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.

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

END SHOWPRESENTATION CREATED BY SIMON PEREZ. All rights reserved

PROBLEM 5 PROBLEM 6

PROBLEM 7 PROBLEM 8

PROBLEM 9 PROBLEM 10

PROBLEM 11

SPECIAL FUNCTIONS

http://www.mrperezonlinemathtutor.com/

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:

PRESENTATION CREATED BY SIMON PEREZ. All rights reserved


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



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



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



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


7m = 3

7

y - 1 = x - 337

+1 +1


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+



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


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+



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)


3m = 4

3


m 43+

+=

( ) = ( ) +b

y= mx + b

+3

4+

Slope Intercept

Form


(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)


2m = 5

2


m 52+

+=

( ) = ( )+b

y= mx + b

+2

5+Slope Intercept

Form


(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


(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


Form


(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


Form


(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


Form


(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


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


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>


y = |x| + 2

Now let’s shift it three units to the right:

y = |x-3| + 2


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>


y = |x| + 2


y = |x-3| + 2

Now let’s graph it upside down

y = – |x-3| + 2




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>


y = |x| + 2


y = |x-3| + 2


y = – |x-3| + 2

Now let’s make it skinner

y = – 6|x-3| + 2PRESENTATION CREATED BY SIMON PEREZ. All rights reserved



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>


y = |x| + 2


y = |x-3| + 2


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.




20

42 6-2-4-6

2

4

6

-2

-4

-6

8 10-8-10

8

-8

10

x

y

Standard 5


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


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


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


m 11+

+=

m 31+

+=


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)



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)




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



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



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:




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.

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