Copyright © 2010 Pearson Education, Inc. 2.1Linear Functions and Models 2.2Equations of Lines 2.3Linear Equations 2.4Linear Inequalities 2.5 Piece-wise Defined Functions Linear Functions and Equations 2

Copyright © 2010 Pearson Education, Inc. Equations of Lines ♦Write the point-slope and slope-intercept forms for a line ♦Find the intercepts of a line ♦Write equations for horizontal, vertical, parallel, and perpendicular lines ♦Model data with lines and linear functions (optional) ♦Use direct variation to solve problems 2.2

Copyright © 2010 Pearson Education, Inc. Point-Slope Form of the Equation of a Line The line with slope m passing through the point (x 1, y 1 ) has equation y = m(x  x 1 ) + y 1 y = m(x  x 1 ) + y 1 or or y  y 1 = m(x  x 1 ) y  y 1 = m(x  x 1 )

Copyright © 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 1Determining a point-slope form Solution (continued) Calculate the slope:

Copyright © 2010 Pearson Education, Inc. Solution (continued) Example 1Determining a point-slope form Substitute (1, 3) for (x 1, y 1 ) and 2 for m y = m(x –x 1 ) + y 1 y = 2(x – 1) + 3 Or substitute (–2, –3) for (x 1, y 1 ) and 2 for m y = m(x –x 1 ) + y 1 y = 2(x – (–2)) + (–3) y = 2(x + 2) – 3

Copyright © 2010 Pearson Education, Inc. Solution (continued) Example 1Determining a point-slope form Here’s the graph:

Copyright © 2010 Pearson Education, Inc. Slope-Intercept Form of the Equation of a Line The line with slope m and y-intercept b is given byThe line with slope m and y-intercept b is given by y = mx + b y = mx + b

Copyright © 2010 Pearson Education, Inc. Find the slope-intercept form for the line passing through the points (–2, 1) and (2, 3). Example 3Finding slope-intercept form Determine m and b in the form y = mx + b Solution Substitute either point to find b, use (2, 3).

Copyright © 2010 Pearson Education, Inc. An equation of a line is in standard form when it is written as ax + by = c where a, b, and c are constants. Some examples are: Standard Form

Copyright © 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 Intercepts

Copyright © 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 5Finding Intercepts To find the x-intercept, let y = 0, solve for x: Solution The x-intercept is 1.5

Copyright © 2010 Pearson Education, Inc. To find the y-intercept, let x = 0, solve for y: Example 5Finding Intercepts Solution (continued) The y-intercept is 2. The graph passes through the points (1.5, 0) and (0, 2).

Copyright © 2010 Pearson Education, Inc. Graph of a constant function fGraph of a constant function f Formula: f (x) = bFormula: f (x) = b Horizontal line with slope 0 and y-intercept b.Horizontal line with slope 0 and y-intercept b. (-3, 3)(3, 3) Horizontal Lines Note that regardless of the value of x, the value of y is always 3.

Copyright © 2010 Pearson Education, Inc. Vertical Lines Cannot be represented by a function Slope is undefined Equation is: x = k Note that regardless of the value of y, the value of x is always 3.Note that regardless of the value of y, the value of x is always 3. Equation is x = 3 (or x + 0y = 3)Equation is x = 3 (or x + 0y = 3) Equation of a vertical line is x = k where k is the x-intercept.Equation of a vertical line is x = k where k is the x-intercept.

Copyright © 2010 Pearson Education, Inc. Equations of Horizontal and Vertical Lines 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.

Copyright © 2010 Pearson Education, Inc. Parallel Lines Two lines with slopes m 1 and m 2, neither of which is vertical, are parallel if and only if their slopes are equal; that is, m 1 = m 2.

Copyright © 2010 Pearson Education, Inc. Perpendicular Lines Two lines with nonzero slopes m 1 and m 2,, are perpendicular if and only if their slopes have a product of –1; that is, m 1 m 2 = –1.

Copyright © 2010 Pearson Education, Inc. Example 8: Finding perpendicular lines Find the slope-intercept form of the line perpendicular to passing through the point (–2, 1). Graph the lines. The line has slope Solution The negative reciprocal is Use the point-slope form of the line...

Copyright © 2010 Pearson Education, Inc. Example 8: Finding perpendicular lines Solution (continued)

Copyright © 2010 Pearson Education, Inc. Example 10: Modeling data The 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 – x 1 ) + y 1, that models the data. Interpret the slope m. (c)Use to estimate tuition and fees in Compare the estimate to the actual value of $14,709. Did your answer involve interpolation or extrapolation?

Copyright © 2010 Pearson Education, Inc. Example 10: Modeling data Solution (a)Make a scatterplot of the data.

Copyright © 2010 Pearson Education, Inc. Example 10: Modeling data Solution (continued) (b)Choose any point, say (1980, 3617) to use for (x 1, y 1 ) and write: Choose two points to estimate the slope, say (1980, 3617) and (2000, 16,233)

Copyright © 2010 Pearson Education, Inc. Example 10: Modeling data Solution (continued) (b)This slope indicates that tuition and fees have risen, on average, $ per year.

Copyright © 2010 Pearson Education, Inc. Example 10: Modeling data Solution (continued) (c)Evaluate f(1998). 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, , 87