1. © The Visual Classroom 2.5a Linear Models Jennifer has a babysitting job for the summer and makes $6 per hour plus an additional $15 per day to cover her traveling expenses. We can model this situation by the following equation. A = 6h + 15 A = amount earned per day h = hours worked

2. © The Visual Classroom HoursAmount 0 1 2 3 4 5 6 7 8 15 21 27 33 39 Make a table of values. A = 6h + 15 45 51 57 63 6h + 15 6(0) + 15 = 15 6(1) + 15 = 21 6(2) + 15 = 27 6(3) + 15 = 33 6(4) + 15 = 39 6(5) + 15 = 45 6(6) + 15 = 51 6(7) + 15 = 57 6(8) + 15 = 63

3. © The Visual Classroom hA 0 1 2 3 4 5 6 7 8 15 21 27 33 39 A = 6h + 15 45 51 57 63 Sketch the graph hours worked amount earned 0 1 2 3 456 7 8 10 20 30 40 50 60 70 9101112 80 90 100

4. © The Visual Classroom A = 6h + 15 Sketch the graph hours worked amount earned 0 1 2 3 456 7 8 10 20 30 40 50 60 70 9101112 80 90 100 Determine the slope of the line. = 6 rise run What is the A-intercept? 15

5. © The Visual Classroom A = 6h + 15 Sketch the graph hours worked amount earned 0 1 2 3 456 7 8 10 20 30 40 50 60 70 9101112 80 90 100 What would happen to the shape of the graph if she made $7/hour but her traveling expenses stayed the same? Steeper slope What would the equation of the new line be? A = 7h + 15

6. © The Visual Classroom A = 6h + 15 Sketch the graph hours worked amount earned 0 1 2 3 456 7 8 10 20 30 40 50 60 70 9101112 80 90 100 What would happen to the shape of the graph if she made $6/hour but her traveling expenses were $20 per day? Same slope but the A-intercept would be higher. What would the equation of the new line be? A = 6h + 20