query_6

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course Mth163

query_6 *********************************************

Question: `qQuery 4 basic function families

What are the four basic functions?

What are the generalized forms of the four basic functions?

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Your solution:

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Y=x, linear, y=mx+b, straight line

Y=x^2, quadratic, y=a x^2 +bx+c,parabola

Y=2^x, exponential, y=A*2^(kx)+c, either increasing of decreasing; horizontal asymptote

Y=x^p, power, y=A(x-h)^p+c, negative: horizontal and vertical asymptotes y=c and h=x , even/ odd: graph symmetric or antisymmertic

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confidence rating #$&*: 3

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Given Solution:

** STUDENT RESPONSE:

Linear is y=mx+b

Quadratic is y=ax^2 + bx +c

Exponential is y= A*2^ (kx)+c

Power = A (x-h)^p+c

INSTRUCTOR COMMENTS:

These are the generalized forms. The basic functions are y = x, y = x^2, y = 2^x and y = x^p. **

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Self-critique (if necessary):

Question: `qFor a function f(x), what is the significance of the function A f(x-h) + k and how does its graph compare to that of f(x)?

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Your solution:

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This is an exponential function.

Difference quotient.

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confidence rating #$&*: 1

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Given Solution:

** STUDENT RESPONSE: A designates the x strectch factor while h affects a y shift & k affects an x shift

INSTRUCTOR COMMENTS:

k is the y shift for a simple enough reason -- if you add k to the y value it raises the graph by k units.

h is the x shift. The reason isn't quite as simple, but not that hard to understand. It's because when x is replaced by x - h the y values on the table shift 'forward' by h units.

A is a multiplier. When all y values are multiplied by A that moves them all A times as far from the x axis, which is what causes the stretch.

Thus A f(x-h) + k is obtained from f(x) by vertical stretch A, horizontal shift h and vertical shift k.

The two aren't the same, but of course they're closely related. **

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Self-critique (if necessary):

K is shift, A is stretch (move A times as far away from x axis)

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Self-critique Rating:2

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Question: `q query introduction to rates and slopes, problem 1 ave vel for function depth(t) = .02t^2 - 5t + 150

give the average rate of depth change with respect to clock time from t = 20 to t = 40

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Your solution:

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2

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confidence rating #$&*: 2

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Given Solution:

** depth(20) = .02(20^2) - 5(20) + 150 = 58

depth(40) = .02(40^2) - 5(40) + 150 = -18

change in depth = depth(40) - depth(20) = -18 - 58 = -76

change in clock time = 40 - 20 = 20.

Ave rate of depth change with respect to clock time = change in depth / change in clock time = -76 / 20 = -3.8 **

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Self-critique (if necessary):

I though we took the absolute value???

Otherwise I got exactly what you did taking abs value away.

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Self-critique Rating:

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Question: `qWhat is the average rate of depth change with respect to clock time from t = 60 to t = 80?

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Your solution:

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-10

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confidence rating #$&*:1

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Given Solution:

** depth(60) = .02(60^2) - 5(60) + 150 = -78

depth(80) = .02(80^2) - 5(80) + 150 = -122

change in depth = depth(80) - depth(60) = -122 - (-78) = -44

change in clock time = 40 - 20 = 20.

Ave rate of depth change with respect to clock time = change in depth / change in clock time = -44 / 20 = -2.2 **

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Self-critique (if necessary):

Im finding change, not slope!

So this would be absolute right???

(-78)-(-122)=-44

Change in clock time (t)= 40-20=20

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Self-critique Rating:3

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The average rate of change of depth with respect to clock time is (change in depth) / (change in clock time).

The change in depth can be positive or negative. The rate of change can therefore be positive or negative.

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Question: `qdescribe your graph of y = .02t^2 - 5t + 150

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Your solution:

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Starting point 150 cm to 0 sec ending point -162.5 cm to 123 seconds

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confidence rating #$&*: 1

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Given Solution:

** The graph is a parabola.

y = .02t^2 - 5t + 150 has vertex at x = -b / (2a) = -(-5) / (2 * .02) = 125, at which point y = .02 (125^2) - 5(125) + 150 = -162.5.

The graph opens upward, intercepting the x axis at about t = 35 and t = 215.

Up to t = 125 the graph is decreasing at a decreasing rate. That is, it's decreasing but the slopes, which are negative, are increasing toward 0.**

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Self-critique (if necessary):

We can use AOS to find vertex to describe our graphs, also to find other points to make our graph.

Does AOS always give us our vertex value???

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Self-critique Rating:3

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The quadratic formula gives you locations of the zeros, typically giving you two zeroes, one corresponding to + sqrt( b^2 - 4 a c) and other to - sqrt( b^2 - 4 a c) as that term appears in the formula.

The t value between the two zeroes is found by leaving off the +- sqrt(b^2 - 4 a c) term of the numerator, leaving you

t = - b / (2 a).

This is the t coordinate of the vertex. It must be so because of the symmetry of the graph.

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Question: `qdescribe the pattern to the depth change rates

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Your solution:

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The coordinates are (125,-162.5)(215,-.5)(35,-.5) usually when you have two of the same y values while the x values are on different sides of the vertex and not on the same line you have a parabola. So this pattern indicates at parabola.

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confidence rating #$&*:1

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Given Solution:

** Rates are -3.8, -3 and -2.2 for the three intervals (20,40), (40,60) and (60,80).

For each interval of `dt = 20 the rate changes by +.8. **

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Self-critique (if necessary):

You meant for me to use the questions above, using (20, 40), (40, 60) and (60, 80)

I do not understand how you found your rates.

Can you please go through the process???

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Self-critique Rating:

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t = 125, 215, 35 values are not relevant to this question.

The coordinates of the graph at t = 20, 40, 60 and 80 are

(20, 58)

(40, -18)

(60, -78)

(80, -122]

What are the slopes between each consecutive pair of points?

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Question: `qquery problem 2. ave rates at midpoint times

what is the average rate of depth change with respect to clock time for the 1-second time interval centered at the 50 sec midpoint?

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Your solution:

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Y=.02(50.5^2)+5(50.5)+150=302.005

Y=.02(49.5^2)+5(49.5)+150=298.005

302.005-298.005=4

4/1=4

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confidence rating #$&*:2

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Given Solution:

** The 1-sec interval centered at t = 50 is 49.5 < t < 50.5.

For depth(t) = .02t^2 - 5t + 150 we have

ave rate = [depth(50.5) - depth(49.5)]/(50.5 - 49.5) = -3 / 1 = -3. **

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Self-critique (if necessary):

How did you get -3???

I see what I did wrong, added the five where I should have subtracted.

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Self-critique Rating:

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Question: `qwhat is the average rate of change with respect to clock time for the six-second time interval centered at the midpoint.

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Your solution:

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There would be 3 on each side of our midpoint to the left and right so fifty plus three equals fifty-three and fifty minus three equals forty-seven .

Plug these in for

Y=.02(53^2)-5(53)+150=-58.82

Y=.02(47^2)-5(47)+150=-40.82

-58.82-(-40.82)=-18

53-47=6

-18/6=-3

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confidence rating #$&*: 3

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Given Solution:

** The 6-sec interval centered at t = 50 is 47 < t < 53.

For depth(t) = .02t^2 - 5t + 150 we have ave rate = [depth(53) - depth(47)]/(53 - 47) = -18 / 6 = -3. **

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Self-critique (if necessary):

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Self-critique Rating:OK

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Question: `qWhat did you observe about your two results?

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Your solution:

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Both of my results ended in negative three, so my graph would be linear???

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confidence rating #$&*: 1

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Given Solution:

** The two rates match each other, and they also match the average rate for the interval 40 < t < 60, which is also centered at t = 50.

For a quadratic function, and only for a quadratic function, the rate is the same for all intervals having the same midpoint. This is a property unique to quadratic functions. **

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Self-critique (if necessary):

The average rate of the intervals has the same midpoint which is t= 50

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Self-critique Rating:2

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You have found the rates across two secants of the parabolic graph, with both intervals centered at t = 50.

You should sketch the graph and these two secants.

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Question: `qquery problem 3. ave rates at midpt times for temperature function Temperature(t) = 75(2^(-.05t)) + 25.

What is the average rate of temperature change with respect to clock time for the 1-second time interval centered at the 50 sec midpoint?

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Your solution:

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= 75(2^(-.05(50.5))) + 25

=38.0305

= 75(2^(-.05(49.5))) + 25

=38.49

38.0305-38.49= -.4595

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confidence rating #$&*:

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Given Solution:

f(50.5) = 38.03048331 deg

f(49.5) = 38.49000231 deg

The change is -0.4595190014 deg. The change in clock time is 1 second.

So the average rate is

• ave rate = change in temperature / change in clock time = -.4595 deg / (1 sec ) = -.4595 deg/sec.

STUDENT RESPONSE: .46 degrees/sec

INSTRUCTOR COMMENT: More precisely -.4595 deg/sec, and this does not agree exactly with the result for the 6-second interval.

Remember that you need to look at enough significant figures to see if there is a difference between apparently identical results.**

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Self-critique (if necessary):

I should have included average rate=change Temp/change in t=-4595 deg/sec

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Self-critique Rating:3

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Question: `qwhat is the average rate of change for the six-second time interval centered at the midpoint.

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Your solution:

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= 75(2^(-.05(53))) + 25

=36.949

= 75(2^(-.05(47))) + 25

=39.711

36.949-39.258= -2.762

Average rate of the change in temperature/ change in time= -2.762 deg/sec

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confidence rating #$&*: 3

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Given Solution:

STUDENT RESPONSE: .46 degrees/minute

INSTRUCTOR COMMENT:

The 1- and 6-second results might possibly be the same to two significant figures, but they aren't the same. Be sure to recalculate these according to my notes above and verify this for yourself.

The average rate for the 6-second interval is -.4603 deg/sec. It differs from the average rate -.4595 deg/min, calculated over the 1-second interval, by more than -.0008 deg/sec.

This differs from the behavior of the quadratic. For a quadratic that the results for all intervals centered at the same point will all agree. This is not the case for the present function, which is exponential. Exact agreement is a characteristic of quadratic functions, and of no other type. **

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Self-critique (if necessary):

Can you tell me where I went wrong???

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Self-critique Rating:0

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The average rate is not the same as the change in the value of the function.

To get the average rate you have to divide one quantity by another.

You know the change in the y value, which is the numerator of the necessary calculation.

What is the denominator and why is it necessary?

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Self-critique (if necessary):

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Self-critique (if necessary):

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Self-critique rating:

#*&!

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See my comments and answers to your questions.

The best way to follow up would be with a revision, according to the instructions below.

&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).

Be sure to include the entire document, including my notes.

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