Transcript Pre-Algebra

Pre-Algebra
Terms: 1. relation – an ordered pair (relationship between x and y)
2. domain – first coordinate of a relation (the “x” value)
3. range – the second coordinate of a relation (the “y” value)
4. Function – a relation in which each member of the domain is
paired with exactly one member of the range.
The Coordinate Plane
Lesson 1-10
Pre-Algebra
Write the coordinates of point G. In which
quadrant is point G located?
Point G is located 2 units to the left of the yaxis.
So the x-coordinate is –2.
The point is 3 units below the x-axis.
So the y-coordinate is –3.
The coordinates of point G are (–2, –3). Point G is located
in Quadrant III.
The Coordinate Plane
Lesson 1-10
Pre-Algebra
Graph point M(–3, 3).
Relations and Functions
Lesson 8-1
Pre-Algebra
Is each relation a function? Explain.
a. {(0, 5), (1, 6), (2, 4), (3, 7)}
List the domain values and the range values
in order.
Draw arrows from the domain values to their
range values.
There is one range value for each domain value. This relation is a
function.
Relations and Functions
Lesson 8-1
Pre-Algebra
(continued)
b. {(0, 5), (1, 5), (2, 6), (3, 7)}
There is one range value for each domain value. This relation is a
function.
Relations and Functions
Lesson 8-1
Pre-Algebra
(continued)
c. {(0, 5), (0, 6), (1, 6), (2, 7)}
There are two range values for the domain value 0.
This relation is not a function.
Relations and Functions
Lesson 8-1
Pre-Algebra
Is the time needed to mow a lawn a function of the
size of the lawn? Explain.
No; two lawns of the same size (domain value)
can require different lengths of time (range values)
for mowing.
Relations and Functions
Lesson 8-1
Pre-Algebra
a. Graph the relation shown in the table.
Domain
Value
–3
–5
3
5
Range
Value
5
3
5
3
Graph the ordered pairs
(–3, 5), (–5, 3), (3, 5), and
(5, 3).
Relations and Functions
Lesson 8-1
Pre-Algebra
(continued)
b. Use the vertical-line test. Is the relation a function? Explain.
Pass a pencil across the graph as shown.
Keep the pencil vertical (parallel to the
y-axis) to represent a vertical line.
The pencil does not pass through two points at any one of its
positions, so the relation is a function.
Equations With Two Variables
Lesson 8-2
Pre-Algebra
Find the solution of y = 4x – 3 for x = 2.
y = 4x – 3
y = 4(2) – 3
y=8–3
y=5
Replace x with 2.
Multiply.
Subtract.
A solution of the equation is (2, 5).
Equations With Two Variables
Lesson 8-2
Pre-Algebra
The equation a = 5 + 3p gives the price for
admission to a park. In the equation, a is the admission
price for one car with p people in it. Find the price of
admission for a car with 4 people in it.
a = 5 + 3p
a = 5 + 3(4)
a = 5 + 12
a = 17
Replace p with 4.
Multiply.
Add.
A solution of the equation is (4, 17). The admission price for one
car with 4 people in it is $17.
Equations With Two Variables
Lesson 8-2
Pre-Algebra
Graph y = 4x – 2.
Make a table of values to show ordered-pair solutions.
x
–2
0
2
4x – 2
4(–2) – 2 = –8 – 2 = –10
4(0) – 2 = 0 – 2 = –2
(x, y)
(–2, –10)
(0, –2)
4(2) – 2 = 8 – 2 = 6
(2, 6)
Graph the ordered pairs.
Draw a line through the points.
Equations With Two Variables
Lesson 8-2
Pre-Algebra
Graph each equation. Is the equation a function?
a. y = –3
b. x = 4
For every value of x, y = –3.
For every value of y, x = 4.
This is a horizontal line.
The equation y = –3 is a
function.
This is a vertical line.
The equation y = 4
is not a function.
Equations With Two Variables
Lesson 8-2
Pre-Algebra
Solve y – 1 x = 3 for y. Then graph the equation.
2
Solve the equation for y.
y–
1
1
x=3
2
1
1
y – 2x + 2x = 3 + 2 x
1
y=2x+3
1
Add 2 x to each side.
Simplify.
Equations With Two Variables
Lesson 8-2
Pre-Algebra
(continued)
Make a table of values.
1
2
1
–2
2
1
0
2
1
2
2
x
Graph.
x+3
(x, y)
(–2) + 3 = –1 + 3 = 2
(–2, 2)
(0) + 3 = 0 + 3 = 3
(0, 3)
(2) + 3 = 1 + 3 = 4
(2, 4)
Slope and y-intercept
Lesson 8-3
Pre-Algebra
Slope and y-intercept
Lesson 8-3
Pre-Algebra
Find the slope of each line.
a.
b.
rise
4
slope = run = 1 = 4
rise
–6
slope = run = 3 = –2
Slope and y-intercept
Lesson 8-3
Pre-Algebra
Find the slope of the line through E(7, 5) and F(–2, 0).
difference in y-coordinates
0–5
–5
5
slope = difference in x-coordinates = –2 – 7 = –9 = 9
Slope and y-intercept
Lesson 8-3
Pre-Algebra
Find the slope of each line.
a.
b.
–3 – (–3)
0
–1 – 3
–4
slope = 4 – (–2) = 6 = 0
slope = –2 – (–2) = 0
Slope is 0 for a horizontal line.
Division by zero is
undefined. Slope is
undefined for a vertical line.
Slope and y-intercept
Lesson 8-3
Pre-Algebra
A ramp slopes from a warehouse door down to a
street. The function y = – 1 x + 4 models the ramp, where x is
5
the distance in feet from the bottom of the door and y is the
height in feet above the street. Graph the equation.
Step 1: Since the y-intercept is 4,
graph (0, 4).
1
Step 2: Since the slope is – 5 , move
1 unit down from (0, 4).
Then move 5 units right to
graph a second point.
Step 3: Draw a line through the
points.
Writing Rules for Linear Functions
Lesson 8-4
Pre-Algebra
A long-distance phone company charges its
customers a monthly fee of $4.95 plus 9¢ for each minute of
a long-distance call.
a.
Write a function rule that relates the total monthly bill to the
number of minutes a customer spent on long-distance calls.
Words
total bill
is
$4.95
plus
9¢
times
number of
minutes
Let m = the number of minutes.
Let t( m ) = total bill, a function of the number of minutes.
Rule
t( m )
=
4.95
+
0.09
A rule for the function is t(m) = 4.95 + 0.09m.
•
m
Writing Rules for Linear Functions
Lesson 8-4
Pre-Algebra
(continued)
b.
Find the total monthly bill if the customer made 90 minutes
of long-distance calls.
t(m) = 4.95 + 0.09m
t(90) = 4.95 + 0.09(90)
Replace m with 90.
t(90) = 4.95 + 8.10
Multiply.
t(90) = 13.05
Add.
The total monthly bill with 90 minutes of long-distance calls
is $13.05.
Writing Rules for Linear Functions
Lesson 8-4
Pre-Algebra
Write a rule for the linear function in the
table below.
–2
–2
–2
x
f(x)
2
0
–2
–4
3
–5
–13
–21
As the x values decrease by 2,
the f(x) values decrease by 8.
–8
–8
–8
–8
So m = –2 = 4.
When x = 0, f(x) = –5. So b = –5.
A rule for the function is f(x) = 4x – 5.
Writing Rules for Linear Functions
Lesson 8-4
Pre-Algebra
Write a rule for the linear function graphed below.
–2 – 2
–4
slope = 0 – 2 = –2 = 2
y-intercept = –2
A rule for the function is f(x) = 2x – 2.
Scatter Plots
Lesson 8-5
Pre-Algebra
The scatter plot shows education and income data.
a.
Describe the person represented by the point
with coordinates (10, 30).
This person has 10 years of education and
earns $30,000 each year.
b.
How many people have exactly 14 years of
education? What are their incomes?
The points (14, 50), (14, 80), and (14, 90)
have education coordinate 14.
The three people they represent earn
$50,000, $80,000, and $90,000, respectively.
Scatter Plots
Lesson 8-5
Pre-Algebra
Use the table to make a scatter plot of the
elevation and precipitation data.
Elevation Above
Sea Level (ft)
1,050
Atlanta, GA
20
Boston, MA
596
Chicago, IL
18
Honolulu, HI
11
Miami, FL
1,072
Phoenix, AZ
75
Portland, ME
40
San Diego, CA
1,305
Wichita, KS
City
Mean Annual
Precipitation (in.)
51
42
36
22
56
8
44
10
29
Scatter Plots
Lesson 8-5
Pre-Algebra
Use the scatter plot below. Is there a positive
correlation, a negative correlation, or no correlation
between temperatures and amounts of precipitation?
Explain.
The values show no relationship.
There is no correlation.
Problem Solving Strategy: Solve by Graphing
Lesson 8-6
Pre-Algebra
Use the data in the table below. Suppose this year
there are 16 wolves on the island. Predict how many moose
are on the island.
Isle Royale Populations
Year
Wolf
1982
1983
1984
14
23
24
1985
1986
1987
22
20
16
Moose
Year
Wolf
Moose
Year
Wolf
Moose
700
900
811
1988
1989
1990
12
11
15
1,653
1,397
1,216
1994
1995
1996
15
16
22
1,800
2,400
1,200
1,062
1,025
1,380
1991
1992
1993
12
12
13
1,313
1,600
1,880
1997
1998
1999
24
14
25
500
700
750
Problem Solving Strategy: Solve by Graphing
Lesson 8-6
Pre-Algebra
(continued)
Step 1: Make a scatter plot by
graphing the (wolf, moose)
ordered pairs. Use the x-axis
for wolves and the y-axis for
moose.
Step 2: Sketch a trend line. The line
should be as close as
possible to each data point.
There should be about as
many points above the trend
line as below it.
Problem Solving Strategy: Solve by Graphing
Lesson 8-6
Pre-Algebra
(continued)
Step 3: To predict the number of
moose when there are 16
wolves, find 16 along the
horizontal axis.
Look up to find the point on
the trend line that
corresponds to 16 wolves.
Then look across to the value
on the vertical axis, which is
about 1,300.
There are about 1,300 moose on the
island.
Solving Systems of Linear Equations
Lesson 8-7
Pre-Algebra
Solve the system y = x – 7 and y = 4x + 2 by
graphing.
Step 1: Graph each line.
Step 2: Find the point of intersection.
The lines intersect at one point, (–3, –10).
The solution is (–3, –10).
Check: See whether (–3, –10) makes both equations
true.
y=x–7
–10
–3 – 7
–10 = –10
Replace x with – 3
and y with –10.
The solution checks.
y = 4x + 2
–10
4(–3) + 2
–10 = –10
Solving Systems of Linear Equations
Lesson 8-7
Pre-Algebra
Solve each system of equations by graphing.
a. 27x + 9y = 36; y = 4 – 3x
The lines are the same line.
There are infinitely
many solutions.
b. 8 = 4x + 2y; 2x + y = 5
The lines are parallel.
They do not intersect.
There is no solution.
Solving Systems of Linear Equations
Lesson 8-7
Pre-Algebra
Find two numbers with a sum of 10 and a
difference of 2.
Step 1: Write equations.
Let x = the greater number.
Let y = the lesser number.
Equation 1 Sum
x+y
is
=
Equation 2 Difference
x–y
10.
10
is
=
Step 2: Graph the equations.
The lines intersect at (6, 4).
The numbers are 6 and 4.
2.
2
Solving Systems of Linear Equations
Lesson 8-7
Pre-Algebra
(continued)
Check: Since the sum of 6 and 4 is 10 and the difference of 6 and 4 is 2,
the answer is correct.
Graphing Linear Inequalities
Lesson 8-8
Pre-Algebra
Graph each inequality on a coordinate plane.
a.
y > 2x + 1
Step 1: Graph the boundary line.
Points on the boundary line do not
make y > 2x + 1 true. Use a dashed
line.
Graphing Linear Inequalities
Lesson 8-8
Pre-Algebra
(continued)
Step 2: Test a point not on the boundary line.
Test (0, 0) in the inequality.
y > 2x + 1
?
0 >? 2(0) + 1
Substitute.
0> 0+1
0>1
false
Since the inequality is false for (0, 0), shade
the region that does not contain (0, 0).
Graphing Linear Inequalities
Lesson 8-8
Pre-Algebra
(continued)
b.
y<
– 3x – 2
Step 1: Graph the boundary line.
Points on the boundary line make
y<
– 3x – 2 true. Use a solid line.
Graphing Linear Inequalities
Lesson 8-8
Pre-Algebra
(continued)
Step 2: Test a point not on the boundary line.
Test (3, 0) in the inequality.
y<
–? 3x – 2
0<
Substitute.
–? 3(3) – 2
0<
–9–2
0<
true
–7
Since the inequality is true for (3, 0), shade
the region containing (3, 0).
Graphing Linear Inequalities
Lesson 8-8
Pre-Algebra
Cashews cost $2/lb. Pecans cost $4/lb. You plan
to spend no more than $20. How many pounds of each can
you buy?
Step 1: Write an inequality.
Words
Inequality
cost of
cashews
plus
cost of
pecans
is at
most
Let
y = number of pounds of cashews.
Let
x = number of pounds of pecans.
2y
+
4x
<
–
twenty
dollars
20
Graphing Linear Inequalities
Lesson 8-8
Pre-Algebra
(continued)
Step 2: Write the equation of the boundary line in slopeintercept form.
2y + 4x <
– 20
y<
– –2x + 10
y = –2x + 10
Step 3: Graph y = –2x + 10 in
Quadrant I since weight is not
negative.
Graphing Linear Inequalities
Lesson 8-8
Pre-Algebra
(continued)
Step 4: Test (1, 1).
y<
–? –2x + 10
1<
– –2(1) + 10
1<
–8
The inequality is true. (1, 1) is a solution.
Step 5: Shade the region containing (1, 1).
The graph shows the possible solutions. For
example, you could buy 1 pound of pecans and 5
pounds of cashews.
Graphing Linear Inequalities
Lesson 8-8
Pre-Algebra
Solve the system y >
– x + 1 and y < 2x + 3 by graphing.
Step 1: Graph y >
– x + 1 on a coordinate
plane. Shade in red.
Step 2: Graph y < 2x + 3 on the same
coordinate plane. Shade in blue.
Graphing Linear Inequalities
Lesson 8-8
Pre-Algebra
(continued)
The solutions are the coordinates of all the points in the region that is
shaded in both colors.
Check: See whether the solution (2, 5) makes both of the inequalities
true.
y>
–? x + 1
5>
Replace x with 2 and y with 5.
–2+1
5>
The solution checks.
– 3y
y < 2x + 3
?
5 < 2(2) + 3
5 < 7y
Replace x with 2 and y with 5.
The solution checks.