Download CH3

Sect 3.1 Reading Graphs
How much money will you earn in a lifetime with an associate’s degree?$1.5 to 1.6 million
What degree must you obtain to earn at least 2 million dollars in a lifetime?
Bachelor’s or higher.
Sect 3.1 Reading Graphs
%
is

100 of
9
x

100 134.8
Find the amount of 134.8 billion
that was given to Pell Grants.
x  9 134.8 100
x  12.132 billion
Avg.  12,132,000,000  5,387,000
Avg.  $2,252.09
Sect 3.1 Reading Graphs
Indicates no visual comparisons!
How many months of regular exercise are required to lower the pulse rate as much as possible?
6 MONTHS
How many months of regular exercise are needed to achieve a pulse rate of 65 beats per min.?
3 MONTHS
Sect 3.1 Reading Graphs
All points are labeled (x, y).
y
P ( x, y )
A(__,
4 __)
3
B(__,
 3 __)
5
C (__,
 4 __)
3
D (__,
__)
4
2 
E (__,
1 __)
5
F (__,
 2 __)
0
G (__,
3
0 __)
x
Sect 3.1 Reading Graphs
These points are
not in a Quadrant!
Sect 3.2 Graphing Linear Equations
Determine if (2, 5) and (-1, -1)
are solutions to
y  2x 1
2,5  y  2x 1
5  22  1
(2, 5)
5  4 1
55
TRUE
Yes (2, 5) is a solution.
1,1  y  2x 1
1  21  1
 1  2  1
 1  1
TRUE
Yes (-1, -1) is a solution.
(-1, -1)
Sect 3.2 Graphing Linear Equations
When the equation is solved for y, y is the dependent variable and x is the independent variable.
This means we can pick values for x and substitute them in to find the y value.
Graph y  3x  2
Start with x = 0. Easiest to multiply by.
x
y  3x  2
(x, y)
0
y  30   2
y2
0,2
1
y  31  2
y  1
1,1
2
y  32   2 2,4
y  4
Use a straight edge to connect
the points to form a straight
line.
Sect 3.2 Graphing Linear Equations
Graph y  2 x  3
Start with x = 0. Easiest to multiply by.
x
y  2x  3
(x, y)
0
y  20   3
y  3
0,3
1
y  21  3
y  1
1,1
2
y  22   3
y 1
2,1
Use a straight edge to connect
the points to form a straight
line.
Sect 3.2 Graphing Linear Equations
2
x 1
3
Graph y 
Start with x = 0, Count by 3’s due to
the denominator in the fraction.
x
y
2
x 1
3
0
y
2
3
(x, y)
0  1 0,1
y  1
3
3  1 +3 +2
3,1
y  2 1
y
2
3
y 1
6
6  1 +3 +2
6,3
y  4 1
y
2
3
y3
Notice a pattern in the points?
Y-coordinates and X-coordinates.
The fraction contains the changes
in the y-coord. and x-coord.!
Sect 3.2 Graphing Linear Equations
1
Graph y   x  3
2
Start with x = 0, Count by 2’s due to
the denominator in the fraction.
x
1
y   x3
2
(x, y)
0
y   12 0  3
0,3
y 3
2
y   12 2  3
y  1  3
+2 -1
2,2
y2
4
y   12 4  3
y  2  3
+2 -1
4,1
y 1
Use a straight edge to connect
the points to form a straight
line.
Sect 3.2 Graphing Linear Equations
Graph 2 x  5 y  5
+ 5 + 5y
Solve the equation for y!
+ 5 + 5y
2x  5  5 y
5
5
5
2
x 1  y
5
x
y
2
x 1
5
0
y
2
5
0  1
y 1
5
5
y
2
5
5  1
y  2 1
y3
y  52  5  1
y  2  1
y  1
(x, y)
0,1
5,3
 5,1
Sect 3.3 Graphing Linear Equations in Standard Form
Ax  By  C ; m 
Graph 2 x  5 y  5
m
x-intercept (
5
____,
0
2
2x  50  5
)
2x  5
x
5
2
1 )
y-intercept ( 0, ____
20  5 y  5
 5 y  5
y 1
2 2

5 5
A
B
Sect 3.3 Graphing Linear Equations in Standard Form
Graph 2 x  4 y  12
 2 1
m

4
2
6 0)
x-intercept ( ____,
2x  40  12
2x  12
x6
y-intercept ( 0, ____
3 )
20  4 y  12
4 y  12
y 3
Sect 3.3 Graphing Linear Equations in Standard Form
Graph 3 x  2 y  12
3 3
m

2 2
4 0)
x-intercept ( ____,
3x  20  12
3x  12
x4
y-intercept ( 0, ____
6 )
30  2 y  12
 2 y  12
y  6
Sect 3.3 Graphing Linear Equations in Standard Form
Graph 4 x  5 y  0
4 4
m

5 5
0 0)
x-intercept ( ____,
4 x  50  0
4x  0
x0
y-intercept ( 0, ____
0 )
40  5 y  0
 5y  0
y0
Sect 3.3 Graphing Linear Equations in Standard Form
Rewrite in standard form: Ax + By = C
Graph 4 x  8  8 y
4 x  8 y  8
Short Cut: Cover-up
technique.
2 0 )
x-intercept ( ____,
To find the x-intercept,
cover-up the y term.
Solve for x.
x  2
1 )
y-intercept ( 0, ____
To find the y-intercept,
cover-up the x term.
Solve for y.
y 1
m
4 1

8 2
Sect 3.3 Graphing Linear Equations
Notice there is no y variable in the equation…the line can’t cross the y-axis. No y-intercept.
2x  3  11
2x  8
x4
4 0)
x-intercept ( ____,
Graph
y-intercept ( 0, ____ )
Sect 3.3 Graphing Linear Equations
Notice there is no x variable in the equation…the line can’t cross the x-axis. No x-intercept.
Graph
3y  6
y2
x-intercept ( ____, 0 )
2 )
y-intercept ( 0, ____
Sect 3.3 Graphing Linear Equations
Graph 5 x  3 y  11
11
x-intercept ( ____,
0)
5
YUCK!
11
y-intercept ( 0, ____
3 )
Lets find clean points. Take the
LARGEST coefficient and take it’s
opposite to the other side.
5 x  3 y  11
5x
 5x
3 y  11  5 x
Start with 11 and keep subtracting
by 5 until you have a number
divisible by 3.
11  5  6
We only subtracted
by one 5, so x = 1.
3 y  6, when x  1
y2
 A 5
1, 2 m  
B
3
Sect 3.3 Graphing Linear Equations
Graph 7 x  5 y  19
19
x-intercept ( ____,
0)
7
19
y-intercept ( 0, ____
5 )
YUCK!
Lets find clean points. Take the
LARGEST coefficient and take it’s
opposite to the other side.
7 x  5 y  19
5 y  19  7 x
7 x
 7 x
Start with 19 and keep subtracting
by 5 until you have a number
divisible by 5.
19  7  12 We subtracted by two
7’s, so x = 2.
12  7  5
5 y  5, when x  2
y  1
 2, 1 m   A  7  7
B 5 5
Sect 3.4 Rates
A Rate is a ratio that indicates how two quantities change with respect to each
other.
Unit rate is when the second quantity is one.
On Jan. 3rd, Joe rented a car with a full tank of gas and 9312 miles on the
odometer. On Jan. 7th, he returned the car with 9630 miles on the odometer. If
Joe had to pay $108 for the total bill which included 12 gallons of gas, find the
following rates. Convert to unit rates. 9630  9312  318miles
Divide for unit rate.
318miles
miles
 26.5
gallons
12 gallons
$108
$21.6

2. Average cost of the rental in dollars per day.
5days
day
rd th th th
th
1. Gas consumption in miles per gallon.
Jan. 3 , 4 , 5 , 6 , and 7 = 5 days
3. Travel rate in miles per day.
318miles
5days

63.6miles
day
Sect 3.5 Slope
The Slope is a ratio that indicates how the change in the y-coordinates change
with the respect to the change in the x-coordinates.
The slope of a line contains two points ( x1, y1 ) and ( x2, y2 ) is given by
rise y2  y1
m


change in x run x2  x1
change in y
Change in x
Change in y
x1 , y1 
x2 , y2 
Sect 3.5 Graphing Linear Equations
Graph ( -4, 3 ) and ( 2, -6 ) and
find the slope.
m
9 3

6
2
rise y2  y1
m

run x2  x1
63
2   4
9 3
m

6
2
m
Change in y = -9
Change in x = 6
Sect 3.5 Graphing Linear Equations
m3 m2
Positive slopes
always have lines that
go uphill.
Slopes > 1 are steep.
0 < Slopes < 1 begin
to flatten out.
m 1
m
1
2
1
m
4
m
2
3
Sect 3.5 Graphing Linear Equations
m  1 m  2 m  3
Negative slopes
always have lines that
go downhill.
Slopes < -1 are steep.
-1 < Slopes < 0 begin
to flatten out.
1
m
4
1
m
2
2
m
3
Sect 3.5 Graphing Linear Equations
Graph ( -4, 3 ) and ( 2, 3 ) and
find the slope.
33
2   4 
0
m 0
6
Horizontal Line m 
Graph ( 5, -1 ) and ( 5, 6 ) and
find the slope.
Vertical Line
6   1
55
7
m   undefined
0
m
Sect 3.5 Slope
The Grade is a slope that is measured as a percent.
7% 
7 vertical ft
100 horizontal ft
Drop 7 feet
for every 100 feet traveled horizontally.
Sect 3.6 Graphing Linear Equations in Slope Intercept Form
Graph
y
2
x4
3
( 0, - 4 )
Starting
point
Directions
2 up & 3 right
Or opposite
2 down & 3 left
Sect 3.6 Graphing Linear Equations in Slope Intercept Form
Graph
5
y   x6
2
( 0, 6 )
Starting
point
Directions
5 down & 2 right
2x  y  3
-3 + y = -3 + y
2x  3  y
y  2x  3
1
( 0, -3 )
Starting
point
Directions
2 up & 1 right
Opposite
2 down & 1 left
Sect 3.6 Graphing Linear Equations in Slope Intercept Form
Solving for y would not be a good decision because it will generate a fraction for a y-intercept.
Graph
5 x  3 y  10
x-intercept and slope.
2,0
 A 5 5
m


B
3 3
5 x  4 y  11
x-intercept and y-intercept
are bad…fractions.
4 y  11  5 x
11  5  6
6 5 1
1  5  4
x = 3 for the three 5’s we subtracted.
4 y  4 y  1;  3,1
m
 A 5 5


B
4 4
Sect 3.6 Graphing lines
Parallel Lines – Lines are parallel when they have the same slopes or the lines
are vertical.
m1  m2
Perpendicular Lines – Lines are perpendicular when their slopes are opposite
reciprocals of each other.
m1 m2   1
I prefer to say, “the perpendicular slopes are opposite reciprocals.”
Sect 3.7 Graphing Linear Equations in Point Slope Form
Consider a line with slope 2 passing through the point ( 4, 1 ).
y 1
m
2
x4
Consider removing the “m” and re-writing
so there is no fraction.
y 1
2
x4
y 1
 2   x  4
 x  4 
x4
( x, y )
( 4, 1 )
y 1  2  x  4
Notice the sign change!
The equation y  y1  mx  x1  is called point-slope form of a linear equation
with the slope m and the point x1 , y1  .
Sect 3.7 Graphing Linear Equations in Point Slope Form
Graph
3
y  5    x  1
4
Start @ (1, 5)
m = -3
4
y2
Start @ (-4, 2)
1
x  4
3
1
m =
3
3
y  5  x  4
2
3
Start @ (-4, -5)
m =
2
y  5  4x  2
Start @ (2, -5)
m =
4
1
Sect 3.7 Graphing lines
y  y1  mx  x1 
5
Write the equation of a line in point-slope form with the slope of
and the
7
point ( -14, 12). Convert to slope-intercept form.
5
5
y  12  x  14
y  12  x  10
7
+ 12 = 7 + 12
5
5 2
y  12  x  14
7
7
5
y  x  22
7
8
Write the equation of a line in point-slope form with the slope of  and the
5
point ( 4, -6). Convert to slope-intercept form.
8
x  4 y  6   8 x  32
5
5
5
–6 =
–6
8
8
y  6   x  4
5
5
y6  
8
32 30
y  x 
5
5
5
8
2
y  x
5
5
Sect 3.7 Graphing lines…ANOTHER APPROACH
8
Write the equation of a line in point-slope form with the slope of  and the
5
point ( 4, -6). Convert to slope-intercept form.
NO FRACTION WORK…Remember Standard Form.
 A 8
m

B
5
A8
B5
Ax  By  C ; m 
8x  5 y  C
Substitute the point ( 4, -6 ) for x and y to solve for C.
84  5 6  C
32 30  C
2C
8x  5 y  2
– 8x
= – 8x
5 y  8 x  2
5
5
Short cut.
b
8x  5 y  C
Solve for y.
A
B
5
C 2

B 5
8
2
y  x
5
5
Sect 3.7 Graphing lines
Ax  By  C ; m 
Same slopes.
A
B
Write the equation of a line that is parallel to 3x + 5y = 11 and contains the
point ( 0, 2). Use slope-intercept form.
 A 3
y-intercept (0, b)
y  mx  b
m
B

5
3
y  x2
5
Opposite and Reciprocal slopes.
Write the equation of a line that is perpendicular to 3x + 5y = 11 and contains
the point ( 0, 2). Use slope-intercept form.
 A 3
m
y-intercept (0, b)
B

Perpendicular slope.
5
y  x2
3
B 5
m 
A 3
5
Sect 3.7 Graphing lines
Must be the same!.
Same slopes.
Write the equation of a line that is parallel to 7x – 4y = 9 and contains the point
A 7
C 47
( 5, -3). Use slope-intercept form.
Ax  By  C ; m 
7x  4 y  C
75  4 3  C
35 12  C  47
7 x  4 y  47
 4 y  7 x  47
4 4 4
7
47
y  x
4
4
B
y

4
b
B

4
7
47
x
4
4
Write the equation of a line that is perpendicular to 8x + 3y = 13 and contains
the point ( -7, 4). Use slope-intercept form.
 A  8 Perpendicular slope.
m
The numbers must flip and change the sign!
8 x  3 y  13
B

3
3 x  8 y  C New. 3x  8 y  53
3 7  84  C
 21  32  C  53
3
53
y  x
8
8
B 3

A 8
C  53 53
b 

B 8
8
m
Sect 9.4 Graphing Linear Inequalities
3
Graph the line y   x  3 .
2
Consider changing the
equation to an inequality.
3
y   x3
2
Determine if the points are
solutions, ( -4, 3 ), (-8, 5 ), and
( 2, -1 ).
3
3
5    8  3  1   2  3
2
2
5  12  3
1  3  3
59
1  6
FALSE
TRUE
Sect 9.4 Graphing Linear Inequalities in Slope Intercept Form
Equal to line means a solid line!
No equal to line means a dashed line!
Graph
5
y   x6
2
The inequality has y > which tells me
to shade all the y coordinates greater
than the line…above the line.
Solve for y.
2x  y  3
– 2x
= – 2x
 y  2 x  3
-1
-1
-1
y  2x  3
Remember to flip the
inequality symbol.
Shade all the y coordinates less
than the line…below the line.
Sect 9.4 Graphing Linear Equations
Graph the line 2 x  4 y  8 .
Find the x and y intercepts.
Consider changing the
equation to an inequality.
2 x  4 y  8
Determine which way to
shade.
Test the Origin (0, 0)
20  40  8
0  8
TRUE
Shade in the direction
of the Origin
Sect 9.4 Graphing Linear Equations in Standard Form
Graph
 4 x  3 y  24
Find the x and y intercepts.
Test the Origin (0, 0)
 40  30  24
0  24 TRUE
2 x  4 y  16
Find the x and y intercepts.
Test the Origin (0, 0)
20  40  16
0  16
FALSE
Sect 9.4 Graphing Linear Equations
Graph
6 y  24
y4
 2 x  12
x  6
Sect 9.4 Graphing Linear Equations
Graph
3 y  2