UNIT 1 ASSESSMENT PROJECT

Skylar Larson

0.00 seconds 1.50 seconds1.00 seconds0.50 seconds

3.50 seconds3.00 seconds2.50 seconds2.00 seconds

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)=

E

displacement equations

Height vs. Time Graph

Velocity equations

f’(x)=

-24.99339x+43.85213912.0381x-5.4 0.50 < x < 2.875-8.6858x+36.62286 2.875 < x < 3.50

0.0 < x < 0.50

Velocity vs. Time Graph

AVERAGE VELOCITY

T=0s

T=3.5s

3.5ft

6.2ft

Average Velocity= (6.2 - 3.5)/(3.5 - 0)

= 0.77143 ft/s

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)

INSTANTANEOUS VELOCITY AT T=3.5 SECONDS

f’(x)= 6.2226 ft/s

f’(x)= -8.6858x+36.62286 2.875 < x < 3.50(From preview page)

3.5 seconds

DID THE BALL EVER TRAVEL AT 5 M/S ( 1 6 . 4 0 4 F T / S ) ?

12.0381x-5.4 = 16.404 0.0 < x < 0.50x=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.

Part 2

Definition of Derivative: lim = (f(x+h) – f(x-h))/(2h)

h 0

ft/s

ft/s

Instantaneous rate of change at 2.0625 seconds

2.0625 seconds

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+

x 4-

They are not equal

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 4Lim (x)= -7 x 4

+

-

• The limit does not exist at x = 4 because the left and right limits don’t equal each other.

Part 4

•Has a limit approaching infinity, as x approaches infinity.

F(x)= (3x^2+4x)/(2x+7)

Horizontal Asymptotes: None

Part 4

•Has a limit approaching 0, as x approaches infinity.

F(x)= (7x^2+3)/(2x^4+x)

Horizontal Asymptotes: y=0

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)

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)