Unit 1 assessment project
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Transcript of Unit 1 assessment project
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UNIT 1 ASSESSMENT PROJECT
Skylar Larson
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0.00 seconds 1.50 seconds1.00 seconds0.50 seconds
3.50 seconds3.00 seconds2.50 seconds2.00 seconds
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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
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Height vs. Time Graph
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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
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Velocity vs. Time Graph
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AVERAGE VELOCITY
T=0s
T=3.5s
3.5ft
6.2ft
Average Velocity= (6.2 - 3.5)/(3.5 - 0)
= 0.77143 ft/s
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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)
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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
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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.
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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
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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
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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.
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Part 4
•Has a limit approaching infinity, as x approaches infinity.
F(x)= (3x^2+4x)/(2x+7)
Horizontal Asymptotes: None
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Part 4
•Has a limit approaching 0, as x approaches infinity.
F(x)= (7x^2+3)/(2x^4+x)
Horizontal Asymptotes: y=0
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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)
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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)