Chapter 3

Quadratic Models

Ch 3 Quadratic Equation

A quadratic equation involves the square of the variable. It has the form

y = ax 2 + bx + c where a, b and c are constants

If a = 0 , there is no x-squared term, so the equation is not quadratic

-3 -2 - 1 0 1 2 + 3

x -3 -2 -1 0 1 2 3

y 13 3 -3 -5 -3 3 13

7

5

3

1

Graph of the quadratic equation y = 2x2 – 5

To solve the equation 2x2 – 5 = 7We first solve for x2 to get 2x2 = 12x2 = 6 x = + = + 2.45 and – 2.45 -

6

Extraction of roots (Ex, 1 Pg 145)

0.5 1 1.5

20

10

t

h

when t = 0.5

a (16, 0.5)

b

= 20 – 16(0.5) 2

= 20 – 16(0.25)

= 20 – 4

= 16ft

When h = 0 the equation to obtain

0 = 20 - 16t 2 16t 2 = 20

t 2 = 20/16 = 1.25

t = + = + 1. 118sec

- -

Hei

ght

Time

The formula h= 20 - 16t2

25.1

Solving Formulas

Volume of Cone V = 1 r2 h

3

3V= r2 h ( Divide both sides by h )

and find square root

r = + 3V

- h

h

r

Volume of Cylinder V= r2 h

h

r

V = r2

hr = + V - h

(Dividing both sides by h )

Compound Interest Formula

• A = P(1 + r) n

Where A = amount, P = Principal, R = rate of Interest, n = No.of years

More Extraction of Roots

Equations of the form

a( x – p) 2 = q

Can also be solved by extraction of roots after isolating the squared expression

( x – p) 2

Pythagorian Formula for Right angled triangle

In a right triangle(Hypotenuse) 2 = (Base) 2 +(Height) 2

Hypotenuse

Base

Height

90 degree

A

B C

8 in

What size of a square can be inscribed in a circle of radius 8 inches ?

16 inches

s

s

s represent the length of a side of the square s 2 + s 2 = 16 2 2s = 256 s 2 = 128 s = 128 = 11.3 inches

8in

Ch 3.2 Some examples of Quadratic Models

Height of a Baseball (Pg 156)

H = -16t 2 + 64t + 4

Evaluate the formula to complete the table of values for the height of the baseball

t 0 1 2 3 4

h 4 52 68 52 4

0 1 2 3 4

70

60

50

40

30

20

10

Highest point

3) After ½ second base ball height h = -16(1/2) 2 + 64(1/2) + 4 = 32 ft4) 3.5 second height will be 32 ft

5) When the base ball height is 64 ft the time will be 1.5 sec and 2.5 sec

6) When 20 ft the time is 0.25 and 3.75 sec7) The ball caught = 4 sec

Example 4 ( Page 156 )

• Using Graphing Calculator• H = - 16 x2 + 64x + 4

Press Y key TblStart = 0 and increment 1 Press 2nd , table

Press graph

3.3 Solving Quadratic Equations by Factoring

Zero Factor Principle

The product of two factors equals zero if and only if one or both of the factors equals zero.

In symbols ab = 0 if and only if a = o or b = 0

Example (x – 6) (x + 2) = 0 x – 6 = 0 or x + 2 = 0 x = 6 or x = -2Check 6, and – 2 are two solutions and satisfy the original equationAnd x-intercepts of the graph are 6, -2

By calculator, draw the graph

Solving Quadratic Equation by factoring

The height h of a baseball t seconds after being hit is given by

h = - 16 t 2 + 64t + 4. When will the baseball reach a height of 64 feet ?

64 = - 16 t 2 + 64t + 4

Standard form 16 t 2 – 64t + 60 = 0

4( 4 t 2 – 16t + 15) = 0 Factor 4 from left side

4(2t – 3)(2t – 5) = 0 Factor the quadratic expression and use zero factor principle

2t – 3 = 0 or 2t – 5 = 0 Solve each equation

t = 3/2 or t = 5/2

h = - 16 t 2 + 64t + 4

0 .5 1 1.5 2 2.5 3 4

7264

48

24

Pg 163

Y1 = x2 – 4x + 3

Y2 = 4(x2 – 4x + 3) Enter window Xmin = -2, Xmax = 8 Ymin = -5 Ymax = 10And enter graph

Quadratic Equations whose solutions are given

Example 31, Page 166

Solutions are – 3 and ½, The equation should be in standard form with integer coefficients

[ x – (-3)] (x – ½) = 0

(x + 3)(x – ½) = 0

x2 – ½ x + 3x – 3/2 = 0

x2 +5x – 3= 0

2 2

2(x2 + 5x –3) = 2(0)

2 2

2 x2 + 5x – 3 = 0

3.3 ,No 48, Page 169I = kCx – k x2

= 0.0002 (6000)x – 0.0002x2 = 1.2x – 0.0002x2

x 0 500 1000

1500

2000

2500

3000

3500

4000

4500

5000

5500

6000

6500

7000

I 0 550 1000

1350

1600

1750

1800

1750

1600

1350

1000

550 0 -650 -1400

Population 2000 will increase by 1600

Population 7000 will decrease by 1400

x – intercept is 6000i.e neither decrease nor increase

Larger Increase

0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000 5500 6000 6500 7000

1800

1750

1600

1350

1000

500

0

3.4 Graphing Parabolas Special cases

• The graph of a quadratic equation is called a parabola

y-intercept

x-intercept

Vertex

x-intercept

Axis of symmetryAxis of symmetry

x-interceptx-intercept

y-intercept

Using Graphing Calculator (Page 171)

Enter Y Y = x2 Y = 3 x2

Y = 0.1 x2Graph

Enter GraphEnter equation

Example 3, Pg 177, Finding the vertex of the graph ofy = -1.8x2– 16.2x

Find the x-intercepts of the graph

a) The x-coordinate of the vertex is xv = -b/2a= -(-16.2)/2(-1.8)

• To find the y-coordinate of the vertex, evaluate y at x = -4.5

• yv = -1.8(-4.5)2 – 16.2(-4.5) = 36.45

• The vertex is (-4.5, 36.45)• b) To find the x-intercepts of the graph, set y = 0 and solve• - 1.8 x2 – 16.2x = 0 Factor• -x(1.8x + 16.2) = 0 Set each factor equal to zero• - x = 0 1.8x + 16.2 = 0 Solve the equation• x = 0 x = -9• The x-intercepts of the graph are (0,0) and (-9,0)

- 10 - 5 0 2

36

24

12

3.6 Quadratic Formula

The solutions of the equation ax 2 + bx + c = 0 with

a = 0 are given by - b + b2 – 4ac

• Complex Numbers i2 = - 1 or i =

• For a > 0, = = i

1

2a

Discriminant D = b2 - 4acIf D > 0, the equation has two unequal real solutionsIf D = 0, the equation has one real solution of multiplicity twoIf D < 0, the equation has two complex (conjugate) solutions

a 1 a a

3.6, Page 195, No. 15Let w represent the width of a pen and l the length of the

enclosure in feet

w

l

Then the amount of chain link fence is given by 4w + 2l = 100b) 4w +2l = 100 2l = 100 – 4w l = 50 – 2wc) The area enclosed is A = wl = w(50 – 2w) = 50w –2w2

The area is 250 feet, so50w – 2w2 = 2500 = w2– 25w + 125Thus a = 1, b = -25 and c = 125W = -(-25) + (- 25) 2 - 4(1)(125) - 2(1) The solutions are 18.09, 6.91 feet

d) l =50 – 2(18.09) = 13.82 feet and l = 50 – 2(6.91) = 36.18 feet

The length of each pen is one third the length of the whole enclosure,so dimensions of each pen are 18.09 feet by 4.61 feet or 6.91 feet by 12.06 feet

Ex 16, Pg 195

r = ½ x

h = x - 2

x

The area of the half circle = 1 r2

2

= 1/2 (1/2 x )2

= 1/8 x2

The area of the rectangle = x2 – 2x

= 1 x2 + x2 - 2x

8

Total area = 120 square feet

120 = = 1 x2 + x2 - 2x 88(120) = 8 ( 1 x2 + x2 - 2x ) 80 = x2 + 8 x2 - 16x – 960 , 0 = ( + 8) x2 - 16x – 9600 = 11.142 x2 – 16x - 960 use quadratic formula , x = 10.03 ft , h = 10.03 – 2 = 8.03ft The overall height of the window is h + r = h + ½ x = 8.03 + ½ (10.03) = 13.05 ft