1. UNIT 1 ASSESSMENT PROJECT Skylar Larson

2. 0.00 seconds 1.50 seconds1.00 seconds 0.50 seconds 3.50 seconds3.00 seconds2.50 seconds2.00 seconds

4. 6.01905x^2-5.4x+3.55950.0 < x < 0.50 -12.496696x^2+43.852139x-21.19176580.50 < x < 2.875 -4.3429x^2+36.62286x-68.632.875 < x < 3.50 f(x)= E displacement equations

5. Height vs. Time Graph

6. Velocity equations f’(x)= -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

7. Velocity vs. Time Graph

8. AVERAGE VELOCITY  T=0s  T=3.5s 3.5ft 6.2ft Average Velocity= (6.2 - 3.5)/(3.5 - 0) = 0.77143 ft/s

9. INSTANTANEOUS VELOCITY AT T=2.0SECONDS f’(2) = -6.135 ft/s f’(x)= -24.99339x+43.852139 0.50 < x < 2.875 2.0 seconds (From preview page)

10. INSTANTANEOUS VELOCITY AT T=3.5 SECONDS f’(x)= 6.2226 ft/s f’(x)= -8.6858x+36.622862.875 < x < 3.50 (From preview page) 3.5 seconds

11. DID THE BALL EVER TRAVEL AT 5 M/S (16.404 FT/S ) ? 12.0381x-5.4 = 16.4040.0 < 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.

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

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

14. Part 3 f(x)= -2x+4 -1 < x < 1 -(x-1)^2+2 1 < x < 4 -0.5|x-8|-5 4 < x < 12 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 + - The limit does not exist at x = 4 because the left and right limits don’t equal each other.

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

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

17. 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)

18. 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)