Section 18-1: Graphical Representation of Linear Equations and Functions Prepare a table of solutions and locate the solutions on a coordinate system: f(x) = 2x 5 Learning Outcome 2 Write x + 3 = 5 as a function and graph the function. x + 3 = 5 Set equation equal to zero. x 2 = 0 Write in function notation. f(x) = x 2 Assign x values and find f(x). f(0) = 0 2 = 2 Let x = 0 f(2) = 2 2 = 0 Let x = 2 f(4) = 4 2 = 2 Let x = 4 Learning Outcome 3 A solution for the equation 2x 3y = 6 is (6, 2). Verify the solution. 2x 3y = 6 Substitute 6 for x and 2 for y. 2(6) 3(2) = 6 12 6 = 6 6 = 6 True Learning Outcome 4 Solve 3x + y = 7 3x + y = 7 Substitute 16 for y. 3x + 16 = 7 Sort terms. 3x = 7 16 Combine terms.
3x = 9 x = 3 Divide. Section 18-2: Graphing Linear Equations with Two Variables Using Alternative Methods Graph by using the intercepts method: 2x 5y = 10 2x 5y = 10 To find x-intercept, let y = 0. 2x 5(0) = 10 2x = 10 The coordinate pair is (5, 0). x = 5 2x 5y = 10 To find the y-intercept, let x = 0. 2(0) 5y = 10 5y = 10 y = 2 The coordinate pair is (0, 2). Locate the two intercepts using the coordinate pairs. Draw the graph through the two points. Learning Outcome 2 2 Use the slope-intercept method to graph: y = x 4 3 The slope-intercept form of the linear equation is y = mx + b where m = slope and b = y-coordinate of the y-intercept. 2 2 y = x 4 m = ; b = 4 3 3 First, locate the y-intercept, (0, 4). Then use the slope, 3 2, and begin at the point (0, 4) and count up 2 (rise) and right 3 (run) to locate the second point. The coordinates of this second point are (4, 2). Draw the line through the two points..
Learning Outcome 3 Use your graphing calculator to graph the equation y = 3x + 2. Before you graph the equation, examine it and predict where it will cross the y-axis and describe its slope. From the equation it is determined the line will cross the y-axis 2 units above the x-axis. Because the equation has a negative slope, it might be expected to slope upward to the left. Learning Outcome 4 Write the equation as a function, graph the function and find the solution from the graph. 3x 5= 7 3x 12= 0 Set the equation equal to zero. f( x) = 3x 12 Write in function notation. f ( 5) = 3( 5) 12 = 15 12 = 27 f ( 0) = 3( 0) 12 = 0 12 = 12 f ( 4) = 3( 4) 12 = 12 12 = 0 f ( 6) = 3( 6) 12 = 18 12 = 6 The line crosses the x axis at x = 4. Thus the solution is x = 4.
Section 18-3: Graphing Linear Inequalities with Two Variables Graph the solution set for y < 3x + 4. Graph the equation y = 3x + 4, which is the boundary. Use 3 the slope-intercept method. The y-intercept is (0, 4) and the slope is. The line is dashed to 1 indicate that it is not included in the solution set. Test (0, 0) to see if it makes a true statement with the inequality. This determines which side of the boundary is shaded as the solution set. y < 3x + 4 0 < 3(0) + 4? 0 < 4 True Therefore, shade the side of boundary that includes the point (0, 0) (see graph). Section 18-4: Graphing Quadratic Equations and Inequalities Identify the linear and quadratic equations from the following. (a) 5x 2 3x = 10 (c) x y = 8 (b) 2xy 3 = y (d) x = y 3 2x + 3 (a) quadratic equation (c) linear equation (b) quadratic equation (d) cubic equation 2xy has degree 2. Prepare a table of solutions of five ordered pairs. Plot the points on a rectangular coordinate system. Connect the points with a smooth curve. 2 y x + 6x+ 8
Learning Outcome 2 Graph: y = x 2 4x 12 b 4 4 x = = = + 2 2 a 2(1) 2 Find the x-coordinate of the vertex by using b. 2a b Axis of symmetry = x =. 2a x = 2 y = (+2) 2 4(+2) 12 Find y-coordinate of the vertex by substituting. y = 4 8 12 y = 16 (+2, 16) Coordinates of the vertex. Find the x-intercepts. 0 = x 2 4x 12 Substitute y = 0. 0 = (x 6) (x + 2) x 6 = 0 x + 2 = 0 x = 6 x = 2 Sketch graph.
Learning Outcome 3 Examine the graph of the equation y = x 2 2x 35 to find the solutions. The graph crosses the x-axis at (6,0) and ( 4, 0). The solutions are x = 4 and x = 6. Learning Outcome 4 Use a calculator to graph the equation: y x 2 3x 10.Clear and initialize the calculator.
Section 18-5: Graphing Other Nonlinear Equations Prepare a table of solutions and graph the function f(x) = 4 x Learning Outcome 2 Use the vertical line test to determine which graph is the graph of a relation and which is the graph of a function. (a) (b) (a) is a relation and not a function because a vertical line intersects the graph in 2 points. (b) is a function because any vertical line intersects the graph in only one point. Learning Outcome 3 Determine the domain and range of the relation. The domain is all values on the graph for the independent variable (x-values). [ 6, 6] The range is all values on the graph for the dependent variable (y-values). [ 3, 3]