1. Chapter 3 Quadratic Models

2. 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. Graph of the quadratic equation y = 2x2 – 5-3 -2 -1 1 2 3 y 13 -5
To solve the equation 2x2 – 5 = 7
We first solve for x2 to get
2x2 = 12
x2 = 6
x = = and – 2.45 - 7 5 3 1

4. Extraction of roots (Ex, 1 Pg 145)
The formula
h= t2 h
when t = 0.5
= 20 – 16(0.5) 2
= 20 – 16(0.25)
= 20 – 4
= 16ft
When h = 0 the equation to obtain
0 = t 2
16t 2 = 20
t 2 = 20/16 = 1.25
t = = sec
a (16, 0.5) 20 10
Height b t
Time

5. Solving Formulas h r r h Volume of Cylinder V= r2 h V = r2 h
Volume of Cone V = r2 h 3
3V= r2 h ( Divide both sides by h )
and find square root
r = V h r r h
Volume of Cylinder V= r2 h
V = r2 h
r = V h
(Dividing both sides
by h )

6. Compound Interest Formula
A = P(1 + r) n
Where A = amount, P = Principal, R = rate of Interest, n = No.of years

7. 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

8. Pythagorian Formula for Right angled triangle
Hypotenuse
Height
90 degree C B
Base
In a right triangle
(Hypotenuse) 2 = (Base) 2 +(Height) 2

9. s represent the length of a side of the square s 2 + s 2 = 16 2
What size of a square can be inscribed in a circle of radius 8 inches ?
8 in
16 inches s
8in s
s represent the length of a side of the square
s 2 + s 2 = 16 2
2s = 256
s 2 = 128
s = = 11.3 inches

10. Ch 3.2 Some examples of Quadratic Models
Height of a Baseball (Pg 156)
H = -16t t + 4
Evaluate the formula to complete the table of values for the height of the baseball t 1 2 3 4 h 52 68 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 ft
4) 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 sec
7) The ball caught = 4 sec

11. Example 4 ( Page 156 ) Press graph Using Graphing Calculator
H = - 16 x2 + 64x + 4
Press Y key TblStart = 0 and increment 1 Press 2nd , table
Press graph

12. 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 = -2
Check 6, and – 2 are two solutions and satisfy the original equation
And x-intercepts of the graph are 6, -2
By calculator, draw the graph

13. Solving Quadratic Equation by factoring
The height h of a baseball t seconds after being hit is given by
h = - 16 t t + 4. When will the baseball reach a height of 64 feet ?
64 = - 16 t t + 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 = Solve each equation
t = 3/2 or t = 5/2
h = - 16 t t + 4 72 64 48 24

14. Pg 163 Enter window Xmin = -2, Xmax = 8 Ymin = -5 Ymax = 10
Y1 = x2 – 4x + 3
Y2 = 4(x2 – 4x + 3)
Enter window Xmin = -2, Xmax = 8
Ymin = -5 Ymax = 10
And enter graph

15. 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(x2 + 5x –3) = 2(0)
2 x2 + 5x – 3 = 0

16. 3.3 ,No 48, Page 169
I = kCx – k x2
= (6000)x – x2 = 1.2x – x2
x – intercept is 6000
i.e neither decrease nor increase
Larger Increase x
500
1000
1500
2000
2500
3000
3500
4000
4500
5000
5500
6000
6500
7000 I
550
1350
1600
1750
1800
-650
-1400
Population 2000 will increase by 1600
1800
1750
1600
1350
1000
500
Population 7000 will decrease by 1400

17. 3.4 Graphing Parabolas Special cases
The graph of a quadratic equation is called a parabola
Vertex
y-intercept
x-intercept
x-intercept
x-intercept
y-intercept
x-intercept
Axis of symmetry
Axis of symmetry

18. Using Graphing Calculator (Page 171)
Enter Y
Y = x2
Y = 3 x2
Y = 0.1 x2
Graph
Enter equation
Enter Graph

19. a) The x-coordinate of the vertex is xv = -b/2a= -(-16.2)/2(-1.8)
Example 3, Pg 177, Finding the vertex of the graph of y = -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 = Factor
-x(1.8x ) = Set each factor equal to zero
- x = x = Solve the equation
x = x = -9
The x-intercepts of the graph are (0,0) and (-9,0) 36 24 12

20. 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 2a
Discriminant D = b2 - 4ac
If D > 0, the equation has two unequal real solutions
If D = 0, the equation has one real solution of multiplicity two
If D < 0, the equation has two complex (conjugate) solutions

21. 3.6, Page 195, No. 15 Let w represent the width of a pen and l the length of the enclosure in feet
Then the amount of chain link fence is given by 4w + 2l = 100
b) 4w +2l = 100
2l = 100 – 4w
l = 50 – 2w
c) The area enclosed is A = wl = w(50 – 2w) = 50w –2w2
The area is 250 feet, so
50w – 2w2 = 250
0 = w2– 25w + 125
Thus a = 1, b = -25 and c = 125
W = -(-25) (- 25) (1)(125) -
2(1) The solutions are 18.09, feet
d) l =50 – 2(18.09) = feet and l = 50 – 2(6.91) = feet
The length of each pen is one third the length of the whole enclosure,
so dimensions of each pen are feet by 4.61 feet or 6.91 feet by feet

22. Ex 16, Pg 195
r = ½ x
The area of the half circle = r2 2
= 1/ (1/2 x )2
= 1/8 x2
The area of the rectangle = x2 – 2x
= x x2 - 2x 8
h = x - 2
Total area = 120 square feet
120 = = x2 + x2 - 2x 8
8(120) = 8 ( x2 + x2 - 2x )
0 = x2 + 8 x2 - 16x – 960 , 0 = ( ) x x – 960
0 = x2 – 16x
use quadratic formula , x = ft , h = – 2 = 8.03ft The overall height of the window is h + r = h + ½ x = ½ (10.03) = ft x