UNIT 1 ASSESSMENT PROJECT Skylar Larson. 0.00 seconds 1.50 seconds1.00 seconds 0.50 seconds 3.50 seconds3.00 seconds2.50 seconds2.00 seconds.

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Presentation transcript:

UNIT 1 ASSESSMENT PROJECT Skylar Larson

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

x^2-5.4x < x < x^ x < x < x^ x < x < 3.50 f(x)= E displacement equations

Height vs. Time Graph

Velocity equations f’(x)= x x < x < x < x < < x < 0.50

Velocity vs. Time Graph

AVERAGE VELOCITY  T=0s  T=3.5s 3.5ft 6.2ft Average Velocity= ( )/( ) = ft/s

INSTANTANEOUS VELOCITY AT T=2.0SECONDS f’(2) = ft/s f’(x)= x < x < seconds (From preview page)

INSTANTANEOUS VELOCITY AT T=3.5 SECONDS f’(x)= ft/s f’(x)= x < x < 3.50 (From preview page) 3.5 seconds

DID THE BALL EVER TRAVEL AT 5 M/S ( FT/S ) ? x-5.4 = < x < 0.50 x= x = < x < x= x = < x < 3.50 x= f’(x)= The ball will reach 5m/s ( ft/s) at 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 Instantaneous rate of change at seconds seconds

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

Part 3 f(x)= -2x+4 -1 < x < 1 -(x-1)^2+2 1 < x < |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 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)