Transcript Section 1.4 - El Camino College

Section 1.4
Meaning of Slope for Equations,
Graphs, and Tables
Finding Slope from a Linear Equation
Finding Slope from a Linear Equation
Example
Find the slope of the line y  2 x  1.
Solution
Create a table
using x = 1, 2, 3.
Then sketch the
graph.
rise 2
m
 2
run 1
Section 1.4
x y
0 1
1 3
2 5
3 7
Lehmann, Intermediate Algebra, 3ed
Slide 2
Finding Slope from a Linear Equation
Finding Slope from a Linear Equation
Observations
Note the following three observations about the slope
of the line y  2 x  1.
1. The coefficient of x is 2, which is
the slope.
2. If the run is 1, then the rise is 2.
3. As the value of x increases by 1,
the value of y increases by 2.
Section 1.4
Lehmann, Intermediate Algebra, 3ed
Slide 3
Finding Slope from a Linear Equation
Finding Slope from a Linear Equation
Example
Find the slope of the line y  3x  8.
Solution
Create a table
using x = 1, 2, 3.
Then sketch the
graph.
rise 3
m

 3
run 1
Section 1.4
x y
0 8
1 5
2 2
3 –1
Lehmann, Intermediate Algebra, 3ed
Slide 4
Finding Slope from a Linear Equation
Finding Slope from a Linear Equation
Property
For a linear equation of the form y  mx  b, m is the
slope of the line.
Example
5
Are the lines y  x  3 and 12 y  10 x  5 parallel,
6
perpendicular, or neither?
Section 1.4
Lehmann, Intermediate Algebra, 3ed
Slide 5
Finding Slope from a Linear Equation
Finding Slope from a Linear Equation
Property
5
5
• For the line y  x  3 the slope is
6
6
• For the other equation we solve for y:
12 y  10 x  5
12 y  10 x  10 x  5  10 x
12 y  10 x  5
12 y 10
5

x
12 12
12
5
5
y  x
6
12
Section 1.4
Original Equation
Add 10x to both sides.
Combine & rearrange terms
Divide both sides by 12.
Simplify.
Lehmann, Intermediate Algebra, 3ed
Slide 6
Finding Slope from a Linear Equation
Finding Slope from a Linear Equation
Solution Continued
5
5
5
• For the line y  x  the slope is
12
6 12
• Since the slopes are the same for both
equations, the lines are parallel
Graphing Calculator
We use ZStandard followed by
ZSquare to draw the line in the same
coordinate system.
Section 1.4
Lehmann, Intermediate Algebra, 3ed
Slide 7
Vertical Change Property
Ver t i ca l C h a n ge P ro p er t y
Property
For the line y  mx  b, if the run is 1, then the rise is
m.
Vertical Change property Vertical Change property
for a positive slope.
for a negative slope.
Section 1.4
Lehmann, Intermediate Algebra, 3ed
Slide 8
Finding the y-intercept of a Linear Line
Finding the y-Intercept of
linear Equation
Sketching Equations:
• It’s helpful to know the y-intercept.
• y-intercept has a x-value of 0.
• Substitute x = 0 gives y  m  0   b  b
Property
For a linear equation of the form y  mx  b, the yintercept is (0, b).
Section 1.4
Lehmann, Intermediate Algebra, 3ed
Slide 9
Finding the y-intercept of a Linear Line
Finding the y-Intercept of
linear Equation
Example
5
What is the y-intercept of y  x  3?
6
Solution
• b is equal to 3, so the y-intercept is (0, 3)
Definition
If an equation of the form y  mx  b, we say that it is
in slope-intercept form.
Section 1.4
Lehmann, Intermediate Algebra, 3ed
Slide 10
Graphing Linear Equations
Graphing Linear Equations
Example
Sketch the graph of y = 3x – 1.
Solution
• The y-intercept is (0, –1) and the slope is
3 rise
3 
1 run
To graph:
1. Plot the y-intercept, (0, 1).
(continued)
Section 1.4
Lehmann, Intermediate Algebra, 3ed
Slide 11
Graphing Linear Equations
Graphing Linear Equations
Solution Continued
2. From (0, –1), look 1 unit to
the right and 3 units up to
plot a second point, which
we see by inspection is (1,
2).
3. Sketch the line that
contains these two points.
Section 1.4
Lehmann, Intermediate Algebra, 3ed
Slide 12
Graphing Linear Equations
Graphing Linear Equations
Guidelines
To sketch the graph of a linear equation of the form
y  mx  b
1.Plot the y-intercept (0, b).
rise
2.Use m =
to plot a second point.
run
3.Sketch the line that passes through the two plotted
points.
Section 1.4
Lehmann, Intermediate Algebra, 3ed
Slide 13
Graphing Linear Equations
Graphing Linear Equations
Example
Sketch the graph of 2x + 3y = 6.
Solution
First we rewrite into slope-intercept form:
2x  3y  6
Original Equation
2 x  3 y  2 x  6  2 x Subtract 2x from both sides.
3 y  2 x  6 Combine & rearrange terms
3 y 2
6 Divide both sides by 3.
 x
3
3
3
Section 1.4
Lehmann, Intermediate Algebra, 3ed
Slide 14
Graphing Linear Equations
Graphing Linear Equations
Solution Continued
2
-a a
y   x  2 Simplfy: =3
b
b
2 2 rise
y-intercept: (0, 2)
Slope:  

3 3 run
1. Plot the y-intercept, (0, 2).
2. From the point (0, 2), look 3 units to the right
and 2 units down to plot a second point, which
we see by inspection is (3, 0).
Section 1.4
Lehmann, Intermediate Algebra, 3ed
Slide 15
Graphing Linear Equations
Graphing Linear Equations
Solution Continued
3. Then sketch the line that contains these two
points. We can verify our result by checking
that both (0, 2) and (3, 0) are solutions.
Section 1.4
Lehmann, Intermediate Algebra, 3ed
Slide 16
Graphing Linear Equations
Graphing Linear Equations
Example
1. Determine the slope and the y-intercept of
ax + by =c, where a, b, and c are constants and
b is nonzero.
2. Find the slope and the y-intercept of the graph
of 3x + 7y = 5.
Solution
First we rewrite into slope-intercept form:
Section 1.4
Lehmann, Intermediate Algebra, 3ed
Slide 17
Graphing Linear Equations
Graphing Linear Equations
Solution Continued
ax  by  c
ax  by  ax  c  ax
by  ax  c
Original equation
Subtract ax from both sides.
Combine and rearrange terms.
by a
c Divide both sides by b.

x
b
b
b
a
c Simplfy: -a = - a
y  x
b
b
b
b
Slope is  ba and the y-intercept is  0, bc  .
Section 1.4
Lehmann, Intermediate Algebra, 3ed
Slide 18
Graphing Linear Equations
Graphing Linear Equations
Solution Continued
Given that ax + by = c in slope-intercept form
c.
is a
y   x  , then given 3x + 7y = 5, we substitute .
b
b
3 for a, 7 for b and 5 for c. Thus, the slope,
.
b
3
   and the y-intercept,
a
7
Section 1.4
 0, c    0, 5  .

 

 b  7
Lehmann, Intermediate Algebra, 3ed
Slide 19
Slope Addition Property
Slope Addition Property
Example
For the following sets, is there a line that passes
through them? If so, find the slope of that line.
Solution
• Value of x increases by 1.
• Value of y changes by –3.
•The slope is –3.
Section 1.4
Lehmann, Intermediate Algebra, 3ed
Slide 20
Slope Addition Property
Slope Addition Property
Solution Continued
Set 2
• Value of x increases by 1.
• Value of y changes by 5.
So, the slope is 5.
Set 3
• Value of x increases by 1.
• Value of y does not change by the same
value. Hence, not a line.
Section 1.4
Lehmann, Intermediate Algebra, 3ed
Slide 21