Transcript Section 7.1 Graphs, Slopes, Inequalities and Applications

Section 7.1
Graphs, Slopes,
Inequalities and
Applications
OBJECTIVES
A
Find and graph an
equation of a line given
its slope and a point on
the line.
OBJECTIVES
B
Find and graph an
equation of a line given
its slope and y-intercept.
OBJECTIVES
C
Find and graph an
equation of a line given
two points on that line.
DEFINITION
Point-slope form
The point-slope form of
the equation of the line
going through (x1, y1), and
having slope m is
y – y1 = m (x – x1)
DEFINITION
Slope-intercept form
The slope-intercept form
of the equation of the line
having slope m and yintercept b is
y = mx + b
PROCEDURE
Finding the Equation of a Line
Given:
a point (x ,y ) and the slope m
1 1
Use:
Point-slope form
y – y = m (x – x )
1
1
PROCEDURE
Finding the Equation of a Line
Given:
slope m and y-intercept b
Use:
Slope-intercept form
y = mx + b
PROCEDURE
Finding the Equation of a Line
Given:
two points (x , y1 ) and (x2 , y2 )
x  x2
1
1
PROCEDURE
Finding the Equation of a Line
Use:
Two-point form
y – y1 = m (x– x1), where
y
–
y
2
1
m = x –x
2
1
NOTE
The Resulting Equation Can
Always Be Written as
Ax + By = C
Chapter
Graphs, Slopes, Inequalities
and Applications
Section 7.1
Exercise #1
Find an equation of the line going through the point (2, – 6)
and with slope – 5. Write the answer in point-slope form
and then graph the line.
(2, – 6), m = – 5
y – y1 = m(x – x1)
y – ( – 6) = – 5(x – 2)
y + 6 = – 5(x – 2)
Find an equation of the line going through the point (2, – 6)
and with slope – 5. Write the answer in point-slope form
and then graph the line.
y + 6 = – 5(x – 2)
y + 6 = – 5x + 10
y
3
5
–5
(1,–1)
(2,– 6)
–7
x
y = – 5x + 4
Chapter 7
Graphs, Slopes, Inequalities
and Applications
Section 7.1
Exercise #2
Find an equation of the line with slope 5 and y-intercept
– 4. Write the answer in slope-intercept form and then
graph the line.
m = 5, b = – 4
y = mx + b
y = 5x – 4
Find an equation of the line with slope 5 and y-intercept
– 4. Write the answer in slope-intercept form and then
graph the line.
3
–5
y = 5x – 4
y
(1, 1)
(0, – 4)
5
x
Chapter 7
Graphs, Slopes, Inequalities
and Applications
Section 7.1
Exercise #3
Find an equation of the line passing through the points
(2, – 8) and ( – 4, – 2). Write the answer in standard form.
Find the slope. (2, – 8) ( – 4, – 2)
y 2 – y1
– 2 – ( – 8)
m=
=
x 2 – x1
–4–2
–2+8
m=
–6
6
m=
–6
m= –1
Find an equation of the line passing through the points
(2, – 8) and ( – 4, – 2). Write the answer in standard form.
Use point slope form.
y – y1 = m(x – x1 )
y – ( – 8) = – 1(x – 2)
y+8= –x+2
y= –x–6
x +y= –6
Section 7.2
Graphs, Slopes,
Inequalities and
Applications
OBJECTIVES
A
Solve applications
involving the point-slope
formula.
OBJECTIVES
B
Solve applications
involving the slopeintercept formula.
OBJECTIVES
C
Solve applications
involving the two-point
formula.
Chapter 7
Graphs, Slopes, Inequalities
and Applications
Section 7.2
Exercise #4
Long-distance rates for m minutes are $10 plus $0.20 for
each minute. If a 10-minute call costs $12, write an
equation for the total cost C and find the cost of a
15-minute call.
Step 1: Read the problem.
Step 2: Select the unknown.
C
Step 3: Translate.
C = 0.20m + 10
Long-distance rates for m minutes are $10 plus $0.20 for
each minute. If a 10-minute call costs $12, write an
equation for the total cost C and find the cost of a
15-minute call.
Step 4: Use Algebra to find the cost C
of an m = 15 minute call.
C = 0.20m + 10
C = 0.20 (15) + 10
C = 3 + 10
C = $13
Chapter 7
Graphs, Slopes, Inequalities
and Applications
Section 7.2
Exercise #5
A cell phone plan costs $40 per month with 500 free
minutes and $0.50 for each additional minute. Find an
equation for the total cost C of the plan when
m minutes are used after the first 500. What is
the cost when 800 total minutes are used?
Step 1: Read the problem.
Step 2: Select the unknown.
C
Step 3: Translate.
C = 0.50m + 40, m > 500
A cell phone plan costs $40 per month with 500 free
minutes and $0.50 for each additional minute. Find an
equation for the total cost C of the plan when
m minutes are used after the first 500. What is
the cost when 800 total minutes are used?
Step 4: Use Algebra to find the cost C
when 800 minutes are used.
C = 0.50(800 – 500) + 40
C = 0.50(300) + 40
A cell phone plan costs $40 per month with 500 free
minutes and $0.50 for each additional minute. Find an
equation for the total cost C of the plan when
m minutes are used after the first 500. What is
the cost when 800 total minutes are used?
C = 0.50(300) + 40
C = 150 + 40
C = $190
Section 7.3
Graphs, Slopes,
Inequalities and
Applications
OBJECTIVES
A
Graph linear inequalities
in two variables.
PROCEDURE
Graphing a Linear Inequality
Determine the line that is the
boundary of the region.
If the inequality involves  or ,
draw the line solid;
PROCEDURE
Graphing a Linear Inequality
If it involves < or >, draw the line
dashed.
The points on a solid line are
part of the solution set.
PROCEDURE
Graphing a Linear Inequality
Use any point (a, b)
as a test point.
Substitute the values of a
and b for x and y in the
inequality.
PROCEDURE
Graphing a Linear Inequality
If a true statement results,
shade the side of the line
containing the test point.
If a false statement results,
shade the other side.
Chapter 7
Graphs, Slopes, Inequalities
and Applications
Section 7.3
Exercise #7
Graph 3x – 2y < – 6.
Solve for y.
y
5
(0, 3)
3x – 2y < – 6
– 2y < – 3x – 6
3
y> x+3
2
5
– 5 (2,0)
–5
x
Chapter 7
Graphs, Slopes, Inequalities
and Applications
Section 7.3
Exercise #8
Graph – y  – 3x + 3.
Solve for y.
– y  – 3x + 3
5
–5
y  3x – 3
y
(1,0)
(0,3)
–5
5
x
Chapter 7
Graphs, Slopes, Inequalities
and Applications
Section 7.3
Exercise #9
Graph 4x – y > 0.
Solve for y.
4x – y > 0
– y > – 4x
5
y
y < 4x
(1, 4)
5
–5
(0, 0)
–5
x
Chapter 7
Graphs, Slopes, Inequalities
and Applications
Section 7.3
Exercise #10
Graph 2y – 8  0
Solve for y.
2y – 8  0
2y  8
y
y4
7
5
–5
–3
x
Chapter 7
Graphs, Slopes, Inequalities
and Applications
Section 7.3
Exercise #12
The maximum weight W that can be supported by a
2-by-4-inch piece of pinewood varies inversely with its
length L. If the maximum weight W that can be
supported by a 10-foot-long 2-by-4
piece of pine is 500 pounds, find
an equation of variation and the
maximum weight W that can be
supported by a 25-foot length of
2-by-4 pine.
Step 1: Read the problem.
The maximum weight W that can be supported by a
2-by-4-inch piece of pinewood varies inversely with its
length L. If the maximum weight W that can be
supported by a 10-foot-long 2-by-4
piece of pine is 500 pounds, find
an equation of variation and the
maximum weight W that can be
supported by a 25-foot length of
2-by-4 pine.
Step 2: Select the unknown.
W and L
The maximum weight W that can be supported by a
2-by-4-inch piece of pinewood varies inversely with its
length L. If the maximum weight W that can be
supported by a 10-foot-long 2-by-4
piece of pine is 500 pounds, find
an equation of variation and the
maximum weight W that can be
supported by a 25-foot length of
2-by-4 pine.
Step 3: Translate.
W=k
L
Step 4: Use Algebra
k
500 =
10
k = 5000
W = 5000
L
W = 5000
25
= 200 pounds
Section 7.4
Graphs, Slopes,
Inequalities and
Applications
OBJECTIVES
A
Find and solve
equations of direct
variation given values of
the variables.
OBJECTIVES
B
Find and solve
equations of inverse
variation given values of
the variables.
OBJECTIVES
C
Solve applications
involving variation.
DEFINITION
y varies directly as x if
There is a constant k such
that y = kx
k is the constant of
variation or proportionality.
Chapter 7
Graphs, Slopes, Inequalities
and Applications
Section 7.4
Exercise #11
Have you raked leaves lately? If you rake leaves for m
minutes, the number C of calories used is proportional
to the time you rake. If 60 calories are used when you
rake for 30 minutes, write an equation of
variation and find the number of calories
used when you rake for 2 21 hours.
Step 1: Read the problem.
Step 2: Select the unknown.
C and m
Have you raked leaves lately? If you rake leaves for m
minutes, the number C of calories used is proportional
to the time you rake. If 60 calories are used when you
rake for 30 minutes, write an equation of
variation and find the number of calories
used when you rake for 2 21 hours.
Step 3: Translate.
The equation of variation is
C = km
Step 4: Use Algebra
60 = k(30)
k=2
C = 2m
(2 21 hours = 150 minutes)
C = 2(150) = 300 calories