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course Mth 163
3/2 around 12:30 am
If water depths of 10, -6.2, -16.9 and -22.2 cm are observed at clock times 21.4, 32.1, 42.8 and 53.5 sec, then at what average rate does the depth change during each time interval?Sketch a graph of this data set and use a sketch to explain why the slope of this graph between 32.1 and 42.8 sec represents the average rate at which depth changes during this time interval.
If f(x) = x2, give the vertex and the three basic points of the graphs of f(x--.75), f(x) - -.15, .5 f(x) and .5 f(x--.75) + -.15. Quickly sketch each graph.
1. 58.1/ 19.4= 2.9
44.7/ 29.1= 1.5
35.6/ 38.8= 0.9
30.8/ 48.5= 0.6
average rate= (2.9+1.5+0.9+0.6)/4= 1.47 cm/s
@& There are four data points and therefore three intervals between them.
There is an average rate for each of the three intervals. The question asked for the rate for each interval, not an average rate that applies to all three.
You don't show how the numbers in your calculation are related to the numbers in the given data.*@
2. slope= (35.6-44.7)/(38.8-29.1)= -0.93cm/s. That is how much the depth decreases each second that passes between 32.1 seconds and 42.8 seconds.
3. f(x--.25)
y=(x--.25)^2= (x+.25)(x+.25)
y=x^2+.5x+0.0625
a=1 b=.5 c= .0625
x= -b/2a =(-.25/2)= -.25
y=(-.25)^2+.5(-.25)+ .0625
y=.0625-.125+.0625
y=0
Vertex= (-.25,0)
@& Very good, but note that the vertex of y = x^2 is at (0, 0) so the vertex of y=(x- -.25)^2, since this function shifts the y = x^2 function -.25 units in the x direction, must be at (-.25, 0).*@
y=(-1)^2+.5(-1)+.0625
y= 1-.5+.0625
y= .56
basic pt (-1,.57)
y= (0^2)+ .5(0)+.0625
y=.0625
basic pt (0,.0625)
@& The basic points of y = x^2 are (-1, 1), (0, 0) and (1, 1).
The function y=(x--.25)^2 shifts every point -.25 units in the x direction.
This does not result in basic points with x components -1, 0 and 1.*@
f(x)--1.65
y=x^2+1.65
x= -b/2a= 0/2 =0
y=0+1.65
y= 1.65
Vertex= (0,1.65)
y= -1^2+1.65
y=2.65
basic pt (-1,2.65)
y=1^2+1.65
y= 2.65
basic pt (1,2.65)
.5f(x)
y=.5(x^2)
x= .5(0/2= 0
y=.5(0)=0
Vertex (0,0)
y= .5(-1^2)
y= .5
basic pt (-1,.5)
y= .5(1^2)
y= .5
basic pt (1,.5)
.5f(x--.25)+-1.65
y=.5(x+.25)^2+-1.65
y=.5(x+.25)(x+.25)+-1.65
y=.5(x^2+.5x+.0625)-1.65
y=.5x^2+.25x-1.62
x=-.25/2(.5)= -.25/1= -.25
y=.5(-.25)^2+ .25(-.25)-1.62
y= .5(.0625-.0625-1.62
y= .03125-.0625-1.62
y= -1.65
Vertex (-.25, -1.65)
y=.5(-1)^2+.25(-1)-1.62
y=.5-.25-1.62
y=-1.37
Basic pt (-1,-1.37)
y= .5(0)+.25(0)-1.62
y= -1.62
basic pt (0,-1.62)
Self-critique (if necessary):
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Self-critique rating:
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@& The problems as stated don't match the solutions you've given. Can you insert your solutions into a copy of the problems you solved?
Must of your work is very good. However there are flaws within your solutions, and my notes should clarify what those are. They shouldn't be difficult for you to remedy.
*@