point:

• Here, the quantity demanded = quantity supplied at the same price

per unit. • The equilibrium point can be obtain by solving both function

simultaneously: - Substitution Method - Elimination Method - Matrix (Inverse or Cramer’s Rule)

Attention !! For this course, the equilibrium point is only considered in the first quarter of the plane!

Price per unit (p) Supply Equilibrium b Point (a,b) Demand a Quantity Demanded/Supplied(q)



Total Revenue (TR) = Price per unit x Quantity of product Sold = p x q

Given two quadratic functions : p = f(q) = -q2 -4q +12 p= g(q) = q + 6

a) Determine which one is the demand/supply function. b) Find the equilibrium point of the two functions.

4.5 TOTAL REVENUE, TOTAL COST, PROFIT AND BREAK-EVENT POINT (B.E.P)

a) Total Revenue : • Total revenue is define by the product between price per unit and

quantity of the product sold.

• Let say the price per unit (p) is determine by the demand (in linear

form). p = -mq + c ---------------demand function where p is the price per unit (RM)when q units of the product were demanded

Example 11:



• Here, the Total Revenue is in quadratic form because: 0

Now, the total revenue is in a form of quadratic!! • Thus, the Total Revenue :

- The Total Revenue function always starts at the origin (0,0). - As the a is negative, therefore the Total Revenue function will have a

maximum point (concave down). - The maximum point (vertex) here represent the quantity that

maximize the total revenue (on x-axis) and the maximum revenue (on y-axis) for the product.

Total Revenue = p x q Therefore, Total Revenue = (-mq + c) q = -mq2 + cq

Where; p = -mq+c (demand)

y = ax2 + bx + c

Total Revenue(RM) Max. Total Revenue=

a

bac4

4 2−

Quantity (units) Quantity that will maximizeTR=

ab

2−



Total Cost = Fixed Cost + Variable Cost

The demand function for a product is given by p = 1200 – 3q where p is the price per unit(RM) and q is the quantity demanded.

a) Determine the Total Revenue function for the product. b) How many product should be produce in order to maximize the

total revenue? c) How much is the maximum total revenue?

b) Total Cost • Total Cost – is the sum of fixed cost and variable cost.

REMEBER!! What is Fixed Cost and what is Varible cost from the previous chapter??

- However, in this chapter, the total cost function is in the form of quadratic

- C(q) = aq2 + bq + c where; Fixed Cost Variable cost

Example 12:



Profit / Loss = Total Revenue – Total Cost

The Total Cost to produce 10 unit of pencils is RM380. Meanwhile the Total Cost to produce 20 unit of the pencils is RM1060. However, if no pencils are produce, the business still needs to pay RM100 for the Total Cost :

Find:

a)Total Cost Function. b)How much is the Fixed Cost? c) The amount of product (quantity) to be produce when the total cost is

RM 740.

c. Profit / Loss • Obtain by substracting the Total Cost from the Total Revenue :

- Profit : Total Revenue > Total Cost - Lost : Total Revenue < Total Cost

Example 13:



Total Revenue = Total Cost OR

Profit/Loss = 0

d.Break-Event Point • Break-event point is thepoint where the Total Cost and Total Revenue

intersect. • Here, the Total Cost = Total Revenue, there are no profit or loss.

ATTENTION!! : For this course, we only considerBEP in the first quarter of the plane.

The total revenue for a product is given by the function R(q) = 2.5q, and the Total Cost function is C(q) = 100 + 2q – 0.01q2

Determine a) Profit Function

b) Profit gain, if 100 unit of the products were sold

c) Break Event Point (BEP)

Example 14:

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(chapter 3) quadratic function