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course Mth163
July 24, 2013 10:46 pm
Precalculus I Practice-Test 2
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Completely document your work and your reasoning.
You will be graded on your documentation, your reasoning, and the correctness of your conclusions.
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Instructions:
Test is to be taken without reference to text or outside notes.
Graphing Calculator is allowed, as is blank paper or testing center paper.
No time limit but test is to be taken in one sitting.
Please place completed test in Dave Smith's folder, OR mail to Dave Smith, Science and Engineering, Va. Highlands CC, Abingdon, Va., 24212-0828 OR email copy of document to dsmith@vhcc.edu, OR fax to 276-739-2590. Test must be returned by individual or agency supervising test. Test is not to be returned to student after it has been taken. Student may, if proctor deems it feasible, make and retain a copy of the test..
Directions for Student:
Completely document your work.
Numerical answers should be correct to 3 significant figures. You may round off given numerical information to a precision consistent with this standard.
Undocumented and unjustified answers may be counted wrong, and in the case of two-choice or limited-choice answers (e.g., true-false or yes-no) will be counted wrong. Undocumented and unjustified answers, if wrong, never get partial credit. So show your work and explain your reasoning.
Due to a scanner malfunction and other errors some test items may be hard to read, incomplete or even illegible. If this is judged by the instructor to be the case you will not be penalized for these items, but if you complete them and if they help your grade they will be counted. Therefore it is to your advantage to attempt to complete them, if necessary sensibly filling in any questionable parts.
Please write on one side of paper only, and staple test pages together.
Test Problems:
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Problem Number 1
For the function y = f(t) = .89 ^ t construct a table of y vs. t for t running from t = 2.3 to t = 2.38 in four equal increments. Using appropriate transformation(s) on the y column, the t column, or both, linearize this data set and demonstrate that the data set has in fact been linearized.
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.x
2.32
2.34
2.36
2.38
Log (x)=y
.365
.369
.373
.377
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The values of .89^2.3, .89^2.32, .89^2.34 etc. are
0.7648865342
0.7631059071
0.7613294252
0.7595570790
0.7577888586
It isn't clear how you got the values .365, .369, etc.. and it isn't completely clear what them mean, since you head that column log(x) = y.
y = .89^t, not log(x). There is no definition of x in the solution or in the problem.
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Now I can take two sets of random x.y coordinates and plug into y=axtb to get the parameters a and b to make my linear model
.365=a (2.32) +b
.373= a(2.36) +b
When you subtract the first form the second you end up with
-.008=a (-.04)
Divide -.04 by both sides
. a=.2
.365= (.2*2.32)+b
Now that I have plugged a in I can solve for b
.365=.464 +b
Subtract .464 from both sides
-.099=b
Now that we have a and b we have our linearized form of y=
.I realize now that I have not applied y=.89^t
Other than that have I applied the steps right at least???( on a test I would redo this question)
And to end this with a wrong linear form would be y=.2x-.099
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Accepting for the moment the x and y coordinates you report the model would be
y = .2 x - .099.
Now, if x stands for log(t) your equation would be
y = .2 log(t) - .099
so that
log(t) = y + .099
and
t = e^(y + .099) = e^.099 * e^y.
This wouldn't help to solve the original problem, but it's the sort of thing you would do if x stood for log(t).
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It isn't clear how you got the numbers
.365
.369
.373
.377
or what they mean.
Can you document every detail of your thinking in order to clarify?
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Problem Number 2
Which of the following sequences s1 or s2 is exponential in nature?
s1 = { 9, 8.063626, 7.264469, 6.578754 }
s2 = { 3, 5.795618, 11.1964, 21.63001 }.
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I start off by trying to find the ratios in the first sequence
8.064/9=.896
7.26/8.064=.900298
6.58/7.26=.906336
If I had rounded the first quotient then there might be a constant pattern here
Now just to be sure I find the ratios of my second sequence
5.76/3=1.93
11.19/5.79=1.93264
21.63/11.19=1.93298
This one is awfuly close to being constant as well
They could both be exponential in nature ( maybe if I were more precise this would show different)
However if only one of these can be exponential in nature then the answer would obviously be sequence 2 they all start off with 1.93 (at least) with no rounding necessary
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You are correct that the second sequence is more likely to be exponential.
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Problem Number 3
A population starts at 9000 and grows at the rate of 6.5 percent per year.
What will be the population 1, 2, 3 and 10 years later?
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9000(1.065^1)=9585
9000(1.065^2)=10208
9000(1.065^3)=10871.5
9000(1.065^10)=16894.2
What function P(t) gives population as a function of number t of years?
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9000(1.065^t)
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What are the growth rate and growth factor of this function?
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.065 growth rate, 1.065 growth factor
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Problem Number 4
Sketch a graph of y = (x + 3.5) ^ 5 (x - 4) ( 2 x^2 + 2 x + 2).
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At (x + 3.5) ^ 5 my graph has a cubic (or odd power function) at x=-3.5 it then goes down through factor (x - 4) at x=4 our last group are comprised of irreducible factors, which I do not think you need to graph right???
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On the test you do need to graph, and you need to show (and explain why) the graph has all the following characteristics:
A zero at x = 4 through which the graph passes in a very nearly straight line when x is close to 4.
A zero at x = -3.5 with the graph leveling off and becoming horizontal for an instant as it passes through the zero, in the same way a y = x^5 power function behaves near the origin.
The graph rapidly approaches +infinity as x approaches +infinity, since for large positive x all the factors (x+3.5), (x-4) and (2 x^2 + 2 x + 1) are large positive numbers whose product is a very large positive numbers.
The graph rapidly approaches +infinity as x approaches -infinity, since for large negative x the factors (x+3.5) and (x-4) are large negative numbers so that (x+3.5)^5 is a very large negative number which, multiplied by the large negative value of (x-4) produces a very large positive result, which multiplied by the very large positive value of (2 x^2 + 2 x + 1) results in a very large positive result.
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Problem Number 5
What are the basic points of the exponential function y = f(x) = 6 * .93^x? Graph the function using these points.
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I know to find the basic points of a quadratic function you use AOS (axis of symmetry)
But I do not recall how to apply this to a power function
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This is an exponential function, not a power function.
You need to review the basic-points pictures of quadratic, exponential and power functions.
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Problem Number 6
Solve the equation log( 4 x) - log(x^ 4) = 4.
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10^(log(4x)- log(x^4) = 10 ^4
(4x)-(x^4)=10^4
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log( 4 x) - log(x^ 4) is not equal to
log( 4 x - x^ 4)
so
10^(log( 4 x) - log(x^ 4))
is not equal to
4x = x^4.
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Wait I believe its done the way I have it below
Log(4x)- 4 log(x) =4
10^(log(4x)) - 4*10^(log(x))= 10^4
4x-4x=10^4
0=10^4
Something is wrong here
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You are not applying the laws of logarithms correctly.
How do you simplify log(4x) - log(x^4) using the laws?
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Problem Number 7
Given the depth vs. clock time function y = f(t) = .02 t2 + -2.02 t + 54, with depth in cm when clock time is in seconds, find the clock time t such that f(t) = 39.995 cm and find f(t) when t = 22 cm. Find the clock time when water depth is 9.995 cm. Using the same function determine the depth at clock time t = 10 sec.
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y=.02t^2 -2.02t+54
40=.02t^2 -2.02+54
Subtract 40 from both sides to get my quad equal to zero, so I can use the quad fn
0=.02t^2 -2.02t+14
Now I put this in the quad formula ([-b+- sqrt(b^2-4ac)]/2a)
. a=.02
.b = -2.02
. c= 14
. t=93.75
or
. t=7.25
y=.02(22^2) -2.02(22)+54
y= 19.24
rounding 9.995 to 10
10=.02(t^2) -2.02(t)+54
Subtract 10 from both sides
0=.02(t^2) -2.02(t)+44
Now I put this in the quad formula ([-b+- sqrt(b^2-4ac)]/2a)
. a=.02
. b= -2.02
. c=44
. clock time=69.55 seconds
or
. clock time =31.45 seconds
Depth=.02(10^2)- 2.02(10) +54
Depth=35.8cm
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Good.
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Problem Number 8
Using DERIVE, find the best fit to the following data, using first a linear fit, then a quadratic fit, then a cubic fit, then a fourth-degree polynomial fit: [ [3, 5], [5, 6.3], [7,9.6], [9, 19.5], [11, 34.1] ]. Give an argument for and against each fit as opposed to the others, then decide which fit is most likely to tell you how the data is really behaving.
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I have tried DRIVE out using the Learning Lab, but it is not available in the testing lab I am not really sure who to contact about that
I believe I can do something similar on my calculator
Data/matrix
Enter my data in
I then hit f5
Calculation type linear regression
Store RegQ to y1(x)
y=3.57x-10.09
this gives me a wide ranges of choices in both neg and positive directions but does not curve when necessary
I then hit f5 again and choose
QuadReg
y=.59x^2-4.73x+14.23
this one gives a more flexible and wider range of (x,y) points but can only go one way or another
CubicReg
y=.0281x^3+ .00223x^2 -.978x +7.25
with cubics they go where they need to go I think this is the best fit because it will try to hit all the necessary values
I do not think I have a fourth- degree polynomial fit
I have a feeling that the higher the power the better the fit ( however I cannot confirm this)
If I were to look for a fourth degree polynomial on my cal, what other name would it have???
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Fourth degree is quartic, but usually you just select a polynomial fit and specify the degree (2 for quadratic, 3 for cubic, 4 for 4th power etc.).
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Problem Number 9
Sketch the graph of a degree 3 polynomial with zeros at x = - 5, 4 and 5, with y taking on increasingly large positive values for large positive values of x.
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My graph starts negative going through x= -5 touching the y axis at 3 then going back down to the x axis through 4 and up again at a number higher than 3 then dips down through the x axis again at 5 and back up with y axis higher and higher number then previous
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Your function in factored form is
y = (x+5)(x-4)(x-5).
y can be zero only at x = -5, 4 and 5.
So it can't come up through the x axis, dip back down, come back up then go back down through the axis, then back up again.
It can come up through the axis at x = -5, back down through at x = 4, and up through again at x = 5.
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Give the y = (x-x1)(x-x2)(x-x3) form of this function has well as the y = ax^3 + bx^2 + cx + d form.
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The factored form is
(x+5)
(x-4)
(x-5)
To find the combined form I first chose the first two and multiply them
. x(x-4)+5(x-4)
. x^2+x -20
I then multiply x^2+x -20 by (x-5)
. x(x^2+x -20) -5(x^2+x -20)
With this anwer you get your finial form
. x^3 -4x^2 -15x +100
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Good.
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Problem Number 10
Explain why the following statement must be true: No polynomial of degree 2 can be the product of three or more polynomials of degree 1.
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If the highest degree is 2 then it cannot be lower than 2 so therefore It cannot be the product of some polynomial whos product is one
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This doesn't answer the question.
Can you explain why the question would be answered if you can show that the product of three or more polynomials of degree 1 would be of degree greater than 3?
Can you show that this is in fact the case?
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Good work on many problems.
See my notes. I do recommend that you submit a revision in which you try to clarify everything and answer my questions.
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