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Looking closely at a function

Looking closely at a function. Chapter 9. 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 + 4 u + 5

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Looking closely at a function

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  1. Looking closely at a function Chapter 9

  2. Quick Review of Parameterization y = 3x2 + 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 + u2) + 24 + 4u + 5 2 + v = 108 +36u + 3u2+ 24 + 4u + 5 2 + v = 108 +36u+ 3u2+ 24 +4u + 5 v = 135 + 40u + 3u2 x -6 = u y - 2 = v y – 2 = 135 + 40 (x - 6) + 3 (x2 – 12x + 36) ....

  3. Linear Approximation using Polynomial Parameterization (1,2) y = 4x2 - 2

  4. y = 4x2 – 2 (1,2) • What do we do first parameterize the function?

  5. Now how about a circle… • x2 + y2 = 25 • What is the equation of the tangent line at (3,4)

  6. Now how about a circle… • x2 + y2 = 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 = 2x2 + 3x + 1

  9. Find the tangent line when x = -2 • y = 2x2 + 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 = 2u2 – 8u + 8 + 3u – 6 + 1 • v = -5u + 2u2 • 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) = x2+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) = x2+5x-24 • at x = 12 • What information can you find using this...

  14. Looking at the tangent line... • f(x) = x2+5x-24 • at x = 12 • Find the tangent line

  15. Looking at the tangent line... • f(12) = x2+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) = x2+5x-24 A(12,190) x=12+u y=190+v • 190+v= (12+u)2+5(12+u)-24 • 190+v=144+24u+u2+70+5u-24 • v-29u-u2=0 x - 12=u y- 190=v

  17. Using parameters... • v-29u-u2=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-u2=0 • y – 190 – 29(x-12) = 0 • y – 190 – 29x + 348=0 • y = 29x - 158 - 158 x-12=u y-190=v

  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) 348 - 158 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) 348 - 158 12

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

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

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