Copyright © 2009 Pearson Education, Inc. CHAPTER 1: Graphs, Functions, and Models 1.1 Introduction to Graphing 1.2 Functions and Graphs 1.3 Linear Functions, Slope, and Applications 1.4 Equations of Lines and Modeling 1.5 Linear Equations, Functions, Zeros and Applications 1.6 Solving Linear Inequalities

Copyright © 2009 Pearson Education, Inc. 1.1 Introduction to Graphing  Plot points.  Determine whether an ordered pair is a solution of an equation.  Find the x-and y-intercepts of an equation of the form Ax + By = C.  Graph equations.  Find the distance between two points in the plane and find the midpoint of a segment.  Find an equation of a circle with a given center and radius, and given an equation of a circle in standard form, find the center and the radius.  Graph equations of circles.

Slide Copyright © 2009 Pearson Education, Inc. Cartesian Coordinate System

Slide Copyright © 2009 Pearson Education, Inc. To graph or plot a point, the first coordinate tells us to move left or right from the origin. The second coordinate tells us to move up or down. Plot (  3, 5). Move 3 units left. Next, we move 5 units up. Plot the point. (–3, 5) Example

Slide Copyright © 2009 Pearson Education, Inc. Solutions of Equations Equations in two variables have solutions (x, y) that are ordered pairs. Example: 2x + 3y = 18 When an ordered pair is substituted into the equation, the result is a true equation. The ordered pair has to be a solution of the equation to receive a true statement.

Slide Copyright © 2009 Pearson Education, Inc. Examples a. Determine whether the ordered pair (  5, 7) is a solution of 2x + 3y = 18. 2(  5) + 3(7) ? 18  ? = 18 FALSE (  5, 7) is not a solution. b. Determine whether the ordered pair (3, 4) is a solution of 2x + 3y = 18. 2(3) + 3(4) ? ? = 18 TRUE (3, 4) is a solution.

Slide Copyright © 2009 Pearson Education, Inc. Graphs of Equations To graph an equation is to make a drawing that represents the solutions of that equation.

Slide Copyright © 2009 Pearson Education, Inc. x-Intercept The point at which the graph crosses the x-axis. An x-intercept is a point (a, 0). To find a, let y = 0 and solve for x. Example: Find the x-intercept of 2x + 3y = 18. 2x + 3(0) = 18 2x = 18 x = 9 The x-intercept is (9, 0).

Slide Copyright © 2009 Pearson Education, Inc. y-Intercept The point at which the graph crosses the y-axis. A y-intercept is a point (0, b). To find b, let x = 0 and solve for y. Example: Find the y-intercept of 2x + 3y = 18. 2(0) + 3y = 18 3y = 18 y = 6 The y-intercept is (0, 6).

Slide Copyright © 2009 Pearson Education, Inc. Example We already found the x-intercept: (9, 0) We already found the y-intercept: (0, 6) We find a third solution as a check. If x is replaced with 5, then Graph 2x + 3y = 18. Thus, is a solution.

Slide Copyright © 2009 Pearson Education, Inc. Example (continued) Graph: 2x + 3y = 18. x-intercept: (9, 0) y-intercept: (0, 6) Third point:

Slide Copyright © 2009 Pearson Education, Inc. Example Graph y = x 2 – 9x – 12. (12, 24)2412 –2  32  26  12 –2 24 y (10, –2)10 (5,  32) 5 (4,  32) 4 (2,  26) 2 (0,  12) 0 (  1, –2) 11 (  3, 24) 33 (x, y)x Make a table of values.

Slide Copyright © 2009 Pearson Education, Inc. The Distance Formula The distance d between any two points (x 1, y 1 ) and (x 2, y 2 ) is given by

Slide Copyright © 2009 Pearson Education, Inc. Example Find the distance between the points (–2, 2) and (3,  6).

Slide Copyright © 2009 Pearson Education, Inc. Midpoint Formula If the endpoints of a segment are (x 1, y 1 ) and (x 2, y 2 ), then the coordinates of the midpoint are

Slide Copyright © 2009 Pearson Education, Inc. Example Find the midpoint of a segment whose endpoints are (  4,  2) and (2, 5).

Slide Copyright © 2009 Pearson Education, Inc. Circles A circle is the set of all points in a plane that are a fixed distance r from a center (h, k). The equation of a circle with center (h, k) and radius r, in standard form, is (x  h) 2 + (y  k) 2 = r 2.

Slide Copyright © 2009 Pearson Education, Inc. Example Find an equation of a circle having radius 5 and center (3,  7). Using the standard form, we have (x  h) 2 + (y  k) 2 = r 2 [x  3] 2 + [y  (  7)] 2 = 5 2 (x  3) 2 + (y + 7) 2 = 25.