Transcript Section 1.2 - El Camino College

Section 1.2
Graphing Linear Equations
Definition of Solution, Satisfy, and Solution Set
D e f i n i t i o n o f S o l u t i o n , S a t i s f y, a n d S o l u t i o n S e t
Consider the equation y  2 x  5. Let’s find y when
x  3.
y  2x  5
Original Equation.
y  2  3  5 Substitute 3 for x.
 65
Multiply before subtracting.
1
Subtract.
So, y  1 when x  3, which cab be represented by
the ordered pair  3,1 .
Section 1.2
Lehmann, Intermediate Algebra, 3ed
Slide 2
Definition of Solution, Satisfy, and Solution Set
D e f i n i t i o n o f S o l u t i o n , S a t i s f y, a n d S o l u t i o n S e t
Definition
For an ordered pair  a, b, we write the value of the
independent variable in the first (left) position and
the value of the dependent variable in the second
(right) position.
• The numbers a and b are called coordinates.
• For  3,1 ,the x-coordinate is 3 and the
y-coordinate is 1.
Section 1.2
Lehmann, Intermediate Algebra, 3ed
Slide 3
Definition of Solution, Satisfy, and Solution Set
D e f i n i t i o n o f S o l u t i o n , S a t i s f y, a n d S o l u t i o n S e t
The equation y  2 x  5 becomes a true
statement when we substitute 3 for xcoordinate and 1 for y-coordinate.
y  2x  5
Original Equation.
?
1  2  3  5 Substitute 3 for x and 1 for y.
?
1 1
true
Section 1.2
Lehmann, Intermediate Algebra, 3ed
Slide 4
Definition of Solution, Satisfy, and Solution Set
D e f i n i t i o n o f S o l u t i o n , S a t i s f y, a n d S o l u t i o n S e t
Definition
• An ordered pair  a, b  is a solution of an equation
in terms of x and y if the equation becomes a true
statement when a is substituted for x and b is
substituted for y.
• We say  a, b  satisfies the equation.
• The solution set of the equation is the set of all
solution of the equation.
Section 1.2
Lehmann, Intermediate Algebra, 3ed
Slide 5
Graphing an Equation
D e f i n i t i o n o f S o l u t i o n , S a t i s f y, a n d S o l u t i o n S e t
Example
Find five solutions to the
equation y  2 x  1, and
plot them in the
coordinate system (on the
right).
Section 1.2
Lehmann, Intermediate Algebra, 3ed
Slide 6
Graphing an Equation
D e f i n i t i o n o f S o l u t i o n , S a t i s f y, a n d S o l u t i o n S e t
Solution
We begin be arbitrarily choosing the values 0, 1, and
2 to substitute for x:
y  2  0   1
 0 1
1
y  2 1  1
y  2  2   1
 2  1
 1
 4  1
 3
Solution:  0,1 Solution: 1, 1 Solution:  2, 3 
The ordered pairs  2,5  and  1,3 are also solutions.
Section 1.2
Lehmann, Intermediate Algebra, 3ed
Slide 7
Graphing an Equation
D e f i n i t i o n o f S o l u t i o n , S a t i s f y, a n d S o l u t i o n S e t
Solution Continued
• Create a table
of solutions
x
-2
-1
0
1
2
Section 1.2
y
5
3
1
-1
-3
• Plot the solutions
• Points form a linear line.
Lehmann, Intermediate Algebra, 3ed
Slide 8
Graphing an Equation
D e f i n i t i o n o f S o l u t i o n , S a t i s f y, a n d S o l u t i o n S e t
• Every point on the
line is a solution to the
equation y  2 x  1
Section 1.2
Lehmann, Intermediate Algebra, 3ed
Slide 9
Graphing an Equation
D e f i n i t i o n o f S o l u t i o n , S a t i s f y, a n d S o l u t i o n S e t
• The point  3, 5  lies
on the line
• Should satisfy the
equations
• Whereas  2, 4  is not
on the line
• Thus should not
satisfy the equation
y  2 x  1
Section 1.2
Lehmann, Intermediate Algebra, 3ed
Slide 10
Graphing an Equation
D e f i n i t i o n o f S o l u t i o n , S a t i s f y, a n d S o l u t i o n S e t
y  2 x  1
Original Equation.
?
4  2  2   1 Substitute 2 for x and 4 for y.
?
4  3
false
• The  2, 4  is not a solution to the equation
Section 1.2
Lehmann, Intermediate Algebra, 3ed
Slide 11
Graphing an Equation
D e f i n i t i o n o f S o l u t i o n , S a t i s f y, a n d S o l u t i o n S e t
Calculator
Use ZDecimal on a graphing calculator.
• To enter y  2 x  1, press (–) 2 X,T,ϴ,n + 1.
The key – is used for subtraction, and the key
. (–) is used for negative numbers as well as
taking the opposite.
Section 1.2
Lehmann, Intermediate Algebra, 3ed
Slide 12
Definition: Graph
D e f i n i t i o n o f S o l u t i o n , S a t i s f y, a n d S o l u t i o n S e t
Definition
The graph of an equation in two variables is the set of
points that correspond to all solutions of the equation.
In the last example we found that the equation
y  2 x  1, is a line. Notice that the equation
y  2 x  1, is of the form y  mx  b (where
m  2 and b  1).
Section 1.2
Lehmann, Intermediate Algebra, 3ed
.
.
Slide 13
Graphs of Linear Equations
Graphs of Linear Equations
Equations of the form y  mx  b
If an equation can be put into the form
y  mx  b
where m and b are constants, then the graph of the
equation is a line.
Example
What is m and b for the equations
3
y  x  2, y  3x and y  3?
2
Section 1.2
Lehmann, Intermediate Algebra, 3ed
Slide 14
Graphing Linear Equations
Graphs of Linear Equations
Definition
3
3
y  x  2 is of the form y  mx  b: m  and b  2
2
2
y  2 x is of the form y  mx  b because we write the
equation as y  2 x  0 (so m  2 and b  0).
y  3 is of the form y  mx  bbecause we write the
equation as y  0 x  3 (so m  0 and b  3).
Example
Sketch the graph of the equation 30 x  6 y  5  0.
Lehmann, Intermediate Algebra, 3ed
Graphing Linear Equations
Graphs of Linear Equations
Solution
First we solve for y
30 x  6 y  12  0
30 x  6 y  12
Original Equation.
Subtract 12 from both sides.
 6 y  30 x  12 Subtract 30x from both sides.
6
30 12
Divide both sides by 6.
y
x
6
6
6
y  5 x  2
Simplify.
Section 1.2
Lehmann, Intermediate Algebra, 3ed
Slide 16
Graphing Linear Equations
Graphs of Linear Equations
Solution Continued
• . y  5 x  2is of the form y  mx  b
• The graph of the equation is a line
• Find 2 points of the line
• Plot the two points
• Sketch the line
• Find a third point
•Verify that the third point lies on the line
Section 1.2
Lehmann, Intermediate Algebra, 3ed
Slide 17
Graphing Linear Equations
Graphs of Linear Equations
Solution Continued
Table of solutions
x
y
0 2  0   3  3
1 2 1  3  1
2 2  2  3  1
Section 1.2
Lehmann, Intermediate Algebra, 3ed
Slide 18
Graphing Linear Equations
Graphs of Linear Equations
Graphing Calculator
• Enter 5x  2 for y1.
• Use Zstandard
followed by Zsquare.
• The graph is correct
assuming that y was
isolated correctly.
Section 1.2
Lehmann, Intermediate Algebra, 3ed
Slide 19
Using the Distributive Law to Help Graph a Linear Equation
Graphs of Linear Equations
Example
Sketch the graph of 3  2 y  5   2 x  3  8 x.
Solution
• Use the distributive property on the left-hand
side.
• Collect like terms.
• Isolate y.
Section 1.2
Lehmann, Intermediate Algebra, 3ed
Slide 20
Using the Distributive Law to Help Graph a Linear Equation
Graphs of Linear Equations
Solution Continued
3 2 y  5  2 x  3  8x
Original equation
Distributive property
6 y  15  6 x  3
6 y  15  15  6 x  3  15 Add 15 to both sides.
6
6 12
Divide both sides by 6.
y  x
6
6
6
Simplify.
y  x  2
Section 1.2
Lehmann, Intermediate Algebra, 3ed
Slide 21
Using the Distributive Law to Help Graph a Linear Equation
Graphs of Linear Equations
Solution Continued
Table of solutions
x
y
0  0  2  2
1  1  2  1
2   2  2  0
Section 1.2
Lehmann, Intermediate Algebra, 3ed
Slide 22
Graphing an Equation That Contains Fractions
Graphs of Linear Equations
Example
1
Sketch a graph of y  x  1.
2
Solution
• Avoid fraction values for y
• Use even values for x
Section 1.2
Lehmann, Intermediate Algebra, 3ed
Slide 23
Graphing an Equation That Contains Fractions
Graphs of Linear Equations
Solution Continued
Table of solutions
x
0
2
4
Section 1.2
y
1
 0   1  1
2
1
 2 1  0
2
1
 4 1  1
2
Lehmann, Intermediate Algebra, 3ed
Slide 24
Graphing an Equation That Contains Fractions
Graphs of Linear Equations
Graphing Calculator
Use Zdecimal to
verify the solution.
Section 1.2
Lehmann, Intermediate Algebra, 3ed
Slide 25
Property
Finding Intercepts of a Graph
Sometimes we find intercepts to graph a line.
• x-intercept is on the y-axis, so y = 0
• y-intercepts in on the x-axis, so x = 0
Directions
• For an equation containing the variables x and y
• x-intercept: Substitute y = 0 and solve for x
• y-intercept: Substitute x = 0 and solve for y
Section 1.2
Lehmann, Intermediate Algebra, 3ed
Slide 26
Using Intercepts to Sketch a Graph
Finding Intercepts of a Graph
Example
Use intercepts to sketch a graph of y  2 x  4.
Solution
x-intercept: Set y = 0.
0  2 x  4
Substitute 0 for y.
4   2 x
Subtract both sides by 4.
4 2
Divide both sides by - 2.

x
2 2
Simplify.
2x
Section 1.2
Lehmann, Intermediate Algebra, 3ed
Slide 27
Using Intercepts to Sketch a Graph
Finding Intercepts of a Graph
Solution Continued
y-intercept: Set x = 0.
y  2  0   4 Set x = 0.
Simplify.
y4
So, the x-intercept is
 2,0  and y-intercept is
 0, 4  .
Section 1.2
Lehmann, Intermediate Algebra, 3ed
Slide 28
Using Intercepts to Sketch a Graph
Finding Intercepts of a Graph
Graphing Calculator
Use ZStandard
followed by ZSquare.
Use TRACE to verify
the y-intercept.
Use “zero” to verify the
x-intercept.
Section 1.2
Lehmann, Intermediate Algebra, 3ed
Slide 29
Graphing a Vertical Line
Finding Intercepts of a Graph
Example
Graph the equation of x  3.
Solution
x
y
3
3
3
3
3
Section 1.2
5
3
1
-1
-3
Notice that the
values of x must be
3, but y can have
any value. Some
solutions are listed
to the left.
Lehmann, Intermediate Algebra, 3ed
Slide 30
Graphing a Horizontal Line
Ve r t i c a l a n d H o r i z o n t a l L i n e s
Example
Graph the equation of y  5.
Solution
x
y
–2
–1
0
1
2
Section 1.2
–5
–5
–5
–5
–5
Notice that the
values of y must be
–5, but x can have
any value. Some
solutions are listed
to the left.
Lehmann, Intermediate Algebra, 3ed
Slide 31
Graphing a Horizontal Line
Ve r t i c a l a n d H o r i z o n t a l L i n e s
Graphing Calculator
Use ZStandard to verify
the graph.
Section 1.2
Lehmann, Intermediate Algebra, 3ed
Slide 32
Vertical and Horizontal Line Property
Ve r t i c a l a n d H o r i z o n t a l L i n e s
Property
If a and b are constants:
• An equation that can be put
into the form x  a. has a
vertical line as its graph
• An equation that can be put
into the form y  b.has a
horizontal line as its graph
Section 1.2
Lehmann, Intermediate Algebra, 3ed
Slide 33
Vertical and Horizontal Line Property
Ve r t i c a l a n d H o r i z o n t a l L i n e s
Property
In an equation can be put into either form
y  mx  b
or
xa
where m, a, and b are constants, then the graph of
the equation is a line. We call such an equation a
linear equation in two variables.
Section 1.2
Lehmann, Intermediate Algebra, 3ed
Slide 34