Chapter 13
Quadratic Relations and Functions
Corresponding
Notes
Date
Assigned
Day 1 & 2
Friday
3/9
Page 507:
Monday
Worksheet: HW Day 3
Consecutive Integer
2
Tuesday
3/13
Worksheet: HW Day 4
Geometric word problems
Worksheet: HW Day 5 More
Geometric word problem
3
Worksheet: Parabolas
5
Solving Quadratic
Equations
Day 3 Notes
Word Problems
Consecutive ntegers
Day 4 Notes
Geometric Word
Problems
Day 5 Notes
More Word
Problems
Day 6 Notes
Graphing
Quadratics
3/12
Wednesday
3/14
Thursday
3/15
Assignment
#
3 – 12 (all) 15 – 42
1
every 3rd ( 1st column)
4
Consecutive Integer HW
1. 7 or –3
5.
6 and 7
2. 5 or –3
6.
8 and 10
3.
–10 and –8
7
4. 7 and 8
– 8 and -7
or
or
7.
10.
2(x – 4)(x + 3)
3 10
Warm - up:
Simplify
 5 72
1. The ratio of the measures of the base and the altitude
of a parallelogram is 3:4. The area of the
parallelogram is 1,200 square centimeters. Find the
measure of the base and altitude of the
parallelogram.
A = bh
1200 = (3x)(4x)
1200 =12x2
4x
12
x2 = 100
x2 – 100 = 0
3x
Let 3x = base
4x = height
12
30 cm
40 cm
(x + 10) (x – 10) = 0
(x + 10) = 0 (x – 10) = 0
x = -10
x = 10
2.
The altitude of a triangle is 5 less than its base.
The area of the triangle is 42 square inches.
Find its base and altitude.
x-5
x
Let x = base
12 in
x – 5 = altitude 7 in
A = ½ bh
42 = ½ x (x – 5 )
2(42 = ½ x (x – 5 ))
84 = x (x – 5 )
84 = x2 - 5x
0 = x2 – 5x – 84
0 =(x – 12)(x +7)
(x - 12) = 0
x = 12
(x + 7) = 0
x = -7
3. The length of a rectangle exceeds its width by 4
inches. Find the dimensions of the rectangle it its
area is 96 square inches.
A =wl
x
96 =x(x+4)
96 =x2 + 4x
0 = x2 + 4x - 96
8 in
0 =(x – 8)(x +12)
X+4
Let x = width
x + 4 = length
12 in
(x - 8) = 0
(x + 12) = 0
x=8
x = -12
4. If the measure of one side of a square is increased
by 2 centimeters and the measure of the adjacent
side is decreased by 2 centimeters, the area of the
resulting rectangle is 32 square centimeters. Find the
measure of one side of the square.
Let x = side
A =lw
32 =(x + 2)(x - 2)
32 =x2 + 2x - 2x - 4
32 =x2 - 4
6 cm
-32
x-2
x+2
-32
0 =x2 – 36
0 =(x – 6)(x + 6)
(x - 6) = 0
x=6
(x + 6) = 0
x = -6
5. Joe’s rectangular garden is 6 meters long and 4
meters wide. He wishes to double the area of his
garden by increasing its length and width by the
same amount. Find the number of meters by which
each dimension must be increased.
6m
A =6(4)
24
4m
A =lw
48 =(x + 6)(x + 4)
48 = x2 + 6x + 4x + 24
48 = x2 + 10x + 24
-48
x+6
L= 8 m
x+4
W= 6 m
-48
0 = x2 + 10x - 24
0 =(x – 2)(x + 12)
(x - 2) = 0
x=2
(x + 12) = 0
x = -12
March 26, 2009
Pick up 1
Homework: Worksheet # 4
Geometric Problems