Section 1 - Montgomery Collegefaculty.montgomerycollege.edu/.../2-SK-Graph-Lines.pdfGraphing Linear...

Graphing Linear Equations

Section 1.2

Lehmann, Intermediate Algebra, 4ed Section 1.2

Consider the equation . Let’s find y when

So, when , which cab be represented by the ordered pair

Slide 2

Definition of Solution, Satisfy, and Solution Set

2 5y x= −3.x =

( )2 52 3 56 51

y xy= −

= −

= −=

Original Equation.

Substitute 3 for x.

Multiply before subtracting.

Subtract.

1y = 3x =( )3,1 .

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


For an ordered pair , 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 the x-coordinate is 3 and the y-coordinate is 1.

Slide 3


( ),a b

( )3,1 ,

Definition



The equation becomes a true statement when we substitute 3 for x-coordinate and 1 for y-coordinate.

Slide 4


2 5y x= −

( )?

?

2 5

1 2 3 5

1 1true

y x= −

= −

=

Original Equation.

Substitute 3 for x and 1 for y.



• An ordered pair 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 satisfies the equation.

• The solution set of the equation is the set of all solution of the equation.

Slide 5


( ),a b

( ),a b

Definition



Find five solutions to the equation and plot them in the coordinate system (on the right).

Slide 6

Graphing an Equation

2 1,y x= − +

Example



We begin be arbitrarily choosing the values 0, 1, and 2 to substitute for x:

The ordered pairs and are also solutions.

Slide 7


( )

( )

( )

( )

( )

( )

2 0 1 2 1 1 2 2 10 1 2 1 4 11 1 3

Solution: 0,1 Solution: 1, 1 Solution: 2, 3

y y y= − + = − + = − +

= + = − + = − += = − = −

− −

( )2,5− ( )1,3−

Solution



• Create a table of solutions

x y -2 5 -1 3 0 1 1 -1 2 -3

Slide 8


• Plot the solutions

• Points form a linear line.

Solution Continued



• Every point on the line is a solution to the equation

Slide 9


2 1y x= − +



• The point lies on the line

• Should satisfy the equations

• Whereas is not on the line

• Thus should not satisfy the equation

Slide 10


( )3, 5−

( )2,4

2 1y x= − +


Lehmann, Intermediate Algebra, 4ed Section 1.2 Slide 11


( )?

?

2 1

4 2 2 1

4 3false

y x= − +

=− +

=−

Original Equation.

Substitute 2 for x and 4 for y.

• The is not a solution to the equation

( )2,4



Use ZDecimal on a graphing calculator.

• To enter 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.

Slide 12


2 1,y x= − +

Calculator



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 . is a line. Notice that the equation . is of the form (where and ).

Slide 13

Definition: Graph

2 1,y x= − +2 1,y x= − +

y mx b= +2m = − 1b =

Definition


Lehmann, Intermediate Algebra, 4ed

Equations of the form

Section 1.2

If an equation can be put into the form where m and b are constants, then the graph of the equation is a line.

Slide 14

Graphs of Linear Equations

y mx b= +

y mx b= +

What is m and b for the equations 3 2, 3 and 3?2

y x y x y= − = − =

Example

G r a p h s o f L i n e a r E q u a t i o n s


is of the form : and

is of the form because we write the equation as (so and ). is of the form because we write the equation as (so and ). Sketch the graph of the equation


3 22

y x= − y mx b= + 2b = −32

m =

2y x= − y mx b= +2 0y x= − + 0b =2m = −

3b =3y =

0 3y x= + 0m =y mx b= +

30 6 5 0.x y− + =

Definition

Example



First we solve for y

Slide 16


30 6 12 030 6 12

6 30 126 30 12

6 6 65 2

x yx y

y x

y x

y x

− + =− = −− = − −− −

= −

= − −

Original Equation.

Subtract 12 from both sides.

Subtract 30x from both sides.

Simplify.

Divide both sides by 6.

Solution



• . is of the form • 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

Slide 17


5 2y x= − − y mx b= +


Solution Continued



Table of solutions

x y

( )( )( )

2 0 3 301 2 1 3 12 2 2 3 1

− = −

− = −

− =

Solution Continued



• Enter for y1.

• Use Zstandard followed by Zsquare.

• The graph is correct assuming that y was isolated correctly.

Slide 19


5 2x− −Graphing Calculator



Sketch the graph of • Use the distributive property on the left-hand

side. • Collect like terms. • Isolate y.

Slide 20

Using the Distributive Law to Help Graph a Linear Equation

( )3 2 5 2 3 8 .y x x− = − −Example

Solution




( )3 2 5 2 3 86 15 6 3

6 15 15 6 3 156 6 126 6 6

2

y x xy x

y x

y x

y x

− = − −

− = − −− + = − − +

−= +

= − +

Original equation

Distributive property

Add 15 to both sides.

Divide both sides by 6.

Simplify.

Solution Continued




Table of solutions

x y

( )( )( )

0 2 201 1 2 12 2 2 0

− + =

− + =

− + =

Solution Continued



Sketch a graph of • Avoid fraction values for y • Use even values for x

Slide 23

Graphing an Equation That Contains Fractions

1 1.2

y x= −Example

Solution



Table of solutions x y 0 2 4

Section 1.2 Slide 24


( )

( )

( )

1 0 1 121 2 1 021 4 1 12

− = −

− =

− =

Solution Continued



Use Zdecimal to verify the solution.

Slide 25


Graphing Calculator



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

Slide 26

Property

• 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

Directions

F i n d i n g I n t e r c e p t s o f a G r a p h


Use intercepts to sketch a graph of

Slide 27

Using Intercepts to Sketch a Graph

2 4.y x= − +

x-intercept: Set y = 0.

0 2 44 24 22 22

xx

x

x

= − +− = −− −=− −

= Simplify.

Substitute 0 for y.

Subtract both sides by 4.

Divide both sides by - 2.

Example

Solution



y-intercept: Set x = 0.

So, the x-intercept is and y-intercept is

Slide 28


( )2 0 44

yy= − +

=

Set x=0.

Simplify.

( )0,4 .( )2,0

Solution Continued



Use ZStandard followed by ZSquare.

Use “zero” to verify the x-intercept.

Slide 29


Use TRACE to verify the y-intercept.

Graphing Calculator



Graph the equation of

Slide 30

Graphing a Vertical Line

3.x =

x y 3 5 3 3 3 1 3 -1 3 -3

Notice that the values of x must be 3, but y can have any value. Some solutions are listed to the left.

Example

Solution



Graph the equation of

Slide 31

Graphing a Horizontal Line

5.y = −

x y –2 –5 –1 –5 0 –5 1 –5 2 –5

Notice that the values of y must be –5, but x can have any value. Some solutions are listed to the left.

Example

Solution

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


Use ZStandard to verify the graph.

Slide 32

Graphing a Horizontal Line

Graphing Calculator



If a and b are constants: • An equation that can be put

into the form . has a vertical line as its graph

• An equation that can be put into the form .has a horizontal line as its graph

Slide 33

Vertical and Horizontal Line Property

y b=

x a=

Property



In an equation can be put into either form 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.

Slide 34

Vertical and Horizontal Line Property

ory mx b x a= + =

Property


Section 1 - Montgomery Collegefaculty.montgomerycollege.edu/.../2-SK-Graph-Lines.pdfGraphing Linear...

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