1. 1 § 1-1 Lines The student will learn about straight lines.

2. Introduction We will begin this course with the study of straight lines. This is, of course, a topic which you should all be very well acquainted. Please rest assured the material will become more challenging quickly. Let’s have a good start.

3. 3 Slope of a Line 1. Def: If P 1 (x 1, y 1 ) and P 2 (x 2, y 2 ) are two points on a line then the slope of that line is given by the formulas: 2. Discuss slope.

4. 4 Slope-Intercept form of a Line Def: The equation y = mx + b is called the slope-intercept form of a line. The slope is m and the y intercept is b. We will use this form to graph a line.

5. 5 Slope-Intercept Form of a Line The slope-intercept form when graphing. y = 2x - 3 We first plot the y intercept (0, - 3). We then plot another point 2 units up and 1 unit over from the y – intercept. (Slope = 2/1)! We finish by connecting those points to form the line.

6. 6 Graphing: Calculator Sketch a graph of y = 2x – 1. By calculator 1. Turn your calculator on and in the y = window enter 2x – 1. 2. On the zoom screen choose “zoom standard” and touch “enter”.

7. 7 Point-Slope Form of a Line Def: The equation y – y 1 = m (x – x 1 ) is called the point-slope form of a line. The slope is m and (x 1, y 1 ) is a point on the line. We will use this form to solve problems that ask for the equation of a line.

8. 8 Special Cases A vertical line has an equation of x = a. x = 2 A horizontal line has an equation of y = b. y = 1

9. 9 X - Intercepts The x intercept is 3 The x intercept of f (x) occurs where and if the graph of the function crosses the x-axis. Algebraically it is the x value where f (x) is zero, or (x, 0). That is, to find the x intercept, let f (x) = 0 and solve for x. Remember f(x) is y! 2x + 3y = 6 2x + 3· 0 = 6 2x = 6 x = 3

10. 10 Y - Intercept The y intercept is 2 That is, to find the y intercept let x = 0 and solve for f (x). 2x + 3y = 6 2· 0 + 3y = 6 3y = 6 The y intercept of f (x) occurs where and if the graph of the function crosses the y-axis. Algebraically it is the f (x) value where x is zero, or (0, f (x)). y = 2

11. 11 Intercepts on a Calculator 1. Consider 2x + 3y = 6. First solve the equation for y. 2. Graph this equation on your calculator. 3. To find the x-intercept use the “zero” function under the “Calc” menu. 4. To find the y-intercept use the “value” function under the “Calc” menu with an x value of 0. Note: x-intercept is 3 or (3,0) and the y-intercept is 2 or (0,2).

12. 12 Review Equations of a Line GeneralAx + By = C Not of much use. Test answers. Slope-Intercept Formy = mx + b Graphing on a calculator. Point-slope formy – y 1 = m (x – x 1 ) “Name that Line”. Horizontal liney = b Vertical linex = a

13. 13 Application Example Linear Depreciation. Office equipment was purchased for $20,000 and is assumed to have a scrap value of $2,000 after 10 years. If its value is depreciated linearly (for tax purposes) from $20,000 to $2,000: 1. Find the linear equation that relates value (V) in dollars to time (t) in years. We know two points. What are they? t = 0 and V = $20,000 (0, $20,000) and t = 10 and V = $2,000 (10, $2,000)

14. 14 Application Example Continued Linear Depreciation. Office equipment was purchased for $20,000 and is assumed to have a scrap value of $2,000 after 10 years. If its value is depreciated linearly for (tax purposes) from $20,000 to $2,000: With points (0, 20000) and (10, 2000) use the point-slope formula. With points (0, 20000) and (10, 2000) use the point-slope formula. (We could also use the slope-intercept formula on this one since we know that b = 20,000.)

15. 15 The slope is  V/  t or m = (2000 – 20000) / (10 – 0) = - 1800 You may use either point. I will choose (10, 2000). Application Example Continued Linear Depreciation. Office equipment was purchased for $20,000 and is assumed to have a scrap value of $2,000 after 10 years. If its value is depreciated linearly for (tax purposes) from $20,000 to $2,000: So V = -1800t + 20000. Points are (10, 2000) and (0, 20000). Substituting into V – V 1 = m (t – t 1 ), yields V – 2000 = - 1800 (t – 10) = - 1800 t + 18000

16. 16 Application Example Continued Linear Depreciation. Office equipment was purchased for $20,000 and is assumed to have a scrap value of $2,000 after 10 years. If its value is depreciated linearly for (tax purposes) from $20,000 to $2,000: V(6) = -1800 6 + 20000 = 2. What would be the value of the equipment after 6 years? V = -1800t + 20000. = $9,200 - 10800 + 20000

17. 17 Application Example Continued 3.Graph the equation V = -1800t + 20000. 0 ≤ t ≤ 10 5. Write a verbal interpretation of the slope of the line found in the first part of this problem. 4. Use “value” to find V (6) on the calculator.

18. 18 Summary. We did an applied problem involving a straight line graph and saw the meaning of the y-intercept and the slope of the graph. We learned about straight lines, slope, and the different forms for straight lines.

19. 19 ASSIGNMENT §1.1; Page 6; 1 to 29 odd problems.