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2.12.1 Linear Functions and ModelsLinear Functions and Models

2.22.2 Equations of LinesEquations of Lines

2.32.3 Linear EquationsLinear Equations

2.42.4 Linear InequalitiesLinear Inequalities

2.5 2.5 Piece-wise Defined FunctionsPiece-wise Defined Functions

Linear Functions Linear Functions and Equationsand Equations

22

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Equations of LinesEquations of Lines

♦ Write the point-slope and slope-intercept forms Write the point-slope and slope-intercept forms for a linefor a line

♦ Find the intercepts of a lineFind the intercepts of a line♦ Write equations for horizontal, vertical, parallel, Write equations for horizontal, vertical, parallel,

and perpendicular linesand perpendicular lines♦ Model data with lines and linear functions Model data with lines and linear functions

(optional)(optional)♦ Use direct variation to solve problemsUse direct variation to solve problems

2.22.2

2.2 - 4Copyright © 2010 Pearson Education, Inc.

Point-Slope Form of the Point-Slope Form of the Equation of a LineEquation of a Line

The line with slope The line with slope mm passing through passing through the point (the point (xx11, , yy11) has equation ) has equation

yy = = mm((xx xx11) + ) + yy11

oror

y y yy11 = = mm((xx xx11))

2.2 - 5Copyright © 2010 Pearson Education, Inc.

Find an equation of the line passing through the points (–2, –3) and (1, 3). Plot the points and graph the line by hand.

Example 1Example 1 Determining a point-slope Determining a point-slope formform

SolutionSolution (continued) (continued)

Calculate the slope:

m 3 3 1 2

6

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2.2 - 6Copyright © 2010 Pearson Education, Inc.

SolutionSolution (continued) (continued)

Example 1Example 1 Determining a point-slope Determining a point-slope formform

Substitute (1, 3) for (x1, y1) and 2 for m

y = m(x –x1) + y1

y = 2(x – 1) + 3

Or substitute (–2, –3) for (x1, y1) and 2 for m

y = m(x –x1) + y1

y = 2(x – (–2)) + (–3)y = 2(x + 2) – 3

2.2 - 7Copyright © 2010 Pearson Education, Inc.

SolutionSolution (continued) (continued)

Example 1Example 1 Determining a point-slope Determining a point-slope formform

Here’s the graph:

2.2 - 8Copyright © 2010 Pearson Education, Inc.

Slope-Intercept Form of the Slope-Intercept Form of the Equation of a LineEquation of a Line

• The line with slope The line with slope mm and and yy-intercept -intercept bb is given byis given by

yy = = mxmx + + bb

2.2 - 9Copyright © 2010 Pearson Education, Inc.

Find the slope-intercept form for the line passing through the points (–2, 1) and (2, 3).

Example 3Example 3 Finding slope-intercept Finding slope-intercept formform

Determine m and b in the form y = mx + bSolutionSolution

m

3 1

2 2 2

4

1

2

Substitute either point to find b, use (2, 3).

3

1

22 b b 2

y

1

2x 2

2.2 - 10Copyright © 2010 Pearson Education, Inc.

An equation of a line is in standard form when it is written as

ax + by = cwhere a, b, and c are constants.

Some examples are:

Standard FormStandard Form

2x 3y 6, y

1

4, x 3, 3x y

1

2

2.2 - 11Copyright © 2010 Pearson Education, Inc.

To find any x-intercepts, let y = 0 in the equation and solve for x.

To find any y-intercepts, let x = 0 in the equation and solve for y.

Finding InterceptsFinding Intercepts

2.2 - 12Copyright © 2010 Pearson Education, Inc.

Locate the x- and y-intercepts for the line whose equation is 4x + 3y = 6. Use the intercepts to graph the equation.

Example 5Example 5 Finding InterceptsFinding Intercepts

To find the x-intercept, let y = 0, solve for x:

SolutionSolution

4x 3 0 6

x 1.5

The x-intercept is 1.5

2.2 - 13Copyright © 2010 Pearson Education, Inc.

To find the y-intercept, let x = 0, solve for y:

Example 5Example 5 Finding InterceptsFinding Intercepts

Solution Solution (continued)(continued)

4(0) 3y 6

y 2

The y-intercept is 2.

The graph passes through the points (1.5, 0) and (0, 2).

2.2 - 14Copyright © 2010 Pearson Education, Inc.

• Graph of a constant function Graph of a constant function ff

• Formula: Formula: f f ((xx) = ) = bb

• Horizontal line with slope 0 and y-intercept Horizontal line with slope 0 and y-intercept bb..

(-3, 3)

(3, 3)

Horizontal LinesHorizontal Lines

Note that regardless of Note that regardless of the value of the value of xx, the , the value of value of yy is always 3. is always 3.

2.2 - 15Copyright © 2010 Pearson Education, Inc.

Vertical LinesVertical Lines• Cannot be represented by a function• Slope is undefined• Equation is: x = k

• Note that regardless Note that regardless of the value of of the value of yy, the , the value of value of xx is always is always 3. 3.

• Equation is Equation is xx = 3 = 3 (or (or xx + 0 + 0yy = 3) = 3)

• Equation of a vertical Equation of a vertical line is line is xx = = kk where where kk is the is the xx-intercept.-intercept.

2.2 - 16Copyright © 2010 Pearson Education, Inc.

Equations of Horizontal and Vertical Equations of Horizontal and Vertical LinesLines

An equation of the horizontal line with y-intercept b is y = b.

An equation of the vertical line with x-intercept k is x = k.

2.2 - 17Copyright © 2010 Pearson Education, Inc.

Parallel LinesParallel Lines

Two lines with slopes m1 and m2, neither of which is vertical, are parallel if and only if their slopes are equal; that is, m1 = m2.

2.2 - 18Copyright © 2010 Pearson Education, Inc.

Perpendicular LinesPerpendicular Lines

Two lines with nonzero slopes m1 and m2,, are perpendicular if and only if their slopes have a product of –1; that is, m1m2 = –1.

2.2 - 19Copyright © 2010 Pearson Education, Inc.

Example 8: Finding perpendicular linesExample 8: Finding perpendicular lines

Find the slope-intercept form of the line

perpendicular to passing

through the point (–2, 1) . Graph the lines. y

2

3x 2,

The line has slope

SolutionSolution

y

2

3x 2

2

3The negative reciprocal is

3

2

Use the point-slope form of the line . . .

2.2 - 20Copyright © 2010 Pearson Education, Inc.

Example 8: Finding perpendicular linesExample 8: Finding perpendicular linesSolutionSolution (continued) (continued)

y m x x1 y

1

y

3

2x 2 1

y

3

2x 3 1

y

3

2x 4

2.2 - 21Copyright © 2010 Pearson Education, Inc.

Example 10: Modeling dataExample 10: Modeling dataThe table lists the average tuition and fees at private colleges for selected years.

(a) Make a scatterplot of the data. (b) Find a linear function, given by

f(x) = m(x – x1) + y1, that models the data. Interpret the slope m.

(c) Use to estimate tuition and fees in 1998. Compare the estimate to the actual value of $14,709. Did your answer involve interpolation or extrapolation?

2.2 - 22Copyright © 2010 Pearson Education, Inc.

Example 10: Modeling dataExample 10: Modeling dataSolutionSolution(a) Make a scatterplot of the data.

2.2 - 23Copyright © 2010 Pearson Education, Inc.

Example 10: Modeling dataExample 10: Modeling dataSolutionSolution (continued) (continued)

f x m x 1980 3617

(b) Choose any point, say (1980, 3617) to use for (x1, y1) and write:

m

16,233 3617

2000 1980630.8

Choose two points to estimate the slope, say (1980, 3617) and (2000, 16,233)

f x 630.8 x 1980 3617

2.2 - 24Copyright © 2010 Pearson Education, Inc.

Example 10: Modeling dataExample 10: Modeling dataSolutionSolution (continued) (continued)

(b) This slope indicates that tuition and fees have risen, on average, $630.80 per year.

2.2 - 25Copyright © 2010 Pearson Education, Inc.

Example 10: Modeling dataExample 10: Modeling dataSolutionSolution (continued) (continued)

(c) Evaluate f(1998).

f 1998 630.8 1998 1980 3617

14,971.40

This value differs from the actual value by less than $300 and involves interpolation.

Book Problems

• P. 108: 29, 35, 38, 45, 47, 50, 52, 75-78, 87