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2.1<br />

Your <strong>Notes</strong><br />

Functions and Their Graphs<br />

Goal p Identify and graph functions.<br />

Vocabulary<br />

Relation A mapping, or pairing, of input values<br />

<strong>with</strong> output values<br />

Domain The set of input values of a relation<br />

Range The set of output values of a relation<br />

Function A relation for which each input has<br />

exactly one output<br />

Equation in two variables An equation that contains<br />

two variables<br />

Independent variable The input variable in an<br />

equation in two variables<br />

Dependent variable The output variable in an<br />

equation in two variables<br />

Example 1 Identify Functions<br />

Identify the domain and range. Then tell whether the<br />

relation is a function. Explain.<br />

a.<br />

Input Output<br />

22 21<br />

21<br />

3<br />

0<br />

6 4<br />

Solution<br />

b.<br />

Input Output<br />

22 22<br />

0<br />

1<br />

2<br />

2 3<br />

a. The domain consists of the inputs: 22, 21, 3 ,<br />

and 6 . The range consists of the outputs: 21, 0 ,<br />

and 4 . The relation is a function because each<br />

input is mapped onto exactly one output.<br />

b. The domain consists of the inputs: 22, 1 , and 2 .<br />

The range consists of the outputs: 22 , 0 , 2,<br />

and 3 . The relation is not a function because the<br />

input 2 is mapped onto both 0 and 3 .<br />

Copyright © McDougal Littell/Houghton Mifflin Company Lesson 2.1 • Algebra 2 C&S Notetaking Guide 25


Your <strong>Notes</strong><br />

VErTical linE TEST For FuncTionS<br />

A relation is a function if and only if no vertical line<br />

intersects the graph at more than one point .<br />

Function Not a function<br />

1<br />

O<br />

y<br />

1<br />

x<br />

Example 2 Apply the Vertical Line Test<br />

Graph the relations from Example 1. Use the vertical<br />

line test to tell whether the relations are functions.<br />

a. First, write the relation<br />

as a set of ordered<br />

pairs.<br />

(–2, –1), (–1, 0) ,<br />

(3, 0), (6, 4)<br />

Then plot the points.<br />

1<br />

O<br />

y<br />

1<br />

No vertical line<br />

contains more<br />

than one point so the<br />

relation is a function .<br />

x<br />

1<br />

O<br />

y<br />

1<br />

b. First, write the relation<br />

as a set of ordered<br />

pairs.<br />

(–2, –2) , (1, 0),<br />

(2, 2) , (2, 3)<br />

Then plot the points.<br />

1<br />

O<br />

y<br />

1<br />

The vertical line at<br />

x 5 2 contains two<br />

different points in the<br />

relation. So, the relation<br />

is not a function.<br />

26 Lesson 2.1 • Algebra 2 C&S Notetaking Guide Copyright © McDougal Littell/Houghton Mifflin Company<br />

x<br />

x


Your <strong>Notes</strong><br />

Homework<br />

GraphinG EquaTionS in Two VariablES<br />

Step 1 Make a table of values. Write the ordered pairs .<br />

Step 2 Plot enough solutions to recognize a pattern .<br />

Step 3 Connect the points <strong>with</strong> a line or a curve .<br />

Example 3 Graph an Equation<br />

Graph y 5 22x 2 2.<br />

Solution<br />

1. Make a table of values . Write the ordered pairs .<br />

Choose x 22 21 0 1 2<br />

Evaluate y 2 0 22 24 26<br />

(x, y): (22, 2 ), (21, 0 ), (0, 22 ), (1, 24 ), (2, 26 )<br />

2. Plot the points. Notice that all<br />

the points lie on a line .<br />

3. Connect the points <strong>with</strong> a<br />

line . Observe that the<br />

graph represents a function .<br />

Checkpoint Complete the following exercises.<br />

1. Graph y 5 x 1 2. 2. Graph y 5 3x 2 1.<br />

1<br />

O<br />

y<br />

1<br />

Copyright © McDougal Littell/Houghton Mifflin Company Lesson 2.1 • Algebra 2 C&S Notetaking Guide 27<br />

x<br />

1<br />

O<br />

y<br />

1<br />

1<br />

O<br />

y<br />

1<br />

x<br />

x


2.2<br />

Your <strong>Notes</strong><br />

linear Functions and Function<br />

notation<br />

Goal p Identify, evaluate, and graph linear functions.<br />

Vocabulary<br />

Linear function A function that can be written in the<br />

form y 5 mx 1 b where m and b are constants<br />

Function notation Using f(x) to represent the<br />

dependent variable of a function<br />

Example 1 Identify a Linear Function<br />

Tell whether the function is linear. Explain.<br />

a. f(x) 5 2x3 1 6 2 x b. g(x) 5 9 1 4x<br />

a. f is not a linear function because it has an<br />

x 3-term .<br />

b. g is a linear function because it can be rewritten as<br />

g(x) 5 4x 1 9 , so m 5 4 and b 5 9 .<br />

Checkpoint Tell whether the function is linear. Explain.<br />

1. f(x) 5 6x 1 10<br />

Yes; it is of the form<br />

y 5 mx 1 b.<br />

Example 2 Evaluate a Function<br />

Evaluate the function when x 5 3.<br />

2. g(x) 5 2x2 1 4x 2 1<br />

No; it is not of the<br />

form y 5 mx 1 b.<br />

a. f(x) 5 22x 1 5 b. g(x) 5 x2 2 x 2 1<br />

a. f (x) 5 22x 1 5 Write original function.<br />

f ( 3 ) 5 22( 3 ) 1 5 Substitute 3 for x.<br />

5 26 1 5 Multiply.<br />

5 21 Simplify.<br />

28 Lesson 2.2 • Algebra 2 C&S Notetaking Guide Copyright © McDougal Littell/Houghton Mifflin Company


Your <strong>Notes</strong><br />

Homework<br />

b. g (x) 5 x 2 2 x 2 1 Write original function.<br />

g( 3 ) 5 ( 3 ) 2 2 3 2 1 Substitute 3 for x.<br />

5 9 2 3 2 1 Multiply.<br />

5 5 Simplify.<br />

Checkpoint Evaluate the function when x 5 24.<br />

3. f(x) 5 2x 2 3 4. g(x) 5 2x 2 1 3x<br />

211 20<br />

Example 3 Graph a Linear Function<br />

Graph f(x) 5 2x 1 2.<br />

Solution<br />

Rewrite the function as y 5 2x 1 2. Find a point on<br />

the graph by substituting a convenient value for x .<br />

y 5 2x 1 2 Write equation.<br />

y<br />

y 5 2( 0 ) 1 2 Substitute 0 for x.<br />

y 5 2 Simplify.<br />

1<br />

O 1<br />

One point is (0, 2) . Find a second<br />

point.<br />

y 5 2(1) 1 2 Substitute 1 for x .<br />

y 5 1 Simplify.<br />

A second point is (1, 1) . Draw a line through the<br />

two points.<br />

Checkpoint Complete the following exercise.<br />

5. Graph h(x) 5 22x 1 1.<br />

Copyright © McDougal Littell/Houghton Mifflin Company Lesson 2.2 • Algebra 2 C&S Notetaking Guide 29<br />

1<br />

O<br />

y<br />

1<br />

x<br />

x


2.3<br />

Your <strong>Notes</strong><br />

When calculating<br />

the slope, be sure<br />

to subtract the<br />

x‑coordinates and<br />

the y‑coordinates<br />

in the correct order.<br />

A vertical line has<br />

“undefined slope”<br />

because for any<br />

two points, the<br />

slope formula’s<br />

denominator<br />

becomes 0, and<br />

division by 0 is<br />

undefined.<br />

Slope and rate of change<br />

Goal p Find and use the slope of a line.<br />

Vocabulary<br />

Slope For a nonvertical line, the ratio of the line’s<br />

vertical change to its horizontal change<br />

ThE SlopE oF a linE<br />

The slope m of the nonvertical line passing through<br />

the points (x1 , y1 ) and (x2 , y2 ) is given by the formula<br />

m 5 y2 2 y1 } .<br />

x2 2 x1 Example 1 Find the Slope of a Line<br />

Find the slope of the line through (1, 3) and (6, 7).<br />

Solution<br />

Let (x 1 , y 1 ) 5 (1, 3) and (x 2 , y 2 ) 5 (6, 7) .<br />

m 5 y2 2 y1 } 5<br />

x2 2 x1 7 2 3<br />

}<br />

6 2 1<br />

The slope of the line is<br />

4<br />

5 }<br />

5<br />

4<br />

} .<br />

5<br />

claSSiFyinG linES by SlopE<br />

Positive Negative Zero Undefined<br />

O<br />

y<br />

x<br />

O<br />

y<br />

Line rises from Line falls from Line is Line is<br />

left to right. left to right . horizontal . vertical .<br />

30 Lesson 2.3 • Algebra 2 C&S Notetaking Guide Copyright © McDougal Littell/Houghton Mifflin Company<br />

x<br />

O<br />

y<br />

x<br />

O<br />

y<br />

x


Your <strong>Notes</strong><br />

Example 2 Classify Lines Using Slope<br />

Without graphing, tell whether the line through the<br />

given points rises, falls, is horizontal, or is vertical.<br />

a. (26, 22), (1, 3) b. (2, 21), (2, 2)<br />

Solution<br />

a. m 5<br />

b. m 5<br />

3 2 (22)<br />

}<br />

1 2 (26)<br />

2 2 (21)<br />

}<br />

2 2 2<br />

5 5<br />

}<br />

7 Because m > 0, the line rises .<br />

5 3<br />

}<br />

0 Because m is undefined , the<br />

line is vertical .<br />

Checkpoint Complete the following exercises.<br />

1. Find the slope of the line through (4, 2) and (7, 9).<br />

7<br />

}<br />

3<br />

2. Without graphing, tell whether the line through the<br />

points (23, 2) and (1, 4) rises, falls, is horizontal, or<br />

is vertical.<br />

rises<br />

Example 3 Compare Steepness of Lines<br />

Tell which line is steeper.<br />

Line 1: through (2, 3) and (21, 4)<br />

Line 2: through (27, 22) and (24, 25)<br />

Solution<br />

Slope of line 1: m 5<br />

Slope of line 2: m 5<br />

4 2 3<br />

}<br />

21 2 2<br />

25 2 (22)<br />

}<br />

24 2 (27)<br />

5 1<br />

}<br />

23<br />

or 2 1<br />

}<br />

3<br />

5 23<br />

}<br />

3<br />

5 21<br />

Both lines have negative slope.<br />

Because |m1 | > |m2 |, line 1 is steeper than<br />

line 2 .<br />

Copyright © McDougal Littell/Houghton Mifflin Company Lesson 2.3 • Algebra 2 C&S Notetaking Guide 31


Your <strong>Notes</strong><br />

Homework<br />

Checkpoint Tell which line is steeper.<br />

3. Line 1: through (3, 5) and (6, 8)<br />

Line 2: through (22, 4) and (10, 12)<br />

line 1<br />

Example 4 Slope as a Rate of Change<br />

census The population of Smalltown was 10,000 in<br />

2001 and 12,500 in 2006. Find the average rate of<br />

change in the population.<br />

Solution<br />

Let (x1 , y1 ) 5 (2001, 10,000 ) and<br />

(x2 , y2 ) 5 ( 2006 , 12,500 ).<br />

Change in population<br />

Average rate of change 5 }}<br />

Change in time<br />

12,500 2 10,000<br />

5 }}<br />

2006 2 2001<br />

5 2500<br />

}<br />

5<br />

5 500<br />

The average rate of change is 500 people per year .<br />

Checkpoint Complete the following exercise.<br />

4. Census Suppose the population of Smalltown will be<br />

20,000 in 2011. Use the information in Example 4<br />

to find the average rate of change in the population<br />

from 2001 to 2011.<br />

1000 people per year<br />

32 Lesson 2.3 • Algebra 2 C&S Notetaking Guide Copyright © McDougal Littell/Houghton Mifflin Company


Your <strong>Notes</strong><br />

2.4 quick Graphs of linear<br />

Equations<br />

Goal p Use slope-intercept form and standard form to<br />

graph equations.<br />

Vocabulary<br />

y-intercept The y-coordinate of a point where a<br />

graph intersects the y-axis<br />

Slope-intercept form A linear equation written in the<br />

form y 5 mx 1 b where m is the slope and b is the<br />

y-intercept of the equation’s graph<br />

x-intercept The x-coordinate of a point where a<br />

graph intersects the x-axis<br />

Standard form of a linear equation A linear equation<br />

written in the form Ax 1 By 5 C where A and B are<br />

not both zero<br />

GraphinG EquaTionS in SlopE-inTErcEpT Form<br />

Step 1 Write the equation in slope-intercept form by<br />

solving for y .<br />

Step 2 Find the y-intercept and plot the<br />

corresponding point.<br />

Step 3 Find the slope and use it to plot a second<br />

point on the line.<br />

Step 4 Draw a line through the two points .<br />

Copyright © McDougal Littell/Houghton Mifflin Company Lesson 2.4 • Algebra 2 C&S Notetaking Guide 33


Your <strong>Notes</strong><br />

Example 1 Use Slope-Intercept Form to Graph a Line<br />

Graph y 1 3 } x 5 1.<br />

2<br />

solution<br />

1. Solve the equation for y by subtracting 3<br />

} x from each<br />

2<br />

side:<br />

y 5 1 2 3<br />

} x or y 5 2<br />

2 3<br />

} x 1 1<br />

2<br />

2. Find the y-intercept. Comparing y 5 2 3<br />

} x 1 1 to<br />

2<br />

y 5 mx 1 b, you can see that b 5 1 . Plot a point<br />

at ( 0 , 1 ).<br />

3. Find the slope. The slope is m 5 2 3<br />

} , so<br />

2 rise<br />

}<br />

run<br />

From (0, 1), move down 3 units and right<br />

2 units. Plot a second point at ( 2 , 22 ).<br />

4. Draw a line through the two points.<br />

1<br />

O<br />

y<br />

1<br />

x<br />

GraphinG Equations in standard Form<br />

Step 1 Write the equation in standard form .<br />

Step 2 Find the x-intercept by letting y 5 0 and<br />

solving for x . Plot the corresponding point.<br />

Step 3 Find the y-intercept by letting x 5 0 and<br />

solving for y . Plot the corresponding point.<br />

Step 4 Draw a line through the two points .<br />

5 23<br />

}<br />

2 .<br />

34 Lesson 2.4 • Algebra 2 C&S Notetaking Guide Copyright © McDougal Littell/Houghton Mifflin Company


Your <strong>Notes</strong><br />

Example 2 Draw Quick Graphs<br />

Graph 4x 1 5y 5 20.<br />

1. The equation is already in standard form .<br />

2. Find the x-intercept. 4x 1 5( 0 ) 5 20 Let y 5 0.<br />

SlopES oF parallEl and pErpEndicular linES<br />

Consider two different<br />

nonvertical lines l 1 and l 2<br />

<strong>with</strong> slopes m 1 and m 2 .<br />

Two lines are parallel if and only<br />

if they have the same slope .<br />

m 1 5 m 2<br />

Two lines are perpendicular if and<br />

only if their slopes are negative<br />

reciprocals of each other.<br />

m 1 5 2 1<br />

}<br />

m 2 or m 1 • m 2 5 –1<br />

x 5 5 Solve for x.<br />

The x-intercept is 5 , so plot the point (5, 0) .<br />

3. Find the y-intercept. 4( 0 ) 1 5y 5 20 Let x 5 0 .<br />

y 5 4 Solve for y.<br />

The y-intercept is 4 , so plot the point (0, 4) .<br />

4. Draw a line through<br />

the intercepts.<br />

Copyright © McDougal Littell/Houghton Mifflin Company Lesson 2.4 • Algebra 2 C&S Notetaking Guide 35<br />

l 1<br />

l 1<br />

1<br />

O<br />

l 2<br />

y<br />

1<br />

O<br />

1<br />

1<br />

O<br />

y<br />

y<br />

1<br />

1<br />

l 2<br />

x<br />

x<br />

x


Your <strong>Notes</strong><br />

Homework<br />

Example 3 Graph Parallel and Perpendicular Lines<br />

a. Draw the graph y 5 2x 2 3.<br />

b. Graph the line that passes through (0, 2) and is<br />

parallel to the graph of y 5 2x 2 3.<br />

c. Graph the line that passes through (0, 2) and is<br />

perpendicular to the graph of y 5 2x 2 3.<br />

Solution<br />

a. Draw the graph y 5 2x 2 3<br />

using the y-intercept 23 and<br />

slope 2 .<br />

b. Graph the point (0, 2). Any<br />

line parallel to the graph of<br />

y 5 2x 2 3 has the same<br />

slope, 2 . From (0, 2), move<br />

up 2 units and right<br />

1 unit. Plot a second point at (1, 4) . Draw a line<br />

through the two points.<br />

c. The slope of any line perpendicular to the<br />

graph of y 5 2x 2 3 is the negative reciprocal<br />

of 2, or 2 1<br />

} . So, from (0, 2), move down 1 unit<br />

2<br />

and right 2 units. Plot a second point at<br />

(2, 1) . Draw the line.<br />

Checkpoint On the same grid, draw each graph.<br />

1. y 5 2 1 }<br />

3 x 1 1<br />

2. The line parallel to<br />

y 5 2 1 }<br />

3 x 1 1 through<br />

the point (0, 4)<br />

3. The line perpendicular<br />

to y 5 2 1 }<br />

3 x 1 1 through<br />

the point (0, 4)<br />

36 Lesson 2.4 • Algebra 2 C&S Notetaking Guide Copyright © McDougal Littell/Houghton Mifflin Company<br />

c<br />

1<br />

O<br />

y<br />

1<br />

1<br />

O<br />

Ex. 3<br />

y<br />

1<br />

b<br />

Ex. 1<br />

Ex. 2<br />

x<br />

a<br />

x


Your <strong>Notes</strong><br />

2.5 writing Equations of lines<br />

Goal p Write linear equations.<br />

Vocabulary<br />

Point-slope form An equation of a line written in<br />

the form y 2 y 1 5 m(x 2 x 1 ) where the line passes<br />

through the point (x 1 , y 1 ) and has a slope of m<br />

wriTinG an EquaTion oF a linE<br />

Slope-Intercept Form Given the slope m and the<br />

y-intercept b , use slope-intercept form: y 5 mx 1 b .<br />

Point-Slope Form Given the slope m and a point<br />

(x1 , y1 ) , use point-slope form: y 2 y1 5 m(x 2 x1 ) .<br />

Example 1 Write an Equation Given Slope and y-Intercept<br />

Write an equation of the<br />

line shown.<br />

y<br />

Solution<br />

1<br />

From the graph, you can see<br />

that the slope is m 5 2<br />

O 2<br />

3<br />

}<br />

2<br />

.<br />

Because the line intersects the<br />

y-axis at ( 0, 1 ), the y-intercept<br />

is b 5 1 .<br />

Use the slope-intercept form to write an equation of<br />

the line.<br />

y 5 mx 1 b Use slope-intercept form.<br />

y 5 2 3<br />

}<br />

2<br />

x 1 1 Substitute 2 3<br />

}<br />

2<br />

for m and 1 for b.<br />

Copyright © McDougal Littell/Houghton Mifflin Company Lesson 2.5 • Algebra 2 C&S Notetaking Guide 37<br />

x


Your <strong>Notes</strong><br />

Example 2 Write an Equation Given Slope and a Point<br />

Write an equation of the line that passes through (2, 1)<br />

and has a slope of 2.<br />

Solution<br />

Because you know the slope and a point on the line,<br />

use the point-slope form <strong>with</strong> (x1 , y1 ) 5 (2, 1) and<br />

m 5 2 .<br />

y 2 y1 5 m(x 2 x1 ) Use point-slope form.<br />

y 2 1 5 2 (x 2 2 ) Substitute for m, x1 , and y1 .<br />

Rewrite the equation in slope-intercept form.<br />

y 2 1 5 2x 2 4 Distributive property<br />

y 5 2x 2 3 Slope-intercept form<br />

Checkpoint Write an equation of the line.<br />

1. y<br />

2. The line through (1, 5)<br />

1<br />

<strong>with</strong> a slope of 22<br />

O<br />

1<br />

x<br />

y 5 2<br />

}<br />

3 x 2 3 y 5 22x 1 7<br />

Example 3 Write Equations of Parallel Lines<br />

Write an equation of the line that passes through<br />

(21, 1) and is parallel to the line y 5 22x 1 3.<br />

Solution<br />

The given line has a slope of 22 . Any line parallel<br />

to this line must also have a slope of 22 . Use<br />

point-slope form <strong>with</strong> (x1 , y1 ) 5 (21, 1) and<br />

m 5 22 to write the equation of the line.<br />

38 Lesson 2.5 • Algebra 2 C&S Notetaking Guide Copyright © McDougal Littell/Houghton Mifflin Company


Your <strong>Notes</strong><br />

Homework<br />

y 2 y 1 5 m(x 2 x 1 ) Use point-slope form.<br />

y 2 1 5 22 [x 2 (21) ] Substitute 1 for y 1 , 22<br />

for m, and 21 for x 1 .<br />

y 2 1 5 22x 2 2 Distributive property<br />

y 5 22x 2 1 Slope-intercept form<br />

Example 4 Write Equations of Perpendicular Lines<br />

Write an equation of the line that passes through<br />

(23, 5) and is perpendicular to the line y 5 4x 2 1.<br />

solution<br />

The given line has a slope of 4 . The slope of any<br />

line perpendicular to this line will be the negative<br />

reciprocal of 4 , which is 2 1<br />

} .<br />

4<br />

Use point-slope form <strong>with</strong> (x1 , y1 ) 5 (23, 5) and<br />

m 5 2 1<br />

} to write the equation of the line.<br />

4<br />

y 2 y 1 5 m(x 2 x 1 ) Use point-slope form.<br />

y 2 5 5 2 1<br />

} [x 2 (23) ]<br />

4<br />

Substitute 5 for y1 , 2 1<br />

}<br />

4<br />

for m, and 23 for x1 .<br />

1<br />

y 2 5 5 2 } x 2<br />

4 3<br />

}<br />

4<br />

Distributive property<br />

1<br />

y 5 2 } x 1<br />

4 17<br />

}<br />

4<br />

Checkpoint Write an equation of the line.<br />

3. The line through<br />

(22, 3) parallel to<br />

y 5 4x 2 6<br />

y 5 4x 1 11<br />

Slope-intercept form<br />

4. The line through<br />

(22, 3) perpendicular<br />

to y 5 4x 2 6<br />

y 5 2 1<br />

} x 1<br />

4 5<br />

}<br />

2<br />

Copyright © McDougal Littell/Houghton Mifflin Company Lesson 2.5 • Algebra 2 C&S Notetaking Guide 39


Your <strong>Notes</strong><br />

2.6 direct Variation<br />

Goal p Write and graph direct variation equations.<br />

Vocabulary<br />

Direct variation Two variables x and y show direct<br />

variation provided that y 5 kx where k is a nonzero<br />

constant.<br />

Constant of variation The nonzero constant k in a<br />

direct variation equation y 5 kx<br />

Example 1 Graph a Direct Variation Equation<br />

Graph y 5 5<br />

} x.<br />

3<br />

Solution<br />

Plot a point at the origin .<br />

Find a second point by substituting<br />

a convenient value for x .<br />

y 5 5 } x<br />

3<br />

Write equation.<br />

y 5 5 } ( 3 )<br />

3<br />

Substitute 3 for x.<br />

y 5 5 Simplify.<br />

A second point is (3, 5) .<br />

Plot the second point. Then draw a line through the<br />

two points.<br />

Checkpoint Graph the equation.<br />

1. y 5 23x<br />

1<br />

O<br />

y<br />

1<br />

40 Lesson 2.6 • Algebra 2 C&S Notetaking Guide Copyright © McDougal Littell/Houghton Mifflin Company<br />

x<br />

1<br />

O<br />

y<br />

1<br />

x


Your <strong>Notes</strong><br />

The value of k is<br />

called the spring<br />

constant. This<br />

constant is unique<br />

for each spring.<br />

Example 2 Write a Direct Variation Equation<br />

physics According to Hooke’s law, the amount that a<br />

spring is stretched varies directly <strong>with</strong> the force that is<br />

applied to the spring. If 64 pounds of force F stretches<br />

a spring a distance d of 8 inches, write an equation<br />

so that F varies directly <strong>with</strong> d. How many pounds of<br />

force are needed to stretch the spring to a distance<br />

of 14 inches?<br />

Solution<br />

Find the constant of variation.<br />

F 5 kd Write a direct variation equation.<br />

64 5 k( 8 ) Substitute 64 for F and 8 for d.<br />

8 5 k Simplify.<br />

An equation is F 5 8d . To find how many pounds<br />

of force is needed to stretch the spring 14 inches,<br />

substitute 14 for d.<br />

F 5 8 ( 14 ) 5 112<br />

A force of 112 pounds will stretch the spring 14 inches.<br />

Checkpoint Complete the following exercise.<br />

2. Suppose that a force of 92 pounds is applied to<br />

the spring discussed in Example 2. How far will<br />

the spring be stretched?<br />

11.5 inches<br />

Copyright © McDougal Littell/Houghton Mifflin Company Lesson 2.6 • Algebra 2 C&S Notetaking Guide 41


Your <strong>Notes</strong><br />

Homework<br />

Example 3 Identify Direct Variation<br />

The table gives sample cell phone bills, showing the<br />

total monthly cost and the number of minutes used<br />

per month. Tell whether total cost and the number of<br />

minutes show direct variation. If so, write an equation<br />

relating c and m.<br />

Total cost, c (dollars) 35 45 80 15 28<br />

Minutes used, m 100 129 229 43 80<br />

solution<br />

Find the ratio of c to m for each data pair.<br />

35<br />

}<br />

100<br />

15<br />

}<br />

43<br />

5 0.35 45<br />

}<br />

129<br />

28<br />

< 0.35 } 5 0.35<br />

30<br />

Checkpoint Complete the following exercise.<br />

3. Tell whether the data (21, 22), (1, 2), (3, 6),<br />

(5, 10), (7, 14) show direct variation. If so, write an<br />

expression relating x and y.<br />

direct variation; y 5 2x<br />

80<br />

< 0.35 } < 0.35<br />

229<br />

The ratio of c to m in each pair is nearly the same , so<br />

the data show direct variation . Substituting 0.35<br />

for k in c 5 km gives the direct variation equation<br />

c 5 0.35 m, which relates the total cost to number of<br />

minutes used.<br />

42 Lesson 2.6 • Algebra 2 C&S Notetaking Guide Copyright © McDougal Littell/Houghton Mifflin Company


2.7<br />

Your <strong>Notes</strong><br />

Scatter plots and correlation<br />

Goal p See correlation in a scatter plot and find a<br />

best-fitting line.<br />

Vocabulary<br />

Scatter plot A graph of a set of data pairs (x, y)<br />

used to determine whether there is a relationship<br />

between the variables x and y<br />

Positive correlation A set of data pairs (x, y) for<br />

which as x increases, y tends to increase<br />

Negative correlation A set of data pairs (x, y) for<br />

which as x increases, y tends to decrease<br />

Relatively no correlation A set of data pairs (x, y) for<br />

which there is no obvious pattern between x and y<br />

Best-fitting line The line that most closely models<br />

the data shown in a scatter plot<br />

Example 1 Identify Correlation<br />

Describe the correlation shown by each plot.<br />

a. y<br />

b.<br />

1<br />

O<br />

1<br />

x<br />

Solution<br />

a. The scatter plot shows a negative correlation: as<br />

x increases, y decreases .<br />

b. The scatter plot shows a positive correlation: as<br />

x increases , y also increases .<br />

Copyright © McDougal Littell/Houghton Mifflin Company Lesson 2.7 • Algebra 2 C&S Notetaking Guide 43<br />

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O<br />

y<br />

1<br />

x


Your <strong>Notes</strong><br />

Be sure that about<br />

the same number<br />

of points lie above<br />

your best-fitting<br />

line as lie below it.<br />

approximatinG a BEst-FittinG LinE<br />

Step 1 Draw a scatter plot of the data .<br />

Step 2 Sketch a line that follows the trend of the<br />

data points.<br />

Step 3 Choose two points that appear to lie on the<br />

line, and estimate their coordinates .<br />

Step 4 Write an equation of the line that passes<br />

through the two points from Step 3. This<br />

equation gives a model for the data.<br />

Example 2 Find a Best-Fitting Line<br />

Football attendance The table gives the number of<br />

people y who attended each of the first seven football<br />

games x of the season. Approximate the best-fitting<br />

line for the data.<br />

x 1 2 3 4 5 6 7<br />

y 722 763 772 826 815 857 897<br />

solution<br />

1. Draw a scatter plot.<br />

950<br />

y<br />

2. Sketch the line that<br />

900<br />

best fits the data.<br />

850<br />

800<br />

3. Choose two points on the<br />

line. For this scatter plot you<br />

750<br />

700<br />

650<br />

might choose (1, 725) and<br />

0<br />

0 1 2 3 4 5 6 7 x<br />

( 2 , 750).<br />

Football game<br />

4. Write an equation of the line. First find the slope using<br />

the two points:<br />

750 2 725 25<br />

m 5 } 5 } 5 25<br />

2 2 1 1<br />

Now use point-slope form. Choose (x1 , y1 ) 5 (1, 725).<br />

y 2 y 1 5 m(x 2 x 1 ) Point-slope form<br />

y 2 725 5 25 (x 2 1 ) Substitute for y 1 , m, and x 1 .<br />

Number of people<br />

y 5 25x 1 700 Solve for y.<br />

44 Lesson 2.7 • Algebra 2 C&S Notetaking Guide Copyright © McDougal Littell/Houghton Mifflin Company


Your <strong>Notes</strong><br />

Homework<br />

Example 3 Use a Best-Fitting Line<br />

Use the equation of the best-fitting line from Example 2<br />

to predict the number of people that will attend the<br />

tenth football game.<br />

Solution<br />

Substitute 10 for x in the equation from Example 2.<br />

y 5 25( 10 ) 1 700 5 250 1 700 5 950<br />

You can predict that 950 people will attend the game.<br />

Checkpoint Complete the following exercises.<br />

For each scatter plot, tell whether the data show a<br />

positive correlation, negative correlation, or relatively no<br />

correlation.<br />

1. y<br />

2. y<br />

1<br />

O<br />

1<br />

positive correlation<br />

x<br />

1<br />

O<br />

1<br />

relatively no correlation<br />

3. School The table gives the average class score y on<br />

each test for the first six chapters x of the textbook.<br />

x 1 2 3 4 5 6<br />

y 84 83 86 88 87 90<br />

a. Approximate the best-fitting<br />

line for the data.<br />

y 5 2x 1 80<br />

b. Use your equation to predict<br />

the test score for the 9th test<br />

that the class will take.<br />

98<br />

Average class score<br />

90<br />

88<br />

86<br />

84<br />

82<br />

y<br />

0<br />

0 1 2 3 4 5<br />

Test<br />

Copyright © McDougal Littell/Houghton Mifflin Company Lesson 2.7 • Algebra 2 C&S Notetaking Guide 45<br />

x<br />

6<br />

x


words to review<br />

use your own words and/or an example to explain<br />

the vocabulary word.<br />

Relation<br />

A mapping, or pairing,<br />

of input values <strong>with</strong><br />

output values<br />

Range<br />

The set of output<br />

values in a relation<br />

Equation in two variables<br />

y 5 2x 2 5<br />

Dependent variable<br />

The dependent variable<br />

in the equation<br />

y 5 2x 2 5 is y.<br />

Function notation<br />

f(x) 5 2x 1 4<br />

y-intercept<br />

The y-intercept of<br />

y 5 2x 1 4 is 4.<br />

x-intercept<br />

The x-intercept of<br />

y 5 2x 1 4 is 22.<br />

Domain<br />

The set of input values<br />

in a relation<br />

Function<br />

A relation for which<br />

each input has exactly<br />

one output<br />

Independent variable<br />

The independent<br />

variable in the equation<br />

y 5 2x 2 5 is x.<br />

Linear function<br />

A function that can be<br />

written in the form<br />

y 5 mx 1 b where m<br />

and b are constants<br />

Slope<br />

The slope of the line<br />

passing through (2, 6)<br />

and (5, 9) is<br />

m 5 y 2 2 y 1<br />

}<br />

x 2 2 x 1<br />

5<br />

9 2 6<br />

}<br />

5 2 2<br />

5 3<br />

}<br />

3 5 1.<br />

Slope-intercept form<br />

y 5 2x 1 4<br />

Standard form of a linear<br />

equation<br />

2x 1 4y 5 7<br />

46 Words to Review • Algebra 2 C&S Notetaking Guide Copyright © McDougal Littell/Houghton Mifflin Company


Point-slope form<br />

y 2 1 5 2(x 2 5)<br />

Constant of variation<br />

The constant of<br />

variation in the direct<br />

variation equation<br />

y 5 6x is 6.<br />

Positive correlation<br />

This scatter plot shows<br />

a positive correlation.<br />

y<br />

Relatively no correlation<br />

This scatter plot<br />

shows relatively no<br />

correlation.<br />

y<br />

x<br />

x<br />

Direct variation<br />

y 5 6x<br />

Scatter plot<br />

A graph of a set of<br />

data pairs (x, y) used<br />

to determine whether<br />

there is a relationship<br />

between the variables<br />

x and y<br />

Negative correlation<br />

This scatter plot shows<br />

a negative correlation.<br />

y<br />

Best-fitting line<br />

The line that most<br />

closely models the data<br />

shown in a scatter plot<br />

review your notes and chapter 2 by using the<br />

chapter review on pages 115–118 of your textbook.<br />

Copyright © McDougal Littell/Houghton Mifflin Company Words to Review • Algebra 2 C&S Notetaking Guide 47<br />

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