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Welcome to MM 212 Welcome to MM 212 Unit 4 Seminar!Unit 4 Seminar!
Graphing and Functions
The Rectangular Coordinate System
4
Rectangular Coordinate System
A rectangular coordinate system consists of a horizontal number line and a vertical number line.
x
y
1234 1 2 3 4
2
34
1
2
3
4
x-axis
y-axisAn ordered pair (x, y) represents a point on the coordinate system.
(3, 4)
The ordered pair (3, 4) means that x = 3 and y = 4.
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Plotting Points
Example:
Plot the points (4, 2), (3, 3), and (1, 0) on a rectangular coordinate system.
x
y
1234 1 2 3 4
2
34
1
2
3
4
(1, 0)
(4, 2)
(3, 3)
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Determining Coordinates
Example:
Write the coordinates of each point plotted in the graph.
(4, 3)
x
y
1234 1 2 3 4
2
34
1
2
3
4A
B
C4 units to the left on the x-axis
3 units up a line parallel to the y-axis
A = (4, 3)
B = (2, 2)
C = (0, 3)
(2, 2)
(0, 3)
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Ordered Pairs for a Linear Equation
A linear equation in two variables is an equation that can be written in the form Ax + By = C where A, B, and C are real numbers but A and B are not both zero.
A linear equation in two variables is an equation that can be written in the form Ax + By = C where A, B, and C are real numbers but A and B are not both zero.
A solution to a linear equation is an ordered pair that makes the equation a true mathematical statement.
2x + y = 5(2, 1) is a solution to the equation 2x + y = 5. 2(2) + (1) = 5
4 + 1 = 5
5 = 5
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Ordered Pairs for a Linear Equation
Example:Find the missing coordinate to complete the following ordered-pair solution for the equation y = 6x + 5.
a. (2, ?) b. (?, 7)
y = 6x + 5
y = 6(2) + 5
y = 12 + 5
y = 17
y = 6x + 5
7 = 6x + 5
12 = 6x
2 = x
(2, 17) is a solution. (2, 7) is a solution.
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Solving a Formula for a Specified Value
Procedure to Solve a Formula for a Specified Value1. Remove any parentheses.2. If fractions exist, multiply all terms on both sides by the LCD of
the fractions.3. Combine like terms, if possible.4. Add or subtract terms on both sides of the equation to get all terms
with the variable on one side of the equation.5. Add or subtract a constant value on both sides of the equation to
get all terms not containing the variable on the other side of the equation.
6. Divide both sides of the equation by the coefficient of the variable.7. Simplify the solution, if possible.
Procedure to Solve a Formula for a Specified Value1. Remove any parentheses.2. If fractions exist, multiply all terms on both sides by the LCD of
the fractions.3. Combine like terms, if possible.4. Add or subtract terms on both sides of the equation to get all terms
with the variable on one side of the equation.5. Add or subtract a constant value on both sides of the equation to
get all terms not containing the variable on the other side of the equation.
6. Divide both sides of the equation by the coefficient of the variable.7. Simplify the solution, if possible.
10
Solving a Formula for a Specified Value
.2
bhA The formula for the area of a triangle is
Example:If the area of a triangle is 66 inches, and the base is 8 inches, find the height of the triangle.
b = 8 in
h = ?
A = 66 in2 66 (28)h
Multiply both sides by 2.132 8h
Divide both sides by 8.16.5 h
The height of the triangle is 16.5 inches.
Substitute in the values.
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Solving a Formula for a Specified Value
( ) .2
h a bA The formula for the area of a trapezoid is
Example:Solve the formula for b.
( )2
h a bA
Multiply both sides by 2.2 ( )A h a b
Divide both sides by h.2A a bh
Subtract a from both sides.2A a bh
The formula is now solved for b.
Graphing Linear Equations
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Graphing by Plotting Points
The graph of any linear equation in two variables is a straight line. The graph of any linear equation in two variables is a straight line.
To Graph a Linear Equation
1. Look for three ordered pairs that are solutions to the equation.
2. Plot the points.
3. Draw a line through the points.
To Graph a Linear Equation
1. Look for three ordered pairs that are solutions to the equation.
2. Plot the points.
3. Draw a line through the points.
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Graphing by Plotting Points
x y
1 2
2 1
1 4
Example:Graph the equation y = –x + 3 by plotting points.
(1, 2) is one of the solutions.
Choose any value for x and find the corresponding y-value.
For x = 1, y = –(1) + 3 = 2.
Find three ordered pairs (solutions) to graph the line.
x
y
1234 1 2 3 4
2
34
1
2
3
4(1, 4)
(2, 1)
(1, 2)
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Graphing by Plotting Intercepts
The x-intercept of a line is the point where the line crosses the x-axis; it has the form (a, 0). The y-intercept of a line is the point where the line crosses the y-axis; it has the form (0, b).
The x-intercept of a line is the point where the line crosses the x-axis; it has the form (a, 0). The y-intercept of a line is the point where the line crosses the y-axis; it has the form (0, b).
Intercept Method of Graphing
1. Find the x-intercept by letting y = 0 and solving for x.
2. Find the y-intercept by letting x = 0 and solving for y.
3. Find one additional ordered pair so that we have three points with which to plot the line.
Intercept Method of Graphing
1. Find the x-intercept by letting y = 0 and solving for x.
2. Find the y-intercept by letting x = 0 and solving for y.
3. Find one additional ordered pair so that we have three points with which to plot the line.
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Graphing by Plotting Intercepts
Let y = 0.
x = 3
Example:Graph the equation –3y – 2x = – 6 by using the x- and y-intercepts.
Let x = 0. –3y – 2(0) = – 6
y = 2
–3(0) – 2x = – 6
(3, 0)
x
y
1234 1 2 3 4
2
34
1
2
3
4
(0, 2)
y-intercept
x-intercept
(3, 4)
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Graphing Horizontal Lines
y = 2
(0, 2)
x
y
1234 1 2 3 4
2
34
1
2
3
4
The graph of the equation y = b, where b is any real number, is a horizontal line through the point (0, b).The graph of the equation y = b, where b is any real number, is a horizontal line through the point (0, b).
The equation of the horizontal line.
y = 3
(0, 3)
x
y
1234 1 2 3 4
2
34
1
2
3
4
18
Graphing Vertical Lines
The graph of the equation x = a, where a is any real number, is a vertical line through the point (a, 0).The graph of the equation x = a, where a is any real number, is a vertical line through the point (a, 0).
x = 4
(4, 0)
x
y
1234 1 2 3 4
2
34
1
2
3
4
The equation of the vertical line.
x = 2
(2, 0)
x
y
1234 1 2 3 4
2
34
1
2
3
4
The Slope of a Line
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Slope of a Line
In a coordinate plane, the slope of a straight line is defined by the change in y divided by the change in x.
change in riseslope = change in run
yx
(3, 0)
x
y
1234 1 2 3 4
2
34
1
2
3
4
(0, 2)Change in y = 2
Change in x = 3
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Definition of Slope of a Line
The slope of any nonvertical line that contains the points with coordinates (x1, y1) and (x2, y2) is defined by the difference ratio
Definition of Slope of a Line
The slope of any nonvertical line that contains the points with coordinates (x1, y1) and (x2, y2) is defined by the difference ratio
Slope of a Line
2 1
2 1slope = =
y ym
x x where x2 x1.
Example:Find the slope of the line that passes through (3, 6) and (1, 2).
2 1
2 1
y ym
x x
6 23 1
4 2 22 1
22
Slope of a Straight LineSlope of a Straight Line
Slope of a Line
Positive SlopeLine goes up to the right
x
y1. Lines with
positive slopes go upward as x increases.
Negative SlopeLine goes downward to the right
x
y 2. Lines with negative slopes go downward as x increases.
Continued.
23
Slope of a Straight Line (continued)Slope of a Straight Line (continued)
Slope of a Line
Zero SlopeHorizontal Line
x
y 3. Horizontal lines have a slope of 0.
Undefined SlopeVertical Line
x
y 4. A vertical line has an undefined slope.
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Slope-Intercept Form of a Line
The slope-intercept form of the equation of a line that has slope m and y-intercept (0, b) is given by
y = mx + b.
Slope-Intercept Form of a Line
The slope-intercept form of the equation of a line that has slope m and y-intercept (0, b) is given by
y = mx + b.
Slope-Intercept Form of a Line
y = 3x – 4
The slope is 3.
The y-intercept is (0, -4).
3 18 4
y x
The slope
is3.8
The y-intercept is
10, .4
25
Slope-Intercept Form of a Line
Example:Find the equation of the line with slope and y-intercept (0, 5).
5)23
(y x
2 53
y x
23
m slope: 5b y-intercept:
y mx b
23
26
Graphing with Slope and y-Intercept
Example:Graph the line with slope and y-intercept (0, 2).
32
m
32
m
-intercept (0, 2)y
x
y
1234 1 2 3 4
2
34
1
2
3
4
rise
run
This is the graph of the line with
slope and y-intercept (0, 2). 32
riserun
3 units down
2 units to the right
(2, 1)
(0, 2)
27
Parallel Lines
Parallel Lines
Parallel lines are two straight lines that never touch.Parallel lines have the same slope but different y-intercepts.
Parallel Lines
Parallel lines are two straight lines that never touch.Parallel lines have the same slope but different y-intercepts.
m1 = m2
x
ySlope m1
Slope m2
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Parallel Lines
Example:
Line c has a slope of
If line d is parallel to line c, what is its slope?
1.2
Parallel lines have the same slope.
x
y
line d
line cLine d has a slope of 1.
2
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Perpendicular Lines
Perpendicular Lines
Perpendicular lines are two lines that meet in a 90° angle.Perpendicular lines have a slope whose product is 1. If m1 and m2 are slopes of perpendicular lines, then
Perpendicular Lines
Perpendicular lines are two lines that meet in a 90° angle.Perpendicular lines have a slope whose product is 1. If m1 and m2 are slopes of perpendicular lines, then
m1m2 = 1
x
y
Slope m1
Slope m2
or
m1 = 2
1m
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Perpendicular Lines
Example:
Line c has a slope of
If line e is perpendicular to line c, what is its slope?
1.2
Perpendicular lines have slopes whose product is 1.
1 2 1m m
x
y
line e
line c2
12
1m
2 2m
2.Line e has a slope of
Writing the Equation of a Line
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To Find the Equation of a Line Given a Point and the Slope
1. Substitute the given values of x, y, and m into the equation y = mx + b.
2. Solve for b.
3. Use the values of b and m to write the equation in the form y = mx + b.
To Find the Equation of a Line Given a Point and the Slope
1. Substitute the given values of x, y, and m into the equation y = mx + b.
2. Solve for b.
3. Use the values of b and m to write the equation in the form y = mx + b.
Finding the Equation Given a Point and the Slope
33
Example:
Find an equation of the line that passes through (4, 3) with
a slope of 5.
m = 5, x = 4, y = 3
The equation of the line is y = 5x 23.
y = mx + b
3 = (5)(4) + b
23 = b
Substitute known values.
Finding the Equation Given a Point and the Slope
34
Finding the Equation Given Two Points
2 1
2 1
y ym
x x
Example:
Find an equation of the line that passes through (2, 1) and
(7, 4).
Find the slope of the line.
42(
17 )
39
13
y = mx + b
)1 23
(1 b
213
b
53
b
The equation of the line is 1 5.3 3
y x
35
Finding the Equation Given a Graph of the Line
change in change in
ymx
Example:
Find the equation of the line for the following graph.
Find the y-intercept.
The equation of the line is2 2.3
y x
(3, 0)
x
y
1234 1 2 3 4
2
34
1
2
3
4
(0, 2)Change in y = 2
Change in x = 3
2b
Find the slope 2
3
Systems of Linear Equations in Two
Variables
37
Systems of EquationsA system of equations or system of inequalities is two or more equations or inequalities in several variables that are considered simultaneously.
x
y
123 1 2 3
2
3
1
2
3
The lines may intersect.
x
y
123 1 2 3
2
3
1
2
3
The lines may be parallel.
x
y
123 1 2 3
2
3
1
2
3
The lines may coincide.
38
Systems of Equations
A solution to a system of two linear equations in two variables is an ordered pair.
Example:
Determine if ( 4, 16) is a solution to the system of equations.
y = 4x
y = 4x 16 = 4( 4)
16 = 16
y = 2x + 8 16 = 2( 4) + 8
16 = 8 + 816 = 16
Yes, it is a solution.
y = 2x + 8
39
Graphing to Solve a System
x y
2 0
0 6
1 9
Example:Solve by graphing.
y = 3x 6y = 2x 1
x y
2 5
0 1
2 3
x
y
2468 2 4 6 8
4
68
2
4
6
8
Continued.
40
Graphing to Solve a SystemExample continued:
x
y
2468 2 4 6 8
4
68
2
4
6
8
The lines intersect at
(1, 3).
Check:y = 3x 6
y = 2x 1
3 = 3(1) 63 = 3 63 = 3
3 = 2(1) 13 = 2 13 = 3
y = 3x 6
y = 2x 1
A system of equations that has one solution is said to be consistent.
41
Inconsistent Systems
Example:Solve by graphing.
3x 2y = 4 9x + 6y = 1
x
y
2468 2 4 6 8
4
68
2
4
6
8
3x 2y = 4
9x + 6y = 1
A system of linear equations that has no solution is called an inconsistent system.
The lines are parallel.
42
Dependent Systems
Example:Solve by graphing.
4x 6y = 8
2x + 3y = 4
x
y
2468 2 4 6 8
4
68
2
4
6
8
4x 6y = 8
2x + 3y = 4
A system of linear equations that has an infinite number of solutions is called a dependent system.
The lines coincide.
43
The Substitution Method
How to Solve a System of Two Linear Equations by the Substitution Method How to Solve a System of Two Linear Equations by the Substitution Method
1. Choose one of the two equations and solve for one variable in terms of the other variable.
2. Substitute the expression from step1 into the other equation.
3. You now have one equation with one variable. Solve this equation for the variable.
4. Substitute this value for the variable into one of the original equations to obtain a value for the second variable.
5. Check the solution in both original equations.
44
The Substitution MethodExample:Find the solution. 2x – y = 13
– 4x – 9y = 7
y = 2x – 13 Solve the first equation for y.
– 4x – 9y = 7 This is the original second equation.
– 4x – 9(2x – 13) = 7
– 4x – 18x + 117 = 7
– 22x = – 110
x = 5
This variable is the easiest to isolate.
Substitute the expression into the other equation and solve.
Continued.
Substitute this value into one of the original equations.
45
The Substitution Method
2x – y = 13– 4x – 9y = 7
First equation
Substitute x = 5 into the first equation.
Second equation
2x – y = 13
2(5) – y = 13
Solve for y.10 – y = 13
– y = 3
y = – 3 The solution is (5, – 3).
Continued.
Example continued:
46
The Substitution Method
Check the solution (5, – 3) in both original equations.
2x – y = 13 – 4x – 9y = 7
2(5) – (3) = 13
10 + 3 = 13
13 = 13
– 4(5) – 9(3) = 7
– 20 + 27 = 7
7 = 7
Example continued:
47
The Addition Method
How to Solve a System of Two Linear Equations by the Addition (Elimination) Method How to Solve a System of Two Linear Equations by the Addition (Elimination) Method 1. Arrange each equation in the form ax + by = c. (Remember
that a, b, and c can be any real number.)2. Multiply one or both equations by the appropriate numbers so
that the coefficients of one of the variables are opposites.3. Add the two equations from step 2 so that one variable is
eliminated.4. Solve the resulting equation for the remaining variable.5. Substitute this value into one of the original equations to find
the value of the other variable.6. Check the solution in both of the original equations.
48
The Addition MethodExample:Solve by addition. 5x – 3y = 14
2x – y = 6
Multiply each term of equation (2) by 3.
x = 4
(3)2x – (3)y = (3)6
Add the two equations.
This equation is equivalent to equation (2). 6x + 3y = 18
5x – 3y = 14 6x + 3y = 18
x = 4 Substitute this value into either equation to find y.
Continued.
49
The Addition Method
5x – 3y = 14 2x – y = 6
8 y = 6
Substitute. 2(4) – y = 6
y = 2
The solution is (4, 2).
Be sure to check the solution in both original equations.
Example continued:
2x – y = 6
y = 2
50
Identifying Inconsistent Systems
Example:Solve the system algebraically.
3x + 6y = 12 x + 2y = 7
x = 2y + 7 Solve equation (2) for x.
3(2y + 7) + 6y = 12 Substitute into equation (1).
6y + 21 + 6y = 12 Simplify.
21 = 12 This results in a false statement.
There is no solution to this system of equations. If graphed, these lines would be parallel.
51
Identifying Dependent SystemsExample:Solve the system algebraically.
(6)6x – (6)4y = (6)8 Multiply each term in (1) by 6.
There are an infinite number of solutions to this system of equations. If graphed, these lines would be the same.
6x – 4y = 8– 9x + 6y = –12
4(–9x) + 4(6y) = 4(–12) Multiply each term in (2) by 4.
36x – 24y = 48 This equation is equivalent to (1). –36x + 24y = –48 This equation is equivalent to (2).
0 = 0 Add the equations.
52
Choosing an Appropriate Method
Often becomes difficult to use if no variable has a coefficient of 1 or 1.
Works well if one or more variable has a coefficient of 1 or 1.
Substitution
Works well if equations have fractional or decimal coefficients, or if no variable has a coefficient of 1 or 1.
NoneAddition
AdvantageAdvantage DisadvantageDisadvantageMethodMethod
53
Choosing an Appropriate Method
Example:Select an appropriate method for solving the system.
3x + 6y = 12 x + 2y = 7
b.)
6x – 4y = 8– 9x + 6y = –12
c.)
5x – 3y = 130.9x + 0.4y = 30
a.)
The substitution method should be used since x has a coefficient of 1.
The addition method should be used since none of the variables have a coefficient of 1 or 1.
The addition method should be used since none of the variables have a coefficient of 1 or 1.
54
Possible Solutions
You obtain an equation that is always true. These equations are dependent.
Infinite number of solutions
You obtain an equation that is inconsistent with known facts. The system is inconsistent.
No solution
You obtain one value for x and one value for y. For example,
x = 3, y = 5.
One unique solution
GraphGraph Algebraic InterpretationAlgebraic InterpretationNumber of SolutionsNumber of Solutions
Two lines intersect at one point.
Parallel lines
Lines coincide
(3, 5)
