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Algebra 2
Chapter 2 Notes
Linear Equations and Functions
1
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A RELATION is a mapping or pairing of input values with output values. The set of input values in the DOMAIN and the set of output values is the RANGE. A relation is a FUNCTION if there is exactly one output for each input. It is not a function if at lest one input has more than one output.
( x , y ) = (domain , range ) = ( input , output) = (independent , dependent)
Not a function Yes a function
- 3 3 - 3 3
1 –2 1 1
1 2
4 4 4 –2
A relation is a function if and only if no vertical line intersects the graph of the relationship at more than one point.
NOT a RELATION YES a RELATION 2
Functions and their Graphs 2.1
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Y axis
Quadrant I
( + , + )
Quadrant II
( – , + )
Quadrant III
( – , – )
Quadrant IV
( + , – )
X axis
( 0 , 0 )
Ordered pairs in form of ( x , y )
Coordinate Plane
x coordinate is firsty coordinate is second 3
2.1
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Graphing equations in 2 variables
1. Construct a table of values
2. Graph enough solutions to recognize a pattern
3. Connect the points with a line or a curve
Graph the function: y = x + 1
y = m x + b LINEAR FUNCTION
f (x) = m x + b FUNCTION NOTATION
x y = x + 1 y
- 2 y = -2 + 1 - 1
- 1 y = -1 + 1 0
0 y = 0 + 1 1
1 y = 1 + 1 2
2 y = 2 + 1 3
y
x
4
2.1Functions and their Graphs
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Are these functions linear?. Evaluate when x = – 2
a) f ( x ) = – x2 – 3 x + 5 Not a function because x is to the 2nd power
f ( – 2 ) = – ( – 2 ) 2 – 3 ( – 2 ) + 5
f ( – 2 ) = 7
b) g ( x ) = 2 x + 6 Yes a function because x is to the 1st power
g ( – 2 ) = 2 ( – 2 ) + 6
g (– 2 ) = 2
5
Functions and their Graphs 2.1
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SLOPE of a non-vertical line is the ratio of a vertical change (RISE) to a horizontal change (RUN).
Slope of a line:m = y2 – y1 = RISE
x2 – x1 = RUN
(differences in y values)(differences in x values)
•
y2 – y1
RISE
x2 – x1
RUNx
y
( x1 , y1 )
( x2 , y2 )•
6
Slope and Rate of Change 2.1
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CLASSIFICATION OF LINES BY SLOPE
A line with a + slope rises from left to right [ m > 0 ]
A line with a – slope falls from left to right [ m < 0 ]
A line with a slope of 0 is horizontal [ m = 0 ]
A line with an undefined slope is vertical [ m = undefined, no slope ]
Positive Slope
Negative Slope
0 Slope
No Slope 7
Slope and Rate of Change 2.2
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SLOPES OF PARALLEL & PERPENDICULAR LINES
PARALLEL LINES: the lines are parallel if and only if they have the SAME SLOPE.
m1 = m2
PERPENDICULAR LINES: the lines are perpendicular if and only if their SLOPES are NEGATIVE RECIPROCALS.
m1 = –1 m2
m1 m2 = – 1
or
8
Slope and Rate of Change 2.2
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Ex 1:
Find the slope of a line passing through ( – 3, 5 ) and ( 2, 1 )
Let ( x1, y1 ) = ( – 3, 5 ) and ( x2, y2 ) = ( 2, 1 )
y
x
( 2, 1 )
( – 3, 5 )
•
•
5
– 4
Slope of a line:m = y2 – y1 = RISE (differences in y values)
x2 – x1 = RUN (differences in x values)
m = 5 – 1 = 4 – 3 – 2 – 5
m = 1 – 5 = – 4 2 + 3 5
OR
9
Slope and Rate of Change 2.2
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Example 2: Without graphing tells if slope rises, falls, horizontal or vertical:
a) ( 3, – 4 ) and ( 1, – 6 ) : m = – 6 – ( – 4 ) = - 2 = 1 m > 0, rises
1 – 3 - 2
b) ( 2, 2 ) and ( – 1, 5 ) : m = 5 – ( – 1 ) = 6 = undefined m = no slope
2 – 2 0
For 2 lines with + slopes, the line with > slope is steeper
For 2 lines with – slopes, the line with slope of > absolute value is steeper:
y
x
m = 1
m = 1/2
m = 3m = - 3m = - 1
m = - 1/2
10
Slope and Rate of Change 2.2
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Classifying Perpendicular Lines
Line 1 through ( – 3 , 3 ) and ( 3 , – 1 )
m 1 = – 1 – 3 = – 4 = – 2
3 – ( - 3) 6 3
Line 2 through ( - 2 , - 3 ) and ( 2 , 3 )
m 2 = 3 – ( – 3 ) = 6 = 3
2 – ( – 2 ) 4 2Because m1 m∙ 2 = – 2 3 ∙ = – 1 3 2are negative reciprocals of each other, the lines are perpendicular.
• •
•
•
( –3 , 3 )
( 2, 3 )
( 3 , – 1 )
( –2 , – 3 )
L1
L2
x
y
11
Classifying Lines Using Slopes 2.2
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Classifying Parallel Lines
Line 1 through ( – 3 , 3 ) and ( 3 , – 1 )
m 1 = 4 – 1 = 3 = 1
3 – ( - 3) 6 2
Line 2 through ( - 2 , - 3 ) and ( 2 , 3 )
m 2 = 1 – ( – 3 ) = 4 = 1
4 – ( – 4 ) 8 2Because m1 = m2 and the lines are different,then the lines are parallel.
•
•
•
•( -3 , 1 )
( 3, 4 )
( 4 , 1 )
( -4 , - 3 )
L1
L2
x
y
12
Classifying Lines Using Slopes 2.2
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Slope Intercept Form of a linear equation is y = m x + b, where m is slope and b is y-intercept
Graphing Equations in slope-intercept form:1. Write the equation in slope-intercept form by solving for y2. Find y-intercept, then plot the point where line crosses the y-axis3. Find the slope and use it to plot a second point on the line.4. Draw a line through the 2 points.
Example 1: Graphing with the slope-intercept
Graph: y = 3 x − 2 4
1. Already in slope-intercept form2. y -intercept is – 2, plot point ( 0 , – 2 )
where the line crosses the y-axis3. Slope is ¾ , so plot 4 units to right,
3 units up, point is ( 4 , 1 )4. Draw a line through the 2 points •
• x
y
3
4( 0 , – 2 )
( 4 , 1 )
13
Quick Graphs of Linear Equations 2.3
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Standard Form of a linear equation is ax + by = c
x-intercept of a line is the x-coordinate of the point where the line intersects the x-axis
Graphing Equations in standard form1. Write the equation in standard form2. Find x-intercept, by letting y = 0, solve for x, plot the point where x crosses the x-axis3. Find y-intercept, by letting x = 0, solve for y, plot the point where y crosses the y-axis4. Draw a line through the 2 points.
Method 1 by using Standard Form: Graph 2 x + 3 y = 12
1. Already using standard form: 2 x2. Let y = 0 2 x + 3(0) = 12
2 x = 12 x = 6
x-intercept is at ( 6 , 0 )
3. Let x = 0 2 (0) + 3 y = 12 3 y = 12 y = 4
y-intercept is at ( 0 , 4 )
( 0 , 4 )
( 6 , 0 )
•
•
2 x + 3 y = 12
14
Quick Graphs of Linear Equations 2.3
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Method 2 by using Slope-Intercept Form: Graph 2 x + 3 y = 12
1. Change from standard form to slope-intercept form: 2 x + 3 y = 12
-2 x - 2 x 3 y = – 2 x + 12 3 3
y = – 2 x + 4 3
2. Identify , plot y-intercept3. Use slope to plot other point4. Draw a line through the points
( 0 , 4 )
( 3 , 2 )
•
•
3
2
2 x + 3 y = 12y-intercept
slope
Horizontal lines --- graph of y = c is a horizontal line through, (0 , c)
Vertical lines --- graph of x = c is a vertical line through (c, 0)
Graph y = 3 and x = -2
y = 3
x = 3
•
•
( 0 , 3 )
( - 2 , 0 ) x
y
15
Quick Graphs of Linear Equations 2.3
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Slope-Intercept Form: y = m x + b
Point-Intercept Form: y – y1 = m ( x – x1 )
Two Points: m = y2 – y1
x2 – x1
* Every non-vertical line has only one slope and one y-intercept
m = slopeb = y-intercept
• Write an equation of line shown• From graph you can see the slope, m = 3
2 • From graph you can see that it intersects at ( 0 ,
−1 ) , so the y-intercepts is b = −1 x
y
( 0 , −1 )
( 2 , 2 )•
• 2
3Equation of the Line is:
y = 3 x − 1 2
16
Writing Equations of Lines 2.4
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m ( x1 , y1 ) Writing equations of the line given slope of − 1 and a point, ( 2 , 3 )
2y – y1 = m ( x – x1 )
y – 3 = – 1 ( x – 2 ) 2
y – 3 = – 1 x + 1 2
+ 3 + 3
y = – 1 x + 4 2
Equation of the Line
17
Writing Equations of Lines 2.4
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Example 3a: Equation of a line that passes through ( 3 , 2 ) that is perpendicular to the line: y = – 3 x + 2
Perpendicular means m2 = – 1 m1
y – y1 = m2 ( x – x1)
y – 2 = 1 ( x – 3) 3y – 2 = 1 x – 1 3 + 2 + 2y = 1 x + 1 3
Example 3b: Equation of a line that passes through ( 3 , 2 ) that is parallel to the line: y = – 3 x + 2
Parallel means m2 = m1 = – 3 and ( x1 , y1 ) = ( 3 , 2 )
y – y1 = m2 ( x – x1)
y – 2 = – 3 ( x – 3) y – 2 = – 3 x + 9 + 2 + 2y = – 3 x + 11
18
Writing Equations of Perpendicular and Parallel Lines
2.4
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Writing Equations given 2 points where ( x1 , y1 ) = ( – 2 , – 1 )
( x2 , y2 ) = ( 3 , 4 )
m = y2 – y1 = 4 – ( – 1 ) = 5 = 1
x2 – x1 3 – ( – 2 ) 5
y – y1 = m ( x – x1)
y – ( – 1 ) = 1 [ x – ( – 2 ) ]
y + 1 = 1 ( x + 2 )
y = x + 1
19
Writing Equations of Lines 2.4
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Writing the Equation of the Line
The form you use depends on the information you have
INFO PROVIDED Name of FORM to be used FORM
Slope = 2y - intercept = – 7 Slope – Intercept Form y = m x + b
y = 2 x – 7
Point ( 2, 3 ) m = 4
Point – Slope FormAlso known as
Point – Intercept Form
( x – x 1 ) m = ( y – y 1 )( x – 2 ) 4 = ( y – 3)
Point ( 2, 3 ) Point (– 1, 4 )
Use m = y – y 1
x – x 1 Then use:
Point – Slope Form
m = 4 – 3 = 1– 1 – (– 2 )
( x – 2 ) 1 = ( y – 3)20
2.4
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Direct Variation is shown by y = k x , where k ≠ 0 and k is a constant, (that is a single value)
Example 1: Variables x and y vary directly, y = 12 and x = 4
• write and graph an equation for y and x
• find y when x = 5
y = k x
(12 ) = k (4)
3 = k
y = 3 x
y
x
x y 0 0 4 12 5 15 •
••
Example 2x = 6 and y = 3
y = k x3 = k 61 = k2
y = 1 x 2
Example 3x = 9 and y = 15
y = k x15 = k 9
5 = k 3
y = 5 x 3
21
Writing Direct Variation Equations 2.5
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A scatter plot is a graph used to determine, whether there is a relationship between paired data.
Positive Correlation Negative Correlation
No Correlation
•
•
••• ••••
•
••• •
••
•
••
••
•
••
•
• • ••
•
•
•••
•
••• •
•
22
Scatter Plots 2.5
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A scatter plot is a graph used to determine, whether there is a relationship between paired data
Approximating a best-fitting line: Graphical approach
Step 1: draw a scatter plot of dataStep 2: sketch the line based on the patternStep 3: choose 2 points on the lineStep 4: find the equation of the line that passes through the 2 points
Example 2: Fitting a line to data
Line through two points:
m = 1 − .6 = .4 = .25 2.5 - .9 1.6
Point-slope form:y – y1 = m ( x – x1)y – .6 = .25 ( x – .91)y – .6 = .25 x - .225 + .6 + .6 y = .25 x + .225
0 1 2 3
1.2
.8
.4
.2
••
••
•
• •
••
•
••
••
23
Scatter Plots 2.5
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Linear Inequalities in Two Variables can be written in one of the following forms:
ax + by < c ax + by ≤ c ax + by > c ax + by ≥ c
An ordered pair ( x , y ) is a solution of a linear inequality if the inequality is true for values of x , y. Example, ( − 6 , 2 ) is a solution of y ≥ 3 x − 9 because 2 ≥ 3 (− 6) − 9 is true as 2 ≥ − 27
Example 1: Checking Solutions of Inequalities
Ordered pairs Substitute Conclusion
( 0 , 1 ) 2 ( 0 ) + 3 ( 1 ) ≥ 5
3 ≥ 5 False, not a solution
( 4 , − 1 ) 2 (4 ) + 3 (− 1 ) ≥ 5
5 ≥ 5 True, is a solution
( 2 , 1 ) 2 ( 2 ) + 3 ( 1 ) ≥ 5
7 ≥ 5 True, is a solution
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Linear Inequalities in Two Variables 2.6
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Graph of a linear inequality in two variables is a half-plane. To graph a linear inequality follow these steps:
Step 1: Graph the boundary line of the inequality. Remember to use a dash line for < or > and a solid line for ≤ or ≥
Step 2: Test a point out that is NOT on the line to see what side to shade.
Graph: y < − 2 Graph: x < 1
y < − 2
y = − 2 x
y y
x
x = 1
x < 1
25
Linear Inequalities in Two Variables 2.6
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x y < 2 x y
0 y < 2 ( 0 ) 0
1 y < 2 ( 1 ) 2
••
y < 2 x
x 2 x – 5 y ≥ 10 y
0 2 ( 0 ) – 5 y ≥ 10 – 5 – 5 y ≤ −2
−2
5 2 x – 5 ( 0 ) ≥ 10 2 2 x ≥ 5
0•
•2 x – 5 y = 10
Test Point( 1, 1 )
•
Test Point( 0, 0 )
•
26
2 x – 5 y > 10
Graphing Linear Inequalities in Two Variables
y =
2 x
2.7
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Piecewise Functions: A combination of equations, each corresponding to part of a domain
Example 1: Evaluating a piecewise function
Evaluate f (x) when x = 0, when x = 2 and when x = 4
Evaluate f (x) = x + 2 , if x < 2
2 x + 1 , if x ≥ 2
Solution:
When x = 0 , f (x) = x + 2
f (0) = 0 + 2 = 2
When x = 2 , f (x) = 2 (x) + 1
f (2) = 2 (2) + 1 = 5
When x = 4 , f (x) = 2 (x) + 1
f (4) = 2 (4) + 1 = 9
{
20•○
27
Piecewise Functions 2.7
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Graphing a Piecewise Functions:
f (x) = 1 x + 3 , if x < 1 2 2
− x + 3 , if x ≥ 1{
f (x) = 1 x + 3 2 2
f (1) = 1 (1 ) + 3 = 4 = 2 2 2 2
f ( −3) = 1 ( −3) + 3 = 0 2 2
f ( x ) = − x + 3
f ( 1 ) = − ( 1 ) + 3 = 2
f (3 ) = − (3) + 3 = 0
•
•
•
•
( 1 , 2 )
2 rays with a common initial point, ( 1, 2 )
28
Graphing Piecewise Functions 2.7
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Graph the following piecewise function, show all work, label graph
f (x) = 2x – 1 , if x ≤ 13 x + 1 , if x > 1{
x y If x ≤ 1,f (x) = 2 x – 1
If x > 1,f (x) = 3 x + 1
x y
1 1 f (x) = 2 (1) – 1f (x) = 2 – 1
f (x) = 1
f (x) = 3 (1) + 1f (x) = 3 + 1
f (x) = 4
1 4
0 –1 f (x) = 2 (0) – 1f (x) = 0 – 1
f (x) = –1
f (x) = 3 (2) + 1f (x) = 6 + 1
f (x) = 7
2 7
•
•
•
○
x
y
29
Graphing Piecewise Functions 2.7
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f (x) = 1 , if 0 ≤ x < 1 2 , if 1 ≤ x < 2 3 , if 2 ≤ x < 3 4 , if 3 ≤ x < 4{
y
x
•
•
•
•
○
○
○
○
It is called a Step function because its graph looks like a set of stair steps
1 2 3 4
1
2
3
4
30
Graphing Piecewise Functions 2.7
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• ••○
( 0 , 2 )( 2 , 2 )
(−2 , 0 )
( 0 , 0 )
m = 0 – 2 = − 2 = 1 − 2 – 0 − 2
y = 1 x + 2
m = 0 – 2 = − 2 = 1 0 – 2 − 2
y = 1 x + 0
(−2 , 0 ) , ( 0 , 2 )
(0, 0 ) , ( 2 , 2 )
31
2.7Writing a Piecewise Function
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x , if x > 0Remember: │x│ = 0, if x = 0
− x, if x < 0{•
••y = x
x
y
( − 2 , 2 ) ( 2 , 2 )
Vertex
Slope – Intercept Formy = m x + b
Absolute Value Function:
y = a │ x – h │ + k
y = │x │
32
Writing a Piecewise Function y = − x
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Absolute Value Function: the a value gives youface up or down and steepness
y = a │ x – h │ + k If a is + , then the graph is face up or
If a is − , then the graph is face down
If a is bigger, then slope is steeper
If a is smaller, then slope is wider
33
Absolute Value Functions 2.8
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Absolute Value Function: the h value gives youRight (+) or left (−) movement of vertex
y = a │ x – h │ + k
The more h is + , the more the vertex moves to the right
The more h is − , the more the vertex moves to the left
y = a │ x – ( − 3 ) │ + k y = a │ x – ( 0 ) │ + k y = a │ x – ( 3 ) │ + k
34
Absolute Value Functions 2.8
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Absolute Value Function: the h value also gives youLine of Symmetry
y = a │ x – h │ + k
y = a │ x – ( − 3 ) │ + k y = a │ x – ( 0 ) │ + k y = a │ x – ( 3 ) │ + k
x = − 3 x = − 3x = 0 35
Absolute Value Functions
The line of symmetry is in the line x = h
2.8
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Absolute Value Function: the k value gives youUp or down movement of the vertex
y = a │ x – h │ + k
The more k is + , the more the vertex moves up
The more k is − , the more the vertex moves down
y = a │ x – h │ + 3 y = a │ x – h │ + 0 y = a │ x – h│ − 3
36
Absolute Value Functions 2.8
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(h, k ) values gives you the Vertex of the Absolute Value Function
y = a │ x – h │ + k
•(h, k )
37
Absolute Value Functions 2.8