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UNIT 1B LESSON 7 USING LIMITS TO FIND TANGENTS

UNIT 1B LESSON 7 USING LIMITS TO FIND TANGENTS. Slopes of Secant Lines. The slope of secant PQ is given by. Slopes of Tangent Lines. 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. .

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UNIT 1B LESSON 7 USING LIMITS TO FIND TANGENTS

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  1. UNIT 1B LESSON 7USING LIMITS TOFIND TANGENTS

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

  3. Slopes of Tangent Lines 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. Lesson 7 EXAMPLE 1 Page 1 We want to find the slope and the equation of any tangent line to the curve y = 2x2 + 4x – 1using thegeneral slope formula and having h (the change in x) approach 0. m = lim [2(x + h)2 + 4(x + h) – 1] – [2x2 + 4x – 1] h→0 (x + h) – x m = lim [2(x2 + 2xh + h2)+ 4(x + h) – 1] – [2x2 + 4x – 1] h→0 (x + h) – x m = lim [2x2 + 4xh + 2h2 + 4x + 4h – 1 – 2x2 – 4x + 1] h→0 h m = lim[ 4xh+ 2h2+ 4h] h→0 h

  5. Lesson 7 EXAMPLE 1 (continued) Page 1 m = lim[ 4xh+ 2h2+ 4h] h→0 h m = limh(4x+ 2h+ 4) h→0 h m = lim [4x + 2h + 4] h→0 m=[4x + 2(0) + 4] m = 4x + 4 The equation for the slope of any tangent line = m = 4x + 4

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

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

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

  9. Practice Question #1 Find the equation for the slope of the tangent line to the parabola y = 2x – x2 m = lim [2(x + h) – (x + h)2] – [2x– x2] h→0 (x + h) – x m = lim2x + 2h – x2 – 2xh – h2– 2x + x2 h→0h m = lim 2h – 2xh – h2 h→0 h m = lim2 – 2x – h = 2 – 2x – 0 = 2 – 2x h→0

  10. Practice Question #1a Find the equation the tangent line to the parabola y = 2x – x2 when x = 2 Slope = m = 2 – 2x = 2 – 2(2) = – 2 Point of tangency (2, 0) 0 = – 2(2) + b b = 4 y = –2 x + 4

  11. Practice Question #1b Find the equation the tangent line to the parabola y = 2x – x2 when x = –3 Slope = m = 2 – 2x = 2 – 2(–3 ) = 8 Point of tangency (-3, -15) –15 = 8(–3) + b b = 9 y = 8x + 9

  12. Practice Question #1c Find the equation the tangent line to the parabola y = 2x – x2 when x = 0 Slope = m = 2 – 2x = 2 – 2(0) = 2 Point of tangency (0, 0) 0= – 2(0) + b b = 0 y = 2x

  13. Practice Question #2a Find the equation for the slope of the tangent line to the parabola y = x2 + 4x – 1 m = lim[(x + h)2 + 4(x + h) – 1 ] – [x2 + 4x – 1] h→0 (x + h) – x m = limx2+ 2xh + h2 + 4x + 4h – 1 – x2 – 4x + 1 h→0 h m = lim2xh + h2+ 4h h→0 h m = lim2x + h + 4 h→0 Slope = m =2x + 0 + 4 = 2x + 4

  14. Practice Question #2 a Find the equation of the tangent line to the parabola y = x2 + 4x – 1 when x = –3 Slope = m = 2x + 4 = 2(–3) + 4 = –2 y = (–3)2 + 4(–3) – 1 = – 4 Point of tangency (-3, –4) – 4 = –2(–3) + b b = – 10 y = –2x – 10

  15. Practice Question #2 b Find the equation of the tangent line to the parabola y = x2 + 4x – 1 when x = –2 Slope = m = 2x + 4 = 2(–2) + 4 = 0 y = (–2)2 + 4(–2) – 1 = – 5 Point of tangency (-2, –5) – 5 = 0(–2) + b b = –5 y = –5

  16. Practice Question #2 c Find the equation of the tangent line to the parabola y = x2 + 4x – 1 when x = 0 Slope = m = 2x + 4 = 2(0) + 4 = 4 y = (0)2 + 4(0) – 1 = – 1 Point of tangency (0, – 1 ) – 1 = 4(0) + b b = – 1 y = 4x – 1

  17. Lesson 7 Page 4 Consider this:

  18. Working with Compound fractions OR OR

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

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

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

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

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

  24. Practice Question 4 con`t continued→

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