1. UNIT 1B LESSON 7 USING LIMITS TO FIND TANGENTS 1

2. Slopes of Secant Lines 2 The slope of secant PQ is given by

3. Slopes of Tangent Lines 3 As the difference in the x values of points P and Q approaches ZERO we can express the slope of a tangent line as the following limit.

4. We want to find the slope and the equation of any tangent line to the curve y = 2x 2 + 4x – 1 using the general slope formula and having h (the change in x) approach 0. 4 m = lim [2(x + h) 2 + 4(x + h) – 1] – [2x 2 + 4x – 1] h→0 (x + h) – x m = lim [2x 2 + 4xh + 2h 2 + 4x + 4h – 1 – 2x 2 – 4x + 1] h→0 h m = lim [2(x 2 + 2xh + h 2) + 4(x + h) – 1] – [2x 2 + 4x – 1] h→0 (x + h) – x Lesson 7 EXAMPLE 1 Page 1 m = lim [ 4xh + 2h 2 + 4h] h→0 h

5. Lesson 7 EXAMPLE 1 (continued) Page 1 5 m = lim [4x + 2h + 4] h→0 m = 4x + 4 The equation for the slope of any tangent line = m = 4x + 4 m = lim [ 4xh + 2h 2 + 4h] h→0 h m = lim h(4x + 2h + 4) h→0 h m =[4x + 2(0) + 4]

6. Lesson 7 Page 1 con’t 6 The equation for the slope of any tangent line m = 4x + 4 Equation of tangent line This equation for the slope of any tangent line can be used for any x value 15 = 12(2) + b b = – 9 y = 12x – 9

7. Lesson 7 Page 1 con’t 7 The equation for the slope of any tangent line = m = 4x + 4 This equation for the slope of any tangent line can be used for any x value Equation of tangent line – 3 = 0(– 1) + b b = – 3 y = – 3

8. Lesson 7 Page 1 con’t 8 The equation for the slope of any tangent line = m = 4x + 4 Equation of tangent line This equation for the slope of any tangent line can be used for any x value 5 = –8(–3) + b b = –19 y = – 8x – 19

9. 9 Practice Question #1 Find the equation for the slope of the tangent line to the parabola y = 2x – x 2 m = lim [2(x + h) – (x + h) 2 ] – [2x – x 2 ] h→0 (x + h) – x m = lim 2x + 2h – x 2 – 2xh – h 2 – 2x + x 2 h→0 h m = lim 2h – 2xh – h 2 h→0 h m = lim 2 – 2x – h = 2 – 2x – 0 = 2 – 2x h→0

10. 10 Practice Question #1a Find the equation the tangent line to the parabola y = 2x – x 2 when x = 2 Point of tangency (2, 0) Slope = m = 2 – 2x = 2 – 2(2) = – 2 0 = – 2(2) + b b = 4 y = –2 x + 4

11. 11 Practice Question #1b Find the equation the tangent line to the parabola y = 2x – x 2 when x = –3 Point of tangency (-3, -15) Slope = m = 2 – 2x = 2 – 2(–3 ) = 8 –15 = 8(–3) + b b = 9 y = 8x + 9

12. 12 Practice Question #1c Find the equation the tangent line to the parabola y = 2x – x 2 when x = 0 Point of tangency (0, 0) Slope = m = 2 – 2x = 2 – 2(0) = 2 0= – 2(0) + b b = 0 y = 2x

13. 13 Practice Question #2a Find the equation for the slope of the tangent line to the parabola y = x 2 + 4x – 1 m = lim [(x + h) 2 + 4(x + h) – 1 ] – [x 2 + 4x – 1] h→0 (x + h) – x m = lim x 2 + 2xh + h 2 + 4x + 4h – 1 – x 2 – 4x + 1 h→0 h m = lim 2xh + h 2 + 4h h→0 h m = lim 2x + h + 4 h→0 Slope = m =2x + 0 + 4 = 2x + 4

14. 14 Practice Question #2 a Find the equation of the tangent line to the parabola y = x 2 + 4x – 1 when x = –3 y = (–3) 2 + 4(–3) – 1 = – 4 Point of tangency (-3, –4) Slope = m = 2x + 4 = 2(–3) + 4 = –2 – 4 = –2(–3) + b b = – 10 y = –2x – 10

15. 15 Practice Question #2 b Find the equation of the tangent line to the parabola y = x 2 + 4x – 1 when x = –2 y = (–2) 2 + 4(–2) – 1 = – 5 Point of tangency (-2, –5) Slope = m = 2x + 4 = 2(–2) + 4 = 0 – 5 = 0(–2) + b b = –5 y = –5

16. 16 Practice Question #2 c Find the equation of the tangent line to the parabola y = x 2 + 4x – 1 when x = 0 y = (0) 2 + 4(0) – 1 = – 1 Point of tangency (0, – 1 ) Slope = m = 2x + 4 = 2(0) + 4 = 4 – 1 = 4(0) + b b = – 1 y = 4x – 1

17. Consider this: Lesson 7 Page 4 17

18. Working with Compound fractions 18 OR

19. Lesson 7 Page 4 Example 2 19 Find the slope of the tangent to at the point where x = 3. continued →

20. so at x = 3 the slope of the tangent is 20 Lesson 7 Page 4 Example 2 con’t

21. = PRACTICE QUESTION 3 Find the slope of the tangent to at the point where x = 5. 21 continued →

22. = so at x = 5 the slope of the tangent is 22 Practice question 3 con’t

23. continued → Practice Question 4 Find the slope of the tangent to at the point where x = 1. 23

24. continued → Practice Question 4 con`t 24