1. SLOPE INTERCEPT FORM AND POINT-SLOPE FORM
Standards 2, 5, 25
SLOPE INTERCEPT FORM AND POINT-SLOPE FORM
PROBLEM 1
PROBLEM 2
PROBLEM 3
PROBLEM 4
PROBLEM 5
PROBLEM 6
PROBLEM 7
PROBLEM 8
PROBLEM 9
PROBLEM 10
PROBLEM 11
SPECIAL FUNCTIONS
SOLVING SYSTEMS OF TWO VARIABLES LINEAR EQUATIONS
BY GRAPHING, ELIMINATION AND SUBSTITUTION.
PROBLEM 12
PROBLEM 13
PROBLEM 14
PROBLEM 15
PROBLEM 16
END SHOW
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2. ALGEBRA II STANDARDS THIS LESSON AIMS:
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.
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3. Standard 5
Write the slope-intercept form for this equation and graph it.
4x + 2y = 10
-4x x 4 2 6 -2 -4 -6 8 10 -8
-10 x y
2y = -4x + 10 +1
y = x + 4 2 10 - 2
y = -2x + 5
Slope Intercept
Form
y= mx + b 2 1 + - =
m= -2
y- intercept = (0,5)
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4. Standard 5
Write the slope-intercept form for this equation and graph it.
-9x + 3y = 3
+9x x 4 2 6 -2 -4 -6 8 10 -8
-10 x y
3y = 9x + 3
y = x + 9 3 3 + +1
y = 3x + 1
Slope Intercept
Form
y= mx + b 3 1 + =
m= 3
y- intercept = (0, 1)
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5. Standard 5
Write the slope-intercept form of the equation that passes through (4,2) and has a slope of , then graph it. 5 4
m = 5 4
Point = (4, 2) y 1 2 x 1 4 4 2 6 -2 -4 -6 8 10 -8
-10 x y
(y – ) = m(x – ) x 1 y
Point-Slope Form 5 4
(y – ) = (x – )
y - 2 = (x – 4) 5 4 5 +
y - 2 = x - 20 4 5 +4
y - 2 = x - 5 5 4
y = x -3 5 4
Slope Intercept
Form
y= mx + b m 5 4 + =
y- intercept = (0,-3)
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6. Standard 5
Write the slope-intercept form of the equation that passes through (7,1) and has a slope of , then graph it. 3 7
m = 3 7
Point = (7, 1) y 1 x 1 7
(y – ) = m(x – ) x 1 y
Point-Slope Form 4 2 6 -2 -4 -6 8 10 -8
-10 x y 3 7
(y – ) = (x – )
y - 1 = (x – 7) 3 7
y - 1 = x - 21 7 3 3 + +7
y - 1 = x - 3 3 7
y = x -2 3 7
Slope Intercept
Form
y= mx + b m 3 7 + =
y- intercept = (0,-2)
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7. Write the slope-intercept form of the equation that passes through points (2,1) and (-3,-4), then graph it.
Standard 5 2 1 3 -1 -2 -3 4 5 -4 -5 x y
First we find the slope:
(2,1) (-3,-4) y 2 1 x 2 y 1 -4 x 1 -3
m = -
1+4 = 5 =
= 1 1 +
2+3 +1
Then using
y - 1 = x - 2
point = (2, 1)
m = 1 y 1 x 1 2
(y – ) = m(x – ) x 1 y
y = x -1
y= mx + b
(y – ) = (x – ) 1 m 1 + =
y - 1 = 1(x – 2)
y- intercept = (0,-1)
y - 1 = x - 2
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8. Standard 5
Write the slope-intercept form of the equation that passes through (6,9) and has a slope of , then graph it. 4 3
m = 4 3
Point = (6, 9) 4 2 6 -2 -4 -6 8 10 -8
-10 x y
(x,y)
y= mx + b
Slope Intercept
Form
( ) = ( ) +b 4 + 4 3 9 6 +3 24 3
9 = b
9 = 8 + b
1 = b 4 3
b = 1
y= x
+ 1 m 4 3 + =
y- intercept = (0,1)
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9. Standard 5
Write the slope-intercept form of the equation that passes through (4,12) and has a slope of , then graph it. 5 2
m = 5 2
Point = (4, 12) 4 2 6 -2 -4 -6 8 10 -8
-10 x y
(x,y)
y= mx + b
Slope Intercept
Form 5 +
( ) = ( )+b 5 2 12 4 +2 20 2
12 = b
12 = 10 + b
b = 2 5 2
y= x
+ 2 m 5 2 + =
y- intercept = (0,2)
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10. Standard 5
Write the standard form of the equation that passes through (2,9) and has a slope of 7 2
m = 7 2
Point = (2, 9)
(x,y)
y= mx + b
Slope Intercept
Form
( ) = ( ) +b 7 2 9 2
9 = 7 + b
2y = 7x + 4
2 = b
-7x -7x
b =2
Standard Form
-7x + 2y = 4 7 2
y= x
+ 2 or
y= x 7 2
+ 2
(2)
7x – 2y = -4
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11. Standard 5
Find the slope-intercept form of the equation that passes through (3,1) and it is parallel to the equation
y = x + 1. 9 2
Two lines are parallel when they have the same slope, therefore the equation must have slope:
m = - 9 2
point = (3, 1)
and
(x,y)
y= mx + b
Slope Intercept
Form
y= x 9 2 - + 29
( ) = ( ) + b 9 2 - 1 3
( ) = b 1 27 2 - 27 2 +
b = 1 27 2 + 2
b = 27 2 +
b = 29 2
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12. Standard 5
Find the slope-intercept form of the equation that passes through (1,2) and it is parallel to the equation
y = x + 3. 4 3
Two lines are parallel when they have the same slope, therefore the equation must have slope:
m = - 4 3
point = (1, 2)
and
(x,y)
y= mx + b
Slope Intercept
Form
y= x 4 3 - + 10
( ) = ( ) + b 4 3 - 2 1
( ) = b 2 4 3 - 4 3 +
b = 2 4 3 + 3
b = 4 3 + 6 3
b = 10 3
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13. Standard 5
Find the slope-intercept form of the equation that passes through (2,3) and it is perpendicular to the equation
y = x + 1. 2 3
Two lines that are perpendicular have slopes whose product is -1. One is the reciprocal of the other: - 2 3
m = 1
and
m = 2 m 1 -1
y= mx + b
Slope Intercept
Form 3 2 - - 2 3
m = -1
( ) = ( ) + b 3 2 3 2 3 2
m =
point = (2, 3)
and
3 = 3 + b
(x,y)
0 = b 3 2
b = 0
y= x
+ 0
y= x 3 2
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14. Means that the function has the same value for all the domain. f(x)= 3
SPECIAL FUNCTIONS
Standard 5
Constant Function:
Means that the function has the same value for all the domain.
f(x)= 3 .5
-.5 1 -1
1.5
-1.5 2 -2
2.5
-2.5 3 -3
3.5
-3.5 4 -4 5 -5 x y
f(x)=3
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15. Graph the following absolute value equation:
SPECIAL FUNCTIONS
Standard 5
Graph the following absolute value equation:
y = |x| 4 2 6 -2 -4 -6 8 10 -8
-10 x y
For x < 0
For x 0
>
y = -x
y = x
Now let’s shift it two units above:
y = |x| + 2
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16. Graph the following absolute value equation:
SPECIAL FUNCTIONS
Standard 5
Graph the following absolute value equation:
y = |x| 4 2 6 -2 -4 -6 8 10 -8
-10 x y
For x < 0
For x 0
>
y = -x
y = x
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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17. Graph the following absolute value equation:
SPECIAL FUNCTIONS
Standard 5
Graph the following absolute value equation:
y = |x| 4 2 6 -2 -4 -6 8 10 -8
-10 x y
For x < 0
For x 0
>
y = -x
y = x
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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18. Graph the following absolute value equation:
SPECIAL FUNCTIONS
Standard 5
Graph the following absolute value equation:
y = |x| 4 2 6 -2 -4 -6 8 10 -8
-10 x y
For x < 0
For x 0
>
y = -x
y = x
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| + 2
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19. Graph the following absolute value equation:
SPECIAL FUNCTIONS
Standard 5
Graph the following absolute value equation:
y = |x| 4 2 6 -2 -4 -6 8 10 -8
-10 x y
For x < 0
For x 0
>
y = -x
y = x
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
So, we have learn how the different parameters in an absolute value equation affect our graph.
y = – |x-3| + 2
Now let’s make it skinner
y = – 2|x-3| + 2
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20. SPECIAL FUNCTIONS Standard 5 Identity function f(x)=x y (6,6)4 2 6 -2 -4 -6 8 10 -8
-10 x y
(6,6)
At all points in the line the x-coordinate and the y-coordinate are the same.
(1,1)
(-6,-6)
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21. y x SPECIAL FUNCTIONS Standard 5 GREATEST INTEGER FUNCTION:
Step function were [x] means the greatest integer not greater than x.
f(x)=[x]
Means: y 2
1.9
1.8
1.7
1.6
1.5
1.4
1.3
1.2
1.1 1 x .5
-.5 1 -1
1.5
-1.5 2 -2
2.5
-2.5 3 -3
3.5
-3.5 4 -4 5 -5 x y 1 1 1 1 1 1 1 1 1 1 2
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22. y x SPECIAL FUNCTIONS Standard 5 GREATEST INTEGER FUNCTION:
Step function were [x] means the greatest integer not greater than x.
f(x)=[x]
Means: y 3
2.9
2.8
2.7
2.6
2.5
2.4
2.3
2.2
2.1 2 x .5
-.5 1 -1
1.5
-1.5 2 -2
2.5
-2.5 3 -3
3.5
-3.5 4 -4 5 -5 x y 2 2 2 2 2 2 2 2 2 2 3
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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: 2 1 3 -1 -2 -3 4 5 -4 -5 x y
2x – 2y = -4
6x –2y = 0 +1
Solution
(1,3) 1 +
2x – 2y = -4 3 +
6x – 2y = 0
-2x x
-6x x
-2y = -2x -4 +1
-2y = -6x
y = x + 2 4
y = 3x
y = x + 2 m 3 1 + = m 1 + =
y- intercept = (0,0)
y- intercept = (0,2)
The system is consistent and independent, with a unique solution at (1,3)
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24. Solve the following system of equations by elimination and verify by graphing:
Standard 2
2x + y = 4
2x + y = 4
5x + y = 7
Multiplying one of the equations by -1 to eliminate y:
2( ) + y = 4 1
2 + y = 4
2x + y = 4 4 2 6 -2 -4 -6 8 10 -8
-10 x y -1
5x + y = 7
y = 2
2x + y = 4
-5x - y = -7 +1 3 -
-3x = -3 2 -
Solution
(1,2) +1
x=1
Changing both equations to Slope Intercept Form:
2x + y = 4
5x + y = 7
-2x x
-5x x
The system is consistent and independent, with a unique solution at (1,2)
y = -2x + 4
y = -5x + 7 +1 +1
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25. Multiplying one of the equations by -2 to eliminate y: 4( ) + 2y = 10
Solve the following system of equations by elimination and verify by graphing:
Standard 2
4x + 2y = 10
4x + 2y = 10
5x + y = 17
Multiplying one of the equations by -2 to eliminate y:
4( ) + 2y = 10 4
16 + 2y = 10
4x + 2y = 10 -2
5x + y = 17 y 8 4 12 -4 -8
-12 16 20
-16
-20 x
2y =-6
4x + 2y = 10
-10x - 2y = -34
y= -3
- 6x = -24
x=4
Solution
(4,-3)
Changing both equations to Slope Intercept Form:
4x + 2y = 10
5x + y = 17
-4x x
-5x x
2y = -4x + 10
y = -5x + 17
The system is consistent and independent, with a unique solution at (4,-3) +1
y = -2x + 5 +1
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26. Solve the following two equations by substitution: Standard 21 2
X – 5 = 2Y + 12
2Y –15 = X
2Y –15 = X 1 2
X= 2Y + 17
(2)
4Y – 30 = X
2Y +17 = 4Y - 30
- 2Y Y
17 = 2Y - 30
47 = 2Y
X= 2Y + 17
Y = 23.5
= 2( ) + 17
23.5
The system is consistent and independent, with a unique solution at (64,23.5) =
X = 64
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27. Solve the following equations by substitution: Standard 2
y + 2 = 4x
2y = 7x
y + 2 = 4x
y = 4x -2
y = 4x -2
= 4( ) -2 4
2y = 7x
= 16 -2
2( ) = 7x
4x-2
y = 14
8x – 4 = 7x
-8x x
(-1)
- 4 = -x
x = 4
The system is consistent and independent, with a unique solution at (4,14)
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