Transcript MTH55A_Lec-07_sec_2-4a_Lines_by_Intercepts

Chabot Mathematics
§2.4a Lines
by Intercepts
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
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[email protected] • MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt
Review §
2.3
MTH 55
 Any QUESTIONS About
• § 2.3 → Algebra of Funtions
 Any QUESTIONS About HomeWork
• § 2.2 → HW-05
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Eqn of a Line  Ax + By = C
 Determine whether each of the following
pairs is a solution of eqn 4y + 3x = 18:
• a) (2, 3);
b) (1, 5).
 Soln-a) We substitute 2 for x and 3 for y
4y + 3x = 18  Since 18 = 18 is
4•3 + 3•2 | 18
true, the pair (2, 3)
12 + 6 | 18
is a solution
18 = 18
True
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Bruce Mayer, PE
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Example  Eqn of a Line
 Soln-b) We substitute 1 for x and 5 for y
4y + 3x = 18
4•5 + 3•1 | 18
20 + 3 | 18
23 = 18 
False
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 Since 23 = 18 is
false, the pair (1, 5)
is not a solution
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To Graph a Linear Equation
1. Select a value for one coordinate and
calculate the corresponding value of the
other coordinate. Form an ordered pair.
This pair is one solution of the equation.
2. Repeat step (1) to find a second ordered
pair. A third ordered pair can be used as
a check.
3. Plot the ordered pairs and draw a straight
line passing through the points. The line
represents ALL solutions of the equation
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Example  Graph y = −4x + 1
 Solution:
Select convenient values for x and
compute y, and form an ordered pair.
• If x = 2, then y = −4(2)+ 1 = −7 so (2,−7)
is a solution
• If x = 0, then y = −4(0) + 1 = 1 so (0, 1)
is a solution
• If x = –2, then y = −4(−2) + 1 = 9 so (−2, 9)
is a solution.
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Example  Graph y = −4x + 1
 Results are often
listed in a table.
x
2
0
–2
y
(x, y)
–7 (2, –7)
1
(0, 1)
9 (–2, 9)
• Choose x
• Compute y.
• Form the pair (x, y).
• Plot the points.
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Example  Graph y = −4x + 1
 Note that all three
points line up. If they
didn’t we would know
that we had made a
mistake
 Finally, use a ruler or
other straightedge to
draw a line
 Every point on the line
represents a solution
of: y = −4x + 1
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Example  Graph x + 2y = 6
 Solution: Select some
convenient x-values and
compute y-values.
• If x = 6, then 6 + 2y = 6,
so y = 0
• If x = 0, then 0 + 2y = 6,
so y = 3
• If x = 2, then 2 + 2y = 6,
so y = 2
x
6
0
y
0
3
(x, y)
(6, 0)
(0, 3)
2
2
(2, 2)
 In Table Form,
Then Plotting
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Example Graph 4y = 3x
 Solution: Begin by
solving for y.
4 y  3x
1
1
 4 y   3x
4
4
3
y  x  0.75 x
4
 To graph the last
Equation we can
select values of x
that are multiples
of 4
• This will allow us to
avoid fractions
when computing the
corresponding
y-values
 Or y is 75% of x
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Bruce Mayer, PE
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Example  Graph 4y = 3x
 Solution: Select some
convenient x-values and
compute y-values.
• If x = 0, then
y = ¾ (0) = 0
• If x = 4, then
y = ¾ (4) = 3
• If x = −4, then
y = ¾ (−4) = −3
x
0
4
y
0
3
(x, y)
(0, 0)
(4, 3)
−4
−3 (4 , 3)
3
y x
4
 In Table Form,
Then Plotting
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Example  Application
 The cost c, in dollars, of shipping a
FedEx Priority Overnight package
weighing 1 lb or more a distance of
1001 to 1400 mi is given by
c = 2.8w + 21.05
• where w is the package’s weight in lbs
 Graph the equation and then use the
graph to estimate the cost of shipping a
10½ pound package
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FedEx Soln: c = 2.8w + 21.05
 Select values for w and then calculate c.
 c = 2.8w + 21.05
• If w = 2, then c = 2.8(2) + 21.05 = 26.65
• If w = 4, then c = 2.8(4) + 21.05 = 32.25
• If w = 8, then c = 2.8(8) + 21.05 = 43.45
 Tabulating
the Results:
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w
2
4
8
c
26.65
32.25
43.45
Bruce Mayer, PE
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 The cost of shipping an
10½ pound package is
about $51.00
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Mail cost (in dollars)
 Plot the points.
 To estimate costs for a
10½ pound package, we
locate the point on the
line that is above 10½
lbs and then find the
value on the c-axis that
corresponds to that point
$51
FedEx Soln: Graph Eqn
10 ½ pounds
Weight (in pounds)
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Finding Intercepts of Lines
 An “Intercept” is the point at which a line
or curve, crosses either the X or Y Axes
 A line with eqn Ax + By = C (A & B ≠ 0)
will cross BOTH the x-axis and y-axis
 The x-CoOrd of the point where the line
intersects the x-axis is called the
x-intercept
 The y-CoOrd of the point where the line
intersects the y-axis is called the
y-intercept
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Bruce Mayer, PE
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Example  Axes Intercepts
 For the graph shown
• a) find the coordinates
of any x-intercepts
• b) find the coordinates
of any y-intercepts
 Solution
• a) The x-intercepts are
(−2, 0) and (2, 0)
• b) The y-intercept is (0,−4)
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Bruce Mayer, PE
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Graph Ax + By = C Using Intercepts
1. Find the x-Intercept  Let y = 0,
then solve for x
2. Find the y-Intercept  Let x = 0,
then solve for y
3. Construct a CheckPoint using any
convenient value for x or y
4. Graph the Equation by drawing a line
thru the 3-points (i.e., connect the dots)
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Bruce Mayer, PE
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To FIND the Intercepts
 To find the y-intercept(s) of an
equation’s graph, replace x with 0
and solve for y.
 To find the x-intercept(s) of an
equation’s graph, replace y with 0
and solve for x.
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Bruce Mayer, PE
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Example  Find Intercepts
 Find the y-intercept and the x-intercept
of the graph of 5x + 2y = 10
 SOLUTION: To find the y-intercept, we
let x = 0 and solve for y
5 • 0 + 2y = 10
2y = 10
y=5
 Thus The y-intercept is (0, 5)
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Bruce Mayer, PE
[email protected] • MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt
Example  Find Intercepts cont.
 Find the y-intercept and the x-intercept
of the graph of 5x + 2y = 10
 SOLUTION: To find the x-intercept,
we let y = 0 and solve for x
5x + 2• 0 = 10
5x = 10
x=2
 Thus The x-intercept is (2, 0)
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Bruce Mayer, PE
[email protected] • MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt
Example  Graph w/ Intercepts
 Graph 5x + 2y = 10 using intercepts
 SOLUTION:
• We found the intercepts
in the previous example.
Before drawing the line,
we plot a third point as a
check. If we let x = 4, then
– 5 • 4 + 2y = 10
– 20 + 2y = 10
–
2y = −10
–
y=−5
5x + 2y = 10
y-intercept (0, 5)
x-intercept (2, 0)
Chk-Pt (4,-5)
• We plot Intercepts (0, 5) & (2, 0), and also (4 ,−5)
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Bruce Mayer, PE
[email protected] • MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt
Example  Graph w/ Intercepts
 Graph 3x − 4y = 8 using intercepts
 SOLUTION: To find the y-intercept, we
let x = 0. This amounts to ignoring the
x-term and then solving.
−4y = 8
y = −2
 Thus The y-intercept is (0, −2)
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Bruce Mayer, PE
[email protected] • MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt
Example  Graph w/ Intercepts
 Graph 3x – 4y = 8 using intercepts
 SOLUTION: To find the x-intercept, we
let y = 0. This amounts to ignoring the
y-term and then solving
3x = 8
x = 8/3
 Thus The x-intercept is (8/3, 0)
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt
Example  Graph w/ Intercepts
 Construct Graph for 3x – 4y = 8
• Find a third point.
If we let x = 4, then
–
–
3•4 – 4y = 8
12 – 4y = 8
–
–
–4y = –4
y=1
Chk-Pt Charlie
x-intercept
y-intercept
• We plot (0, −2),
3x  4y = 8
(8/3, 0), and (4, 1)
and Connect the Dots
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Bruce Mayer, PE
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Example  Graph y = 2
 SOLUTION: We regard the equation
y = 2 as the equivalent eqn: 0•x + y = 2.
• No matter what number we choose for x,
we find that y must equal 2.
y=2
Choose any number for x.
x
0
4
−4
y
2
2
2
(x, y)
(0, 2)
(4, 2)
(−4 , 2)
y must be 2.
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Bruce Mayer, PE
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Example  Graph y = 2
 Next plot the ordered pairs (0, 2), (4, 2) &
(−4, 2) and connect the points to obtain a
horizontal line.
 Any ordered pair
y=2
(0, 2)
of the form (x, 2)
(4, 2)
(4, 2)
is a solution, so
the line is parallel
to the x-axis with
y-intercept (0, 2)
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Bruce Mayer, PE
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Example  Graph x = −2
 SOLUTION: We regard the equation
x = −2 as x + 0•y = −2. We build a table
with all −2’s in the x-column.
x = −2
x must be 2.
x
−2
−2
−2
y
(x, y)
4 (−2, 4)
1 (−2, 1)
−4 (−2, −4)
Any number can be used for y.
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Bruce Mayer, PE
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Example  Graph x = −2
 When we plot the ordered pairs (−2,4),
(−2,1) & (−2, −4) and connect them, we
x = 2
obtain a vertical line
(2, 4)
 Any ordered pair
of the form (−2,y)
(2, 1)
is a solution. The
line is parallel to
the y-axis with
(2, 4)
x-intercept (−2,0)
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Bruce Mayer, PE
[email protected] • MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt
Linear Eqns of ONE Variable
 The Graph of y = b
is a Horizontal Line,
with y-intercept (0,b)
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 The Graph of x = a
is a Vertical Line,
with x-intercept (a,0)
Bruce Mayer, PE
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Example  Horiz or Vert Line
 Write an equation
for the graph
 SOLUTION: Note
that every point on
the horizontal line
passing through
(0,−3) has −3 as the
y-coordinate.
 Thus The equation
of the line is y = −3
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Bruce Mayer, PE
[email protected] • MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt
Example  Horiz or Vert Line
 Write an equation
for the graph
 SOLUTION: Note
that every point on
the vertical line
passing through
(4, 0) has 4 as the
x-coordinate.
 Thus The equation
of the line is x = 4
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Bruce Mayer, PE
[email protected] • MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt
SLOPE Defined
 The SLOPE, m, of the line
containing points (x1, y1) and (x2, y2)
is given by
Change in y
m
Change in x
rise y2  y1


run x2  x1
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Bruce Mayer, PE
[email protected] • MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt
 Graph the line
containing the points
(−4, 5) and (4, −1)
& find the slope, m
 SOLUTION
Change in y
m
Change in x
rise y2  y1


run x2  x1
m
 1  5  6

4   4 8
Chabot College Mathematics
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Change in y = −6
Example  Slope City
Change in x = 8
 Thus Slope
m = −3/4
Bruce Mayer, PE
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Example  ZERO Slope
 Find the slope of the
line y = 3
 SOLUTION: Find
Two Pts on the Line
(3, 3)
(2, 3)
• Then the Slope, m
rise
33
m

run 2   3
0
m 0
5
 A Horizontal Line has ZERO Slope
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Bruce Mayer, PE
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Example  UNdefined Slope
 Find the slope of
the line x = 2
 SOLUTION: Find
Two Pts on the Line
(2, 4)
• Then the Slope, m
rise 4   2
m

run
22
6
m   ??
0
(2, 2)
 A Vertical Line has an UNDEFINED Slope
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Bruce Mayer, PE
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Applications of Slope = Grade
 Some applications use slope to
measure the steepness.
 For example, numbers like 2%, 3%, and
6% are often used to represent the
grade of a road, a measure of a road’s
steepness.
• That is, a 3% grade means
that for every horizontal
distance of 100 ft, the road
rises or falls 3 ft.
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Bruce Mayer, PE
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Grade Example
 Find the slope
(or grade) of the
treadmill
rise 0.42 ft
m

 0.0764
run
5.5 ft
0.42 ft
 SOLUTION: Noting
the Rise & Run
5.5 ft
 In %-Grade for Treadmill
100%
m  Grade  0.0764 
 7.64%
1
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Bruce Mayer, PE
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Slope Symmetry
 We can Call
EITHER Point No.1
or No.2 and Get the
Same Slope
(−4,5) Pt1
 Example, LET
• (x1,y1) = (−4,5)
m
rise y2  y1

run x2  x1
 1  5  6
3
m


4   4 8
4
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(4,−1)
 Moving L→R
Bruce Mayer, PE
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Slope Symmetry cont
 Now LET
(−4,5)
• (x1,y1) = (4,−1)
m
m
rise y2  y1

run x2  x1
5   1
6
3


 4  4  8
4
(4,−1)
Pt1
 Moving R→L
 Thus
Chg in y y2  y1 y1  y2
m


Chg in x x2  x1 x1  x2
Chabot College Mathematics
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
y2  y1
y1  y2
or
x1  x2
x2  x1
Bruce Mayer, PE
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Slopes Summarized
 POSITIVE Slope
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 NEGATIVE Slope
Bruce Mayer, PE
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Slopes Summarized
 ZERO Slope
slope = 0
• Note that when a line is
horizontal the slope is 0
Chabot College Mathematics
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 UNDEFINED Slope
slope =
undefined
• Note that when the line is
vertical the slope is undefined
Bruce Mayer, PE
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WhiteBoard Work
 Problems From §2.4 Exercise Set
• 26 (PPT), 12, 24, 52, 56

More Lines
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Bruce Mayer, PE
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P2.4-26  Find Slope for Lines
 Recall
rise y2  y1
m

run x2  x1
rise 1  2
m1 

1
run 1  2
rise 2  2
m2 

2 3
run 2  3
rise 3  4
m3 

 2
run 3  2
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Bruce Mayer, PE
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All Done for Today
Some
Slope
Calcs
rise y2  y1 y
m


run x2  x1 x
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10
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file =XY_Plot_0211.xls
file =XY_Plot_0211.xls
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-10
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20x20 Grid
5
Chabot Mathematics
Appendix
r  s  r  s r  s 
2
2
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
–
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-07_sec_2-3a_Lines_by_Intercepts.ppt