1. MATH 31 LESSONS Chapter 1: Limits 1. Linear Functions and Tangents

2. Section 1.1: Linear Functions and The Tangent Problem Read Textbook pp. 5 - 9

3. A. Linear Functions Recall y = m x + b where m = slope b = y-intercept x y b

4. x y (x 1, y 1 ) (x 2, y 2 )  x = x 2 - x 1  y = y 2 - y 1

5. Note:  is called the rate of change of y with respect to x i.e. how quickly y changes relative to changes in x

6.  is called the rate of change of y with respect to x i.e. how quickly y changes relative to changes in x  If  y is positive, then y is increasing If  y is negative, then y is decreasing

7. e.g. How can this be interpreted? x y  x = 2  y = 5

8. When x increases by 2 units, y increases by 5 units. x y  x = 2  y = 5

9. Clearly, the greater the slope the greater the rate of change of y (with respect to x) the “faster” y changes Low rate of change High rate of change

10. Note: 

11. Ex. 1Find the equation of the linear function that passes through A(2, 1) and B(-4, -9). Answer in general form. Try this example on your own first. Then, check out the solution.

12. A(2, 1) and B(-4, -9)  Find slope x1x1 y1y1 x2x2 y2y2

13. A(2, 1)  Use point-slope formula: x1x1 y1y1

14. Multiply both sides by 3 to remove the denominator

15. Ax + By + C = 0 A > 0 A, B, and C are integers

16. Ex. 2For the linear function 32x + 4y - 19 = 0, if x decreases by 2, how does y change? Try this example on your own first. Then, check out the solution.

17.  Find slope

18. Put in slope-intercept form: y = mx + b

19. The slope is -8

20.  Find the change in y (  y) :

22. So, y increases by 16

23. B. The Tangent Problem What is a tangent line? The question is more challenging than it appears, and as we will see, only calculus can truly answer it.

24. For circles, the tangent line is readily defined. A tangent is a straight line that touches the circle only once. Tangent, t Not tangent

25. A tangent line is a straight line that touches the circle only once However, the definition above is not adequate for general curves. To show why, consider the next two illustrations.

26. Consider the following parabola: y = f (x)

27. Both l and t touch the curve only once. However, only t is a tangent. l t

28. This line t is a tangent. t

29. For an even more interesting example, consider the cubic function below: y = f (x)

30. The line t is a tangent line, even though it crosses the curve twice! t

31. So, the statement... “a tangent line is a straight line that touches a curve only once” is clearly inadequate.

32. C. An Approach to the Tangent Problem Much of this unit is devoted to solving the tangent problem. To illustrate the approach we will take, consider the following problem.

33. Question:Find the slope of the tangent line to the parabola y = x 2 at the point P(2, 4). y = x 2 y x 2 4 t P(2, 4)

34. Problem: In order to find the slope of the tangent line, we need to know two points on the tangent line. However, we only know one. y = x 2 y x2 4 t P(2, 4)

35. Solution: We introduce a second point on the curve y = x 2, somewhere close to P. We will call this point Q, and it will have the coordinates (x, x 2 ). y = x 2 x2 4 t P(2, 4) Q(x, x 2 ) x2x2

36. The line l that connects P and Q is called a secant line. It can be thought of as an approximation of the tangent line. y = x 2 x2 4 t P(2, 4) Q(x, x 2 ) x2x2 l

37. The slope of the secant line l is given by y = x 2 x2 4 t P(2, 4) Q(x, x 2 ) x2x2 l

38. The slope of the secant line l is given by y = x 2 x2 4 t P(2, 4) Q(x, x 2 ) x2x2 l

39. The slope of the secant line l is given by y = x 2 x2 4 t P(2, 4) Q(x, x 2 ) x2x2 l

40. The slope of the secant line l is given by y = x 2 x2 4 t P(2, 4) Q(x, x 2 ) x2x2 l

41. How can we use the secant slope to find the tangent slope? The main idea is to start moving Q closer and closer to the point P. y = x 2 x2 4 t P(2, 4) Q(x, x 2 ) x2x2 l

42. Notice that as Q approaches P... x approaches 2 l approaches t m PQ approaches m t y = x 2 x2 4P(2, 4) Q(x, x 2 ) x2x2 tl

43. In fact, if Q gets really close to P, the secant line l is almost identical to the tangent line t It follows that the slope of the secant line would be almost identical to the slope of the tangent. y = x 2 x2 4P(2, 4) Q(x, x 2 ) x2x2 tl

44. In calculus, we attempt to bring Q right to P, so that the line l actually becomes the tangent line t. In this way, the slope is exactly the same. But how do we do this? We use the idea of limits. y = x 2 2 4P(2, 4)Q tl

45. We will now find the pattern of slopes as Q approaches P (i.e. as x approaches 2 “from the left”) y = x 2 x2 4 t P(2, 4) Q(x, x 2 ) x2x2 l xm t = x + 2 1 1.5 1.9 1.99 1.999

46. Slope of the secant as x approaches 2 from the left: xm t = x + 2 1 1.5 1.9 1.99 1.999 3 3.5 3.9 3.99 3.999

47. Slope of the secant as x approaches 2 from the left: xm t = x + 2 1 1.5 1.9 1.99 1.999 3 3.5 3.9 3.99 3.999 24 As x approaches 2 from the left, the slopes appear to be approaching 4

48. We should also find the slope of the secants as x approaches 2 from the right as well: xm t = x + 2 3 2.5 2.1 2.01 2.001

49. So, as x gets extremely close to 2, the slope of the secant lines get extremely close to 4. In fact, in calculus we state that if x gets infinitely close to 2 (e.g. x = 1.9 ), the slopes would become so close to 4 (e.g. slope = 3.9 ) as to become actually equal to 4.

50. So, as x gets extremely close to 2, the slope of the secant lines get extremely close to 4. In fact, in calculus we state that if x gets infinitely close to 2 (e.g. x = 1.9 ), the slopes would become so close to 4 (e.g. slope = 3.9 ) as to become actually equal to 4. This value of 4 would represent the true slope of the tangent. We say that the slope of the tangent is the limit of the secant slopes, because it represents the limit (endpoint) of slopes as x gets infinitely close to 2.

51. Notation: If the tangent slope is the limit of the secant slopes, then we write: We say, “The limit as Q approaches P of the slopes of secant line PQ is the slope of the tangent.”

52. If the tangent slope is the limit of the secant slopes, then we write: or

53. Note: We could also find the equation of the tangent, since we now know the slope and one point. y = x 2 2 4P(2, 4) t m t = 4

54. Equation of tangent:P(2, 4)m t = 4 or

55. Ex. 3For the function y = x 2 - 5x, determine the equation of the tangent line at the point P(2, -6). Be certain to: - include a sketch - find the formula for the secant slope - see what the slopes approach as x  -2 from both sides to get the tangent slope Try this example on your own first. Then, check out the solution.

56. Sketch: y = x 2 - 5x y x 2 -6 t P(2, -6)

57. We introduce a second point Q (x, x 2 - 5x) on the curve. The point PQ is the secant line. As Q approaches P, the secant approaches the tangent. y = x 2 - 5x y x 2 -6 t Q(x, x 2 - 5x) P(2, -6)

58. Slope of secant: y = x 2 - 5x y x 2 -6 t P(2, -6) Q(x, x 2 - 5x)

59. Slope of secant:

61. Slope of the secant as x approaches 2 from the left: xm t = x - 3 1 1.5 1.9 1.99 1.999

62. Slope of the secant as x approaches 2 from the left: xm t = x - 3 1 1.5 1.9 1.99 1.999 -2 -1.5 -1.1 -1.01 -1.001 2 As x approaches 2 from the left, the slopes appear to be approaching -1

63. Slope of the secant as x approaches 2 from the right: xm t = x - 3 3 2.5 2.1 2.01 2.001

64. Slope of the secant as x approaches 2 from the left: xm t = x - 3 3 2.5 2.1 2.01 2.001 0 -0.5 -0.9 -0.99 -0.999 2 As x approaches 2 from the right, the slopes appear to be approaching -1 as well

65. The tangent slope is the limit of the secant slopes. So, it follows that

66. Equation of tangent:P(2, -6)m t = -1

67. or