2­2 Day 1 Linear Equations
September 09, 2008
2.2 Linear Equations
day one
Objective: To graph linear equations.
Jul 2­6:29 PM
1
2­2 Day 1 Linear Equations
September 09, 2008
Check skills you'll need
Evaluate each expression for x = ­2
2
1. x + 7
3
3. 3x + 1
1
4. x ­ 8
2
3
2. x ­ 2
5
Jul 2­6:29 PM
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2­2 Day 1 Linear Equations
September 09, 2008
A funtion whose graph is a line is a Linear Function.
You can represent a linear funtion with a Linear Equation.
Example: y = 3x + 2
The solution of a linear equation is any ordered pair (x, y) that makes the equation true.
Notice that y depends on the value of x, so y is called the Dependent Variable and the x is called the Indepenent Variable.
Jul 2­6:29 PM
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2­2 Day 1 Linear Equations
September 09, 2008
Graphing a Linear Equation
2
Graph the equation y = x + 3
3
Choose two values for x
Plug in your two x values
Find your two y values
Write your two solutions as an ordered pair.
2
x x+3
3
y
(x,y)
2
­3 (­3)+3
3
1
(­3,1)
2
3 (3)+3
3
5
(3,5)
Notice that there are an infite number of solutions.
Jul 2­6:29 PM
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2­2 Day 1 Linear Equations
September 09, 2008
The y­intercept of a line is the point at which the line crosses the y­axis.
The x­intercept of a line is the point at which the line crosses the x­axis.
The y­intercept always has a 0 for the x­coordinate.
You can write the y­intercept as an ordered pair ( 0, b)
or "y­intercept is b"
(0,b) y­intercept
(a,0) x­intercept
You can write the x­intercept as an ordered pair ( a, 0)
or "x­intercept is a"
The x­intercept always has a 0 for the y­coordinate.
Jul 2­6:29 PM
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2­2 Day 1 Linear Equations
September 09, 2008
The Standard form of a Linear Equation is Ax + By = C, where A, B, are not both zero. It is easiest to graph using the x and y interceps.
Let,s graph the line 3x +2y = 120
using the x and y intercepts.
The x­intercept always has a 1
0 for the y­coordinate.
So we have the two intercepts
(40,0) (0,60)
lets graph them
3x +2y = 120
3x +2(0) = 120
3x = 120
x = 40
The y­intercept always has a 0 for the x­coordinate.
graph here!
60
3x +2y = 120
3(0) +2y = 120
2y = 120
y = 60
10
10
40
Jul 2­6:29 PM
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2­2 Day 1 Linear Equations
September 09, 2008
Find the slope of the line through the points (3,2) and (­9,6).
Slope =
6 ­ 2
­9 ­ 3
y2­y1
x2­x1
Subsitute (3,2) for (x1,y1) and (­9,6) for (x2,y2)
4
­12
­
1
3
Subtract
Simplify
The slope of the line is ­13
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2­2 Day 1 Linear Equations
September 09, 2008
HOMEWORK
pg 67
#1­ 19 and 54­59
Jul 2­6:29 PM
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2­2 Day 1 Linear Equations
September 09, 2008
Jul 2­6:29 PM
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