1 standards 2, 5, 25 slope intercept form and point-slope form problem 1 problem 3 problem 4 problem...
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1
Standards 2, 5, 25
SLOPE INTERCEPT FORM AND POINT-SLOPE FORM
PROBLEM 1
PROBLEM 3 PROBLEM 4
PROBLEM 2
PROBLEM 12
SOLVING SYSTEMS OF TWO VARIABLES LINEAR EQUATIONS
BY GRAPHING, ELIMINATION AND SUBSTITUTION.
PROBLEM 13
PROBLEM 15 PROBLEM 16
PROBLEM 14
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PROBLEM 5 PROBLEM 6
PROBLEM 7 PROBLEM 8
PROBLEM 9 PROBLEM 10
PROBLEM 11
SPECIAL FUNCTIONS
2
Standard 2:
Students solve systems of linear equations and inequalities (in two or three variables) by substitution, with graphs, or with matrices.
Estándar 2:
Los estudiantes resuelven sistemas de ecuaciones lineares y desigualdades (en 2 o tres variables) por substitución, con gráficas o con matrices.
Standard 5:
Students demonstrate knowledge of how real and complex numbers are related both arithmetically and graphically. In particular, they can plot complex numbers as points in the plane.
Estándar 5:
Los estudiantes demuestran conocimiento de como los números reales y complejos se relacionan tanto aritméticamente como gráficamente. En particular, ellos grafican números como puntos en el plano.
Standard 25:
Students use properties from number systems to justify steps in combining and simplifying functions.
Estándar 25:
Los estudiantes usan propiedades de sistemas numéricos para justificar pasos en combinar y simplificar funciones.
ALGEBRA II STANDARDS THIS LESSON AIMS:
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3
Standard 5
Write the slope-intercept form for this equation and graph it.
4x + 2y = 10-4x -4x
2y = -4x + 102 2
y = - x +42
10 2
y = -2x + 5
m= -2
y- intercept = (0,5)
42 6-2-4-6
2
4
6
-2
-4
-6
8 10-8-10
8
-8
10
x
y
+1
21+
-=
-
2
y= mx + b
Slope Intercept
Form
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4
Standard 5
Write the slope-intercept form for this equation and graph it.
-9x + 3y = 3+9x +9x
3y = 9x + 33 3
y = x +93
3 3
y = 3x + 1
m= 3
y- intercept = (0, 1)
42 6-2-4-6
2
4
6
-2
-4
-6
8 10-8-10
8
-8
10
x
y
31+
+=
y= mx + b
3++1
Slope Intercept
Form
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5
Standard 5
42 6-2-4-6
2
4
6
-2
-4
-6
8 10-8-10
8
-8
10
x
y
(y – ) = (x – )54
y - 2 = (x – 4)54
y - 2 = x - 20 4
54
y = x -3 54
(y – ) = m(x – )x1y1
Point = (4, 2)x1
4
y1
2
Write the slope-intercept form of the equation that passes through (4,2) and has a slope of , then graph it.5
4m = 5
4
y - 2 = x - 554
+2 +2
y- intercept = (0,-3)
m 54+
+=
y= mx + b
+4
5+
Point-Slope Form
Slope Intercept
Form
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6
Standard 5
(y – ) = (x – )37
y - 1 = (x – 7)37
y - 1 = x - 21 7
37
y = x -2 37
(y – ) = m(x – )x1y1
Point = (7, 1)x1
7
y1
1
Write the slope-intercept form of the equation that passes through (7,1) and has a slope of , then graph it.3
7m = 3
7
y - 1 = x - 337
+1 +1
y- intercept = (0,-2)
m 37+
+=
y= mx + b
Point-Slope Form
Slope Intercept
Form
42 6-2-4-6
2
4
6
-2
-4
-6
8 10-8-10
8
-8
10
x
y
+7
3+
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7
Standard 5
(y – ) = (x – )
y - 1 = 1(x – 2)
y - 1 = x - 2
y = x -1 (y – ) = m(x – )x1y1
point = (2, 1)x1
2
y1
1
Write the slope-intercept form of the equation that passes through points (2,1) and (-3,-4), then graph it.
m = 1 +1 +1
y- intercept = (0,-1)
m 11+
+=
y= mx + b
First we find the slope:
= 1
(2,1) (-3,-4)x2
2
x1
-3
y2
1
y1
-4=
5 5
m =--
= 1+4
2+3
Then using
1
y - 1 = x - 2
21 3-1-2-3
1
2
3
-1
-2
-3
4 5-4-5
4
-4
5
x
y
+11+
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8
Standard 5
42 6-2-4-6
2
4
6
-2
-4
-6
8 10-8-10
8
-8
10
x
y
43
Point = (6, 9)
Write the slope-intercept form of the equation that passes through (6,9) and has a slope of , then graph it.4
3m = 4
3
y- intercept = (0,1)
m 43+
+=
( ) = ( ) +b
y= mx + b
+3
4+
Slope Intercept
Form
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(x,y)
69
24 3
9 = +b
9 = 8 + b -8 -8
1 = b
b = 1 y= x 43
+ 1
9
Standard 5
42 6-2-4-6
2
4
6
-2
-4
-6
8 10-8-10
8
-8
10
x
y
52
Point = (4, 12)
Write the slope-intercept form of the equation that passes through (4,12) and has a slope of , then graph it.5
2m = 5
2
y- intercept = (0,2)
m 52+
+=
( ) = ( )+b
y= mx + b
+2
5+Slope Intercept
Form
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(x,y)
412
20 2
12 = + b
12 = 10 + b -10 -10
b = 2y= x
52
+ 2
10
Standard 5
72
Point = (2, 9)
Write the standard form of the equation that passes through (2,9) and has a slope of .7
2m = 7
2
( ) = ( ) +b
y= mx + b Slope Intercept
Form
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(x,y)
29
9 = 7 + b -7 -7
2 = b
b =2
y= x 72
+ 2
y= x 72
+ 2 (2)
2y = 7x + 4
-7x -7x
-7x + 2y = 4
or
7x – 2y = -4
Standard Form
11
Standard 5
Find the slope-intercept form of the equation that passes through (3,1) and it is parallel to the equation
m = - 92
( ) = ( ) + b
y= mx + b Slope Intercept
Form
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(x,y)
3 1
y = - x + 1. 92
Two lines are parallel when they have the same slope, therefore the equation must have slope: point = (3, 1)and
92
-
( ) = + b 1 27
2-
27 2
+ 27 2
+
b = 1 27 2
+
b = 27 2
+22
22
b = 29 2
y= x 92
- + 29 2
12
Standard 5
Find the slope-intercept form of the equation that passes through (1,2) and it is parallel to the equation
m = - 43
( ) = ( ) + b
y= mx + b Slope Intercept
Form
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(x,y)
1 2
y = - x + 3. 43
Two lines are parallel when they have the same slope, therefore the equation must have slope: point = (1, 2)and
43
-
( ) = + b 2 4
3-
43
+43
+
b = 2 43
+
b = 43
+63
33
b = 10 3
y= x 43
- + 10 3
13
Standard 5
Find the slope-intercept form of the equation that passes through (2,3) and it is perpendicular to the equation
( ) = ( ) + b
y= mx + b Slope Intercept
Form
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(x,y)
23
y = - x + 1. 23
Two lines that are perpendicular have slopes whose product is -1. One is the reciprocal of the other:
point = (2, 3)and
32
m =2
m1
-1
- 23
m =2
-132
- 32
-
32
m =2
- 23
m =1
and
3 = 3 + b -3 -3
0 = b
b = 0 y= x 32
+ 0
y= x
32
14
Standard 5
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SPECIAL FUNCTIONS
Constant Function:
f(x)= 3 Means that the function has the same value for all the domain.
.5-.5 1-1 1.5-1.5 2-2 2.5-2.5 3-3 3.5-3.5
1
-1
2
-2
3
-3
4
-4
5
-5
x
y
f(x)=3
15
Standard 5
42 6-2-4-6
2
4
6
-2
-4
-6
8 10-8-10
8
-8
10
x
y
Graph the following absolute value equation:
y = |x|
For x < 0
y = -x y = x
For x 0>
Now let’s shift it two units above:
y = |x| + 2
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SPECIAL FUNCTIONS
16
Standard 5
42 6-2-4-6
2
4
6
-2
-4
-6
8 10-8-10
8
-8
10
x
y
y = |x|
For x < 0
y = -x y = x
For x 0>
Now let’s shift it two units above:
y = |x| + 2
Now let’s shift it three units to the right:
y = |x-3| + 2
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Graph the following absolute value equation:SPECIAL FUNCTIONS
17
Standard 5
42 6-2-4-6
2
4
6
-2
-4
-6
8 10-8-10
8
-8
10
x
y
y = |x|
For x < 0
y = -x y = x
For x 0>
Now let’s shift it two units above:
y = |x| + 2
Now let’s shift it three units to the right:
y = |x-3| + 2
Now let’s graph it upside down
y = – |x-3| + 2
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Graph the following absolute value equation:SPECIAL FUNCTIONS
18
Standard 5
42 6-2-4-6
2
4
6
-2
-4
-6
8 10-8-10
8
-8
10
x
y
y = |x|
For x < 0
y = -x y = x
For x 0>
Now let’s shift it two units above:
y = |x| + 2
Now let’s shift it three units to the right:
y = |x-3| + 2
Now let’s graph it upside down
y = – |x-3| + 2
Now let’s make it skinner
y = – 6|x-3| + 2PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
Graph the following absolute value equation:SPECIAL FUNCTIONS
19
Standard 5
42 6-2-4-6
2
4
6
-2
-4
-6
8 10-8-10
8
-8
10
x
y
y = |x|
For x < 0
y = -x y = x
For x 0>
Now let’s shift it two units above:
y = |x| + 2
Now let’s shift it three units to the right:
y = |x-3| + 2
Now let’s graph it upside down
y = – |x-3| + 2
Now let’s make it skinner
y = – 2|x-3| + 2
So, we have learn how the different parameters in an absolute value equation affect our graph.
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Graph the following absolute value equation:SPECIAL FUNCTIONS
20
42 6-2-4-6
2
4
6
-2
-4
-6
8 10-8-10
8
-8
10
x
y
Standard 5
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SPECIAL FUNCTIONS
f(x)=x Identity function
(1,1)
(6,6)
(-6,-6)
At all points in the line the x-coordinate and the y-coordinate are the same.
21
Standard 5
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SPECIAL FUNCTIONS
GREATEST INTEGER FUNCTION:
f(x)=[x] Step function were [x] means the greatest integer not greater than x.
.5-.5 1-1 1.5-1.5 2-2 2.5-2.5 3-3 3.5-3.5
1
-1
2
-2
3
-3
4
-4
5
-5
x
y
2
1
1
1
1
1
1
1
1
1
1
Means:
y
2
1.9
1.8
1.7
1.6
1.5
1.4
1.3
1.2
1.1
1
x
22
Standard 5
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SPECIAL FUNCTIONS
GREATEST INTEGER FUNCTION:
f(x)=[x] Step function were [x] means the greatest integer not greater than x.
.5-.5 1-1 1.5-1.5 2-2 2.5-2.5 3-3 3.5-3.5
1
-1
2
-2
3
-3
4
-4
5
-5
x
y
3
2
2
2
2
2
2
2
2
2
2
Means:
y
3
2.9
2.8
2.7
2.6
2.5
2.4
2.3
2.2
2.1
2
x
23
Standard 2
Graph the following system of equations, find the solution if it exist, and state if it is consistent and independent, consistent and dependent or inconsistent:
6x –2y = 02x – 2y = -4
6x – 2y = 0
-6x -6x
-2y = -6x-2 -2
y = 3x
2x – 2y = -4
-2x -2x
-2y = -2x -4-2 - 2
y = x + 2 2
4 2
y = x + 2
y- intercept = (0,2)
m 11+
+=
m 31+
+=
y- intercept = (0,0)
The system is consistent and independent, with a unique solution at (1,3)
21 3-1-2-3
1
2
3
-1
-2
-3
4 5-4-5
4
-4
5
x
y
+11+
3+
+1
Solution(1,3)
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24
Standard 2Solve the following system of equations by elimination and verify by graphing:
2x + y = 45x + y = 7
2x + y = 45x + y = 7
Multiplying one of the equations by -1 to eliminate y:
-1 -1
2x + y = 4-5x - y = -7
-3x = -3-3 -3
x=1
2x + y = 4
2( ) + y = 41
2 + y = 4
-2 -2
y = 2
Changing both equations to Slope Intercept Form:
2x + y = 4 5x + y = 7
-2x -2x
y = -2x + 4
-5x -5x
y = -5x + 7
+1 +1
42 6-2-4-6
2
4
6
-2
-4
-6
8 10-8-10
8
-8
10
x
y
+1
2-3-
+1Solution
(1,2)
The system is consistent and independent, with a unique solution at (1,2)
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25
Standard 2Solve the following system of equations by elimination and verify by graphing:
4x + 2y = 105x + y = 17
4x + 2y = 105x + y = 17
Multiplying one of the equations by -2 to eliminate y:
-2 -2
4x + 2y = 10-10x - 2y = -34
- 6x = -24-6 -6
x=4
4x + 2y = 10
4( ) + 2y = 104
16 + 2y = 10
-16 -16
2y =-6
Changing both equations to Slope Intercept Form:
4x + 2y = 10 5x + y = 17
-4x -4x
2y = -4x + 10
-5x -5x
y = -5x + 17
+1
+1The system is consistent and independent, with a unique solution at (4,-3)
2 2
y= -3
y
84 12-4-8-12
4
8
12
-4
-8
-12
16 20-16-20
16
-16
20
xSolution(4,-3)
2 2y = -2x + 5
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26
X – 5 = 2Y + 12+5 +5
X= 2Y + 17 (2)(2)
4Y – 30 = X
2Y +17 = 4Y - 30
- 2Y -2Y
17 = 2Y - 30
+30 +30
47 = 2Y 2 2
Y = 23.5X= 2Y + 17= 2( ) + 1723.5= 47 + 17
X = 64
Standard 2
2Y –15 = X
Solve the following two equations by substitution:
The system is consistent and independent, with a unique solution at (64,23.5)
12
2Y –15 = X12
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27
y + 2 = 4x
2y = 7x
-2 -2
y = 4x -2
2( ) = 7x4x-2
8x – 4 = 7x
-8x -8x
- 4 = -x (-1)(-1)
x = 4
y = 4x -2
= 4( ) -24
= 16 -2
y = 14
Standard 2
y + 2 = 4x
2y = 7x
Solve the following equations by substitution:
The system is consistent and independent, with a unique solution at (4,14)
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