1. Looking closely at a function Chapter 9

2. Quick Review of Parameterization y = 3x 2 + 4x + 5 Point P(6,2) x = 6 + u y = 2 + v 2 + v = 3 (6 + u) 2 + 4(6 + u) + 5 2 + v = 3 (36 +12u + u 2 ) + 24 + 4u + 5 2 + v = 108 +36u + 3u 2 + 24 + 4u + 5 v = 135 + 40u + 3u 2 x -6 = u y - 2 = v y – 2 = 135 + 40 (x - 6) + 3 (x 2 – 12x + 36)....

3. Linear Approximation using Polynomial Parameterization y = 4x 2 - 2 (1,2)

4. y = 4x 2 – 2 (1,2) What do we do first parameterize the function?

5. Now how about a circle… x 2 + y 2 = 25 What is the equation of the tangent line at (3,4)

6. Now how about a circle… x 2 + y 2 = 25 (3,4)

7. Writing the equation of a line… What do you need to know? Then what do you do?

8. Find the tangent line when x = -2 y = 2x 2 + 3x + 1

9. Find the tangent line when x = -2 y = 2x 2 + 3x + 1 if x = -2 y = 2(-2) 2 + 3(-2) + 1 y = 8 – 6 + 1 y = 3 (-2, 3) x = u – 2 y = v + 3 v + 3 = 2(u - 2) 2 + 3(u – 2) + 1 v + 3 = 2u 2 – 8u + 8 + 3u – 6 + 1 v = -5u + 2u 2 y – 3 = -5 (x + 2) y = -5x - 7

10. y As indicated in the diagram (which is not to scale) the tangent line to the graph of f(x) = x 2 +5x-24 at x = 12 meets the x-axis at an angle AOB whose tangent is. The angle AOB measures radians.

11. How can you find angle measures?

12. Use trigonometry... What do you need to use trig? How can you find it given this situation?

13. Looking at the tangent line... f(x) = x 2 +5x-24 at x = 12 What information can you find using this...

14. Looking at the tangent line... f(x) = x 2 +5x-24 at x = 12 Find the tangent line

15. Looking at the tangent line... f(12) = x 2 +5x-24 f(12) = (12) 2 +5(12)-24 f(12) = 144 + 70 – 24 f(12) = 190 A(12,190)

16. Using parameters... f(x) = x 2 +5x-24 A(12,190) x=12+u y=190+v 190+v= (12+u) 2 +5(12+u)-24 190+v=144+24u+u 2 +70+5u-24 v-29u-u 2 =0 x - 12=u y- 190=v

17. Using parameters... v-29u-u 2 =0 y – 190 – 29(x-12) = 0 y – 190 – 29x + 348=0 y = 29x - 158 x-12=u y-190=v

18. Using parameters... v-29u-u 2 =0 y – 190 – 29(x-12) = 0 y – 190 – 29x + 348=0 y = 29x - 158 x-12=u y-190=v - 158

19. Looking at the tangent line... Diagram is not to scale. Now what do you know? (12,190) - 158

20. Looking at the tangent line... Diagram is not to scale. (12,190) - 158 348 12

21. Looking at the tangent line... Diagram is not to scale. Now you have 2 sides – you can find the desired angle. (12,190) - 158 348 12

22. Looking at the tangent line... Diagram is not to scale. 348/12 tan  = 348/12 (must find degree to change to radians) tan  = tan -1 (348/12). (12,190) - 158 348 12

23. Another way to find the tangent line? Derivative? What is the Derivative?