Transcript Document

Welcome to MM 212
Unit 4 Seminar!
Graphing and Functions
The Rectangular
Coordinate System
Rectangular Coordinate System
A rectangular coordinate system consists of a
horizontal number line and a vertical number line.
y
y-axis
4
3
x-axis
2
An ordered pair (x, y)
represents a point on
the coordinate system.
1
x
4
3 2 1
2
3
1
2
3
4
(3, 4)
The ordered pair (3, 4)
means that x = 3 and y = 4.
4
4
Plotting Points
Example:
Plot the points (4, 2), (3, 3), and (1, 0) on a rectangular
coordinate system.
y
4
(4, 2)
3
(1, 0)
2
1
x
4
3 2 1
1
2
3
4
2
3
4
(3, 3)
5
Determining Coordinates
Example:
Write the coordinates of each point plotted in the graph.
y
(4, 3)
A
4
3
3 units up a
line parallel to
the y-axis
B (2, 2)
2
1
x
4
3 2 1
4 units to
the left on
the x-axis
2
3
4
1
2
3
C (0, 3)
4
A = (4, 3)
B = (2, 2)
C = (0, 3)
6
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 solution to a linear equation is an ordered pair that
makes the equation a true mathematical statement.
(2, 1) is a solution to
the equation 2x + y = 5.
2x + y = 5
2(2) + (1) = 5
4+1=5
5=5
7
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, ?)
y = 6x + 5
y = 6(2) + 5
y = 12 + 5
y = 17
(2, 17) is a solution.
b. (?, 7)
y = 6x + 5
 7 = 6x + 5
 12 = 6x
2=x
(2, 7) is a solution.
8
Solving a Formula for a Specified Value
Procedure to Solve a Formula for a Specified Value
1.
2.
3.
4.
5.
6.
7.
Remove any parentheses.
If fractions exist, multiply all terms on both sides by the LCD of
the fractions.
Combine like terms, if possible.
Add or subtract terms on both sides of the equation to get all terms
with the variable on one side of the equation.
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.
Divide both sides of the equation by the coefficient of the variable.
Simplify the solution, if possible.
9
Solving a Formula for a Specified Value
The formula for the area of a triangle is A  bh .
2
Example:
If the area of a triangle is 66 inches, and the base is 8
inches, find the height of the triangle.
A = 66
in2
h=?
(8)h
66 
2
132  8h
16.5  h
b = 8 in
Substitute in the values.
Multiply both sides by 2.
Divide both sides by 8.
The height of the triangle is 16.5 inches.
10
Solving a Formula for a Specified Value
The formula for the area of a trapezoid is
A
h(a  b)
.
2
Example:
Solve the formula for b.
h(a  b)
A
2
2 A  h(a  b)
2A
ab
h
2A
a b
h
Multiply both sides by 2.
Divide both sides by h.
Subtract a from both sides.
The formula is now solved for b.
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Graphing Linear
Equations
Graphing by Plotting Points
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.
13
Graphing by Plotting Points
Example:
Graph the equation y = –x + 3 by plotting points.
y
Choose any value for x and find
the corresponding y-value.
(1, 4)
4
3
2
For x = 1, y = –(1) + 3 = 2.
(1, 2) is one of the solutions.
x
y
1
2
2
1
1
4
(2, 1)
1
x
4
Find three ordered
pairs (solutions) to
graph the line.
(1, 2)
3 2 1
1
2
3
4
2
3
4
14
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).
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.
15
Graphing by Plotting Intercepts
Example:
Graph the equation –3y – 2x = – 6 by using the x- and
y-intercepts.
y
Let x = 0.
–3y – 2(0) = – 6
y=2
4
(3, 4)
y-intercept
3
2
(0, 2)
1
Let y = 0.
–3(0) – 2x = – 6
x=3
(3, 0)
x
4
3 2 1
2
3
1
2
3
4
x-intercept
4
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Graphing Horizontal Lines
The graph of the equation y = b, where b is any real
number, is a horizontal line through the point (0, b).
y
y
4
3
2
4
3
2
(0, 2)
1
4 3 2 1
1
x
1
2
3
4
4 3 2 1
2
3
4
y=2
2
The equation of the
horizontal line.
3
4
x
1
2
3
4
(0, 3)
y = 3
17
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).
y
(4, 0)
y
4
4
3
2
3
2
1
1
4 3 2 1
x
1
2
3
4
4 3 2 1
2
3
4
x = 4
(2, 0)
x
1
2
3
4
2
The equation of the
vertical line.
3
4
x=2
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The Slope of a Line
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.
y
change in y rise
slope =

change in x run
4
3
2
Change in y = 2 1
(0, 2)
(3, 0)
x
4
3 2 1
2
3
1
2
3
4
Change in
x=3
4
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Slope of a Line
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
y2  y1
slope = m =
x2  x1
where x2  x1.
Example:
Find the slope of the line that passes through (3, 6) and (1, 2).
y2  y1 6  2 4 2
m

  2
x2  x1 3  1 2 1
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Slope of a Line
Slope of a Straight Line
Positive Slope
Line goes up to the right
y
y
Negative Slope
Line goes downward to the right
1. Lines with
positive slopes go
upward as x
x
increases.
2. Lines with
negative slopes
go downward as
x x increases.
Continued.
22
Slope of a Line
Slope of a Straight Line (continued)
Zero Slope
Horizontal Line
y
3. Horizontal lines
have a slope of 0.
x
Undefined Slope
Vertical Line
y
4. A vertical line
has an undefined
slope.
x
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Slope-Intercept Form of a Line
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.
y = 3x – 4
The
The slope
y-intercept
is 3.
is (0, -4).
3
1
y  x
8
4
The slope
is 3 .
8
The
y-intercept
is 0, 1 .
 4
24
Slope-Intercept Form of a Line
Example:
Find the equation of the line with slope 2 and y-intercept
3
(0, 5).
2
slope: m 
3
y-intercept: b  5
y  mx  b
2
y  x  (5)
3
2
y  x 5
3
25
Graphing with Slope and y-Intercept
Example:
3
Graph the line with slope m   and y-intercept (0, 2).
2
3
m
2
y
y-intercept  (0, 2)
4
3
rise
run
3 units down
2 units to the right
(0, 2)
2
rise1
x
4
3 2 1
2
This is the graph of the line with
slope  3 and y-intercept (0, 2).
2
1
run
2
3
4
(2, 1)
3
4
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Parallel Lines
Parallel Lines
Parallel lines are two straight lines that never touch.
Parallel lines have the same slope but different
y-intercepts.
y
m1 = m2
Slope m1
x
Slope m2
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Parallel Lines
Example:
1
Line c has a slope of 2 .
If line d is parallel to line c, what is its slope?
Parallel lines have the same slope.
Line d has a slope of 1 .
2
y
line c
x
line d
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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
y
m1m2 =  1
or
1
m1 =  m
2
Slope m2
x
Slope m1
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Perpendicular Lines
Example:
1
Line c has a slope of .
2
If line e is perpendicular to line c, what is its slope?
Perpendicular lines have slopes whose product is 1.
m1m2  1
1
m2  1
2
m2  2
Line e has a slope of 2.
y
line c
x
line e
30
Writing the Equation
of a Line
Finding the Equation Given a Point
and the Slope
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.
32
Finding the Equation Given a Point
and the Slope
Example:
Find an equation of the line that passes through (4, 3)
with a slope of 5.
m = 5, x = 4, y = 3
y = mx + b
3 = (5)(4) + b
Substitute known values.
23 = b
The equation of the line is y = 5x  23.
33
Finding the Equation Given Two
Points
Example:
Find an equation of the line that passes through (2, 1) and
(7, 4).
Find the slope of the line.
y2  y1
m
x2  x1
4 1
3 1

 
7  (2) 9 3
1
5
The equation of the line is y  x  .
3
3
y = mx + b
1
1  (2)  b
3
2
1
b
3
5
b
3
34
Finding the Equation Given a
Graph of the Line
Example:
Find the equation of the line for the following graph.
y
Find the y-intercept.
b2
4
3
2
Find the slope
m
change in y
2

change in x
3
Change in y = 2 1
(0, 2)
(3, 0)
x
4
3 2 1
2
3
2
The equation of the line is y   3 x  2.
1
2
3
4
Change in
x=3
4
35
Systems of Linear
Equations in Two
Variables
Systems of Equations
A system of equations or system of inequalities is two or more
equations or inequalities in several variables that are considered
simultaneously.
y
y
y
3
2
3
2
3
2
1
1
1
x
3 2 1
1
2
3
x
3 2 1
1
2
3
x
3 2 1
2
2
2
3
3
3
The lines may intersect.
The lines may be parallel.
1
2
3
The lines may coincide.
37
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 =  2x + 8
y =  4x
16 =  4( 4)
16 = 16
Yes, it is a
solution.
y =  2x + 8
16 =  2( 4) + 8
16 = 8 + 8
16 = 16
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Graphing to Solve a System
Example:
Solve by graphing.
y =  3x  6
y = 2x  1
y
x
y
2
0
0
6
6
1
9
4
8
2
x
8
6 4 2
x
y
2
5
4
0
1
6
2
3
2
4
6
8
8
Continued.
39
Graphing to Solve a System
Example continued:
y
y = 2x  1
8
6
y =  3x  6
4
2
8
The lines
intersect at
(1, 3).
6 4 2
2
4
6
8
Check:
y = 3x  6
3 =  3(1)  6
3 = 3  6
x 3 = 3 
4
6
8
A system of equations that has one
solution is said to be consistent.
y = 2x  1
3 = 2(1)  1
3 = 2  1
3 = 3 
40
Inconsistent Systems
Example:
Solve by graphing.
y
3x  2y = 4
8
3x  2y = 4
 9x + 6y = 1
6
4
 9x + 6y = 1
2
x
8
The lines are parallel.
6 4 2
2
4
6
8
4
6
8
A system of linear equations that has no
solution is called an inconsistent system.
41
Dependent Systems
Example:
Solve by graphing.
y
The lines coincide.
8
4x  6y = 8
 2x + 3y =  4
6
4
4x  6y = 8
2
x
8
6 4 2
 2x + 3y = 4
2
4
6
8
4
6
8
A system of linear equations that has an infinite
number of solutions is called a dependent system.
42
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.
43
The Substitution Method
Example:
Find the solution.
2x – y = 13
– 4x – 9y = 7
y = 2x – 13
– 4x – 9y = 7
– 4x – 9(2x – 13) = 7
– 4x – 18x + 117 = 7
This variable is the
easiest to isolate.
Solve the first equation for y.
This is the original second equation.
Substitute the expression into the
other equation and solve.
– 22x = – 110
x =5
Substitute this value into one of the
original equations.
Continued.
44
The Substitution Method
Example continued:
2x – y = 13
– 4x – 9y = 7
First equation
Second equation
2x – y = 13
2(5) – y = 13
10 – y = 13
Substitute x = 5 into the first equation.
Solve for y.
–y=3
y=–3
The solution is (5, – 3).
Continued.
45
The Substitution Method
Example continued:
Check the solution (5, – 3) in both original equations.
2x – y = 13
– 4x – 9y = 7
2(5) – (3) = 13
– 4(5) – 9(3) = 7
10 + 3 = 13
– 20 + 27 = 7
13 = 13

7=7

46
The Addition 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.
47
The Addition Method
Example:
Solve by addition. 5x – 3y = 14
2x – y = 6
(3)2x – (3)y = (3)6 Multiply each term of equation (2) by 3.
 6x + 3y = 18
This equation is equivalent to equation (2).
5x – 3y = 14
 6x + 3y = 18
x=4
Add the two equations.
x=4
Substitute this value into
either equation to find y.
Continued.
48
The Addition Method
Example continued:
5x – 3y = 14
2x – y = 6
2x – y = 6
2(4) – y = 6
Substitute.
8y=6
y=2
y=2
The solution is (4, 2).
Be sure to check the solution in both original equations.
49
Identifying Inconsistent Systems
Example:
Solve the system algebraically.
3x + 6y = 12
x + 2y = 7
x = 2y + 7
3(2y + 7) + 6y = 12
6y + 21 + 6y = 12
21 = 12
Solve equation (2) for x.
Substitute into equation (1).
Simplify.
This results in a false statement.
There is no solution to this system of equations. If graphed,
these lines would be parallel.
50
Identifying Dependent Systems
Example:
Solve the system algebraically.
6x – 4y = 8
– 9x + 6y = –12
(6)6x – (6)4y = (6)8
4(–9x) + 4(6y) = 4(–12)
36x – 24y = 48
–36x + 24y = –48
0=0
Multiply each term in (1) by 6.
Multiply each term in (2) by 4.
This equation is equivalent to (1).
This equation is equivalent to (2).
Add the equations.
There are an infinite number of solutions to this system of
equations. If graphed, these lines would be the same.
51
Choosing an Appropriate Method
Method
Advantage
Disadvantage
Substitution
Works well if one or more
Often becomes difficult to
variable has a coefficient of 1 use if no variable has a
or 1.
coefficient of 1 or 1.
Addition
Works well if equations have None
fractional or decimal
coefficients, or if no variable
has a coefficient of 1 or 1.
52
Choosing an Appropriate Method
Example:
Select an appropriate method for solving the system.
a.)
5x – 3y = 13
0.9x + 0.4y = 30
b.)
3x + 6y = 12
x + 2y = 7
c.)
6x – 4y = 8
– 9x + 6y = –12
The addition method should be used
since none of the variables have a
coefficient of 1 or 1.
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.
53
Possible Solutions
Graph
Number of Solutions Algebraic Interpretation
One unique solution
You obtain one value for x
and one value for y. For
example,
x = 3, y = 5.
No solution
You obtain an equation
that is inconsistent with
known facts. The system
is inconsistent.
Infinite number of
solutions
You obtain an equation
that is always true. These
equations are dependent.
(3, 5)
Two lines intersect at one point.
Parallel lines
Lines coincide
54