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Unit 1 assessment project

Unit 1 assessment project. Skylar Larson. 0.00 seconds. 0.50 seconds. 1.00 seconds. 1.50 seconds. 2 .00 seconds. 2.50 seconds. 3.00 seconds. 3.50 seconds. E. displacement equations. 6.01905x^2-5.4x+3.5595 0.0 < x < 0.50 -12.496696x^2+43.852139x-21.1917658 0.50 < x < 2.875

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Unit 1 assessment project

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  1. Unit 1 assessment project Skylar Larson

  2. 0.00 seconds 0.50 seconds 1.00 seconds 1.50 seconds 2.00 seconds 2.50 seconds 3.00 seconds 3.50 seconds

  3. E displacement equations 6.01905x^2-5.4x+3.5595 0.0 < x < 0.50 -12.496696x^2+43.852139x-21.1917658 0.50 < x< 2.875 -4.3429x^2+36.62286x-68.63 2.875 < x < 3.50 f(x)=

  4. Height vs. Time Graph

  5. Velocity equations -24.99339x+43.852139 12.0381x-5.4 0.50 < x < 2.875 -8.6858x+36.62286 2.875 < x < 3.50 0.0 < x < 0.50 f’(x)=

  6. Velocity vs. Time Graph

  7. Average Velocity • T=0s • T=3.5s 3.5ft 6.2ft Average Velocity= (6.2 - 3.5)/(3.5 - 0) = 0.77143 ft/s

  8. Instantaneous Velocity at t=2.0seconds f’(x)= -24.99339x+43.852139 0.50 < x < 2.875 (From preview page) 2.0 seconds f’(2) = -6.135 ft/s

  9. Instantaneous velocity at T=3.5 seconds f’(x)= -8.6858x+36.62286 2.875 < x < 3.50 (From preview page) 3.5 seconds f’(x)= 6.2226 ft/s

  10. Did the ball ever travel at 5 m/s (16.404 ft/s)? 12.0381x-5.4 = 16.404 0.0 < x < 0.50 x=1.81125 -24.99339x+43.852139 = 16.404 0.50 < x < 2.875 x=1.09822 -8.6858x+36.62286 =16.404 2.875 < x <3.50 x=2.3278 f’(x)= The ball will reach 5m/s (16.404 ft/s) at 1.09822 seconds. The other two x values are not in the domain.

  11. Part 2 Instantaneous rate of change at 2.0625 seconds Definition of Derivative: lim = (f(x+h) – f(x-h))/(2h) h 0 ft/s ft/s 2.0625 seconds

  12. Part 3 -2x+4 -1<x < 1 -(x-1)^2+2 1 < x< 4 -0.5|x-8|+6 4 < x < 12 f(x)= They are not equal Not continuous at x=4 because lim f(x)=4 And lim f(x)=7 - x 4 x 4 +

  13. Part 3 • The limit does not exist at x = 4 because the left and right limits don’t equal each other. -2x+4 -1<x < 1 -(x-1)^2+2 1 < x< 4 -0.5|x-8|-5 4 < x < 12 f(x)= • Change function so there is a limit: move the absolute value equation/line down 11 units. Lim f(x) = -7 x 4 Lim (x)= -7 x 4 + -

  14. Part 4 • Has a limit approaching infinity, as x approaches infinity. F(x)= (3x^2+4x)/(2x+7) Horizontal Asymptotes: None

  15. Part 4 • Has a limit approaching 0, as x approaches infinity. F(x)= (7x^2+3)/(2x^4+x) Horizontal Asymptotes: y=0

  16. Part 4 • Has a limit approaching a line which is not 0, as x approaches infinity. F(x)= (3x^2+4x)/(4x^2) Horizontal Asymptotes: y= 0.75 (Found using coefficients)

  17. Part 4 • Has a limit approaching two separate lines as x approaches positive or negative infinity. F(x)= (|2x|)/(3x) Horizontal Asymptotes: y=2/3 and y= -2/3 (Found using coefficients)

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