#$&*
course Mth163
July 14,2013 1:16 pm
http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
023. `query 23
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Question: `qQuery problem 2.
Describe the sum of the two graphs.
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Your solution:
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Parabola graph points at
(-3,8)
(-2,3)
(-1,0)
(0,-1)
(1,0)
(2,3)
(3,8)
Points on somewhat linear/power function graph
(-3,1.5)
(-2,1)
(-1,0)
(0,-.1)
(1,-1.5)
(2,-1.8)
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confidence rating #$&*: 3
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Given Solution:
`a** The 'black' graph takes values 8, 3, 0, -1, 0, 3, 8 at x = -3, -2, -1, 0, 1, 2, 3.
The 'blue' graph takes approximate values 1.7, .8, .2, -.1, -.4, -.6, -.8 at the same x values.
The 'blue' graph takes value zero at approximately x = -.4.
The sum of the two graphs will coincide with the 'blue' graph where the 'black' graph is zero, which occurs at x = -1 and x
= 1.
The sum will coincide with the 'black' graph where the 'blue' graph is zero, which occurs at about x = -.4. **
STUDENT QUESTION:
I don't understand where you are getting these numbers.
INSTRUCTOR RESPONSE: The graph in the stated problem appears below. The x and y scales are marked in units. You should begin by orienting yourself to these graphs: convince yourself that the points marked on the graphs have the coordinates quoted in the given solution.
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Self-critique (if necessary):
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Self-critique rating:
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The sum will coincide with the 'black' graph where the 'blue' graph is zero, which occurs at about x = -.4.
I thought both met at (-1,0) ???
@&
Where the graphs meet does not determine where their sum is 0, unless they happen to meet at a point on the x axis.
At the point where the graphs meet, the y value of the sum is double the common y value of the two graphs.
The 'blue' graph is still clearly above the x axis at x = -1. It doesn't cross until somewhere between x = -1 and x = 0, and it appears that the crossing is closer to x = 0. So the estimate made in the given solution is that the 'blue' graph has y value 0 when x = -.4.
*@
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Question: `qWhere it is the sum graph higher than the 'black' graph, and where is it lower? Answer by giving specific intervals.
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Your solution:
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Higher and/or level with
(-1,0)
(0,-.1)
(1,-1.5)
Lower
(-3,1.5)
(-2,1)
(1,-1.5)
(2,-1.8)
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confidence rating #$&*: 2
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Given Solution:
`a** The sum of the graphs is higher than the 'black' graph where the 'blue' graph is positive, lower where the 'blue' graph is negative.
The 'blue' graph is positive on the interval from x = -3 to x = -.4, approx.. This interval can be written [-3, -.4), or -3 <= x < -.4. **
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Self-critique (if necessary):
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So using my data would the interval be at x=1,x=-1
-1<= x < 1
“Written [-3, -.4)” are you saying the neg 3 is included while the neg point 4 is not??? I’m not all to familiar with interval notation in Algebra two out teacher said one is for closed and the other is for open dots, but all the dots on the graph are closed so I don’t understand why your [ ) would deferrer ???
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The first thing you need to understand is that the sum graph is higher than the 'black' graph when the 'blue' graph is positive, and lower when the 'blue' graph is negative.
If the 'blue' graph is zero at x = -.4, which is the estimate used, then it isn't positive at x = -.4. So .4 would not be included in the interval where the 'blue' graph is positive.
It certainly appears from the graph that the 'blue' graph is positive for x = -3, so -3 is included in the interval.
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Self-critique rating:
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Question: `qWhere it is the sum graph higher than the 'blue' graph, and where is it lower? Answer by giving
specific intervals.
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Your solution:
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-3 <= x < 3
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confidence rating #$&*:
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Given Solution:
`a** The sum of the graphs is higher than the 'blue' graph where the 'black' graph is positive, lower where the 'black'
graph is negative.
The 'black' graph is positive on the interval from x = -1 to x = 1, not including the endpoints of the interval. This interval
can be written (-1, 1) or -1 < x < 1. **
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Self-critique (if necessary):
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Okay I think I have something backwards here
I am taking about the highest and lowest points on the black graph
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Self-critique rating:
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Question: `qWhere does thus sum graph coincide with the 'black' graph, and why? Give your estimate of the
specific coordinates of the point or points where this occurs.
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Your solution:
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Yes, you can observe it with your eyes
(-1,0)
@&
At x = -1 the 'blue' graph is positive, so the sum graph would be higher than the 'black' graph.
At x = -1 the 'black' graph is zero, so the sum graph would for x = - 1 coincide with the 'blue' graph.
The sum graph coincides with the 'black' graph where the 'blue' graph is zero, which occurs at x = -.4.
*@
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confidence rating #$&*: 3
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Given Solution:
`a** The sum coincides with the 'black' graph where the 'blue' graph is zero, which occurs at about x = -.4. The
coordinates would be about (-.4, -.7), on the 'black' graph. **
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Self-critique (if necessary):
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With the coordinates I used does my answer seem probable????
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At x = -1 the sum graph coincides with the 'blue' graph, which appears to have y coordinate approximately .3.
However the point where the sum graph coincides with the 'black' graph is about (-.4, -.7).
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Self-critique rating:
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Question: `qWhere does thus sum graph coincide with the 'blue' graph, and why? Give your estimate of the
specific coordinates of the point or points where this occurs.
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Your solution:
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The only other place they interact would be at an un-dotted spot on the graph (.8,-.3)
@&
The graphs intersect at a point which is plausibly (.8, -.3).
So assume that at x = .8, the y coordinate of each graph is -.3.
The 'sum' graph would therefore take value -.3 + -.3 = -.6 at x = .8. The point (.8, -.6) would be a point of the 'sum' graph.
This question, however, asks where the sum graph coincides with the 'blue' graph. See my preceding notes regarding this distiction.
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confidence rating #$&*: 2
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Given Solution:
`a** The sum coincides with the 'blue' graph where the 'black' graph is zero, which occurs at x = -1 and x = 1. The
coordinates would be about (-1, .2) and (1, -.4), on the 'blue' graph. **
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Self-critique (if necessary):
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Why do you keep giving two sets of coordinates when they cross somewhere shouldn’t they share the same set of coordinates???
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The sum graph coincides with one of the two graphs when the other is zero, not where the two graphs meet.
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Self-critique rating:
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Question: `qQuery problem 3
Describe the quotient graph obtained by dividing the 'black' graph by the 'blue' graph. You should answer the following
questions:
Where it is the quotient graph further from the x axis than the 'black' graph, and where is it closer? Answer by giving
specific intervals, and explaining why you believe these to be the correct intervals.
Where it is the quotient graph on the same side of the x axis as the 'black' graph, and where is it on the opposite side,
and why? Answer by giving specific intervals.
Where does thus quotient graph coincide with the 'black' graph, and why? Give your estimate of the specific coordinates
of the point or points where this occurs.
Where does the quotient graph have vertical asymptote(s), and why? Describe the graph at each vertical asymptote.
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Your solution:
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Where it is the quotient graph further from the x axis than the 'black' graph, and where is it closer? Answer by giving specific intervals, and explaining why you believe these to be the correct intervals.
(1,1.5) where the black graph has peaked above x axis the blue graph dips down below it and this is when they’re farthest apart
The intervals would be 0
Where it is the quotient graph on the same side of the x axis as the 'black' graph, and where is it on the opposite side, and why? Answer by giving specific intervals.
When they are both above the x axis the intervals are 5.5
Where does thus quotient graph coincide with the 'black' graph, and why? Give your estimate of the specific coordinates of the point or points where this occurs.
(1.2,7.7) is exactly where they intersect
Where does the quotient graph have vertical asymptote(s), and why? Describe the graph at each vertical asymptote.
The graph does not go into the second quadrant so there may be a vertical asymptote there at (-.1,.1) the graph also does not seem to go past (0, -1.5)
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confidence rating #$&*: 2
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Given Solution:
`a** The 'black' graph is periodic, passing through 0 at approximately x = -3.1, 0, 3.1, 6.3. This graph has peaks with y
= 1.5, approx., at x = 1.6 and 7.8, approx., and valleys with y = -1.5 at x = -1.6 and x = 4.7 approx.
The 'blue' graph appears to be parabolic, passing thru the y axis at x = -1 and reaching a minimum value around y = -1.1
somewhere near x = 1. This graph passes thru the x axis at x = 5.5, approx., and first exceeds y = 1 around x = 7.5.
The quotient will be further from the x axis than the 'black' graph wherever the 'blue' graph is within 1 unit of the origin,
since division by a number whose magnitude is less than 1 gives a result whose magnitude is greater than the number being
divided. This will occur to the left of x = 1, and between about x = 2 and x = 7.5.
Between about x = 0 and x = 1 the 'blue' graph is more than 1 unit from the x axis and the quotient graph will be closer to
the x axis than the 'black' graph. The same is true for x > 7.5, approx..
The 'black' graph is zero at or near x = -3.1, 0, 3.1, 6.3. At both of these points the 'blue' graph is nonzero so the
quotient will be zero.
The 'blue' graph is negative for x < 5.5, approx.. Since division by a negative number gives us the opposite sign as the
number being divided, on this interval the quotient graph will be on the opposite side of the x axis from the 'black' graph.
The 'blue' graph is positive for x > 5.5, approx.. Since division by a positive number gives us the same sign as the number
being divided, on this interval the quotient graph will be on the same side of the x axis as the 'black' graph.
The quotient graph will therefore start at the left with positive y values, about 3 times as far from the x axis as the 'black'
graph (this since the value of the 'blue' graph is about -1/3, and division by -1/3 reverses the sign and gives us a result with
3 times the magnitude of the divisor).
The quotient graph will have y value about 2.5 at x = -1.6, where the 'black' graph 'peaks', but the quotient graph will
'peak' slightly to the left of this point due to the increasing magnitude of the 'blue' graph.
The quotient graph will then reach y = 0 / (-1) = 0 at x = 0 and, since the 'black' graph then becomes positive while the
'blue' graph remains negative, the quotient graph will become negative.
Between x = 0 and x = 2 the magnitude of the 'blue' graph is a little greater than 1, so the quotient graph will be a little
closer to the x axis than the 'black' graph (while remaining on the other side of the x axis).
At x = 3.1 approx. the 'black graph is again zero, so the quotient graph will meet the x axis at this point.
Past x = 3.1 the quotient graph will become positive, since the signs of both graphs are negative. As we approach x =
5.5, where the value of the 'blue' graph is zero, the quotient will increase more and more rapidly in magnitude (this since
the result of dividing a negative number by a negative number near zero is a large positive number, larger the closer the
divisor is to zero). The result will be a vertical asymptote at x = 5.5, with the y value approaching +infinity as x
approaches 5.5 from the left.
Just past x = 5.5 the 'blue' values become positive. Dividing a negative number by a positive number near zero results in a
very large negative value, so that on this side of x = 5.5 the asymptote will rise up from -infinity.
The quotient graph passes through the x axis near x = 6.3, where the 'black' graph is again zero. To the right of this point
both graphs have positive values and the quotient graph will be positive.
Around x = 7.5, where the 'blue' value is 1, the graph will coincide with the 'black' graph, giving us a point near (7.5, 1.3).
Past this point the 'blue' value is greater than 1 so that the quotient graph will become nearer the x axis than the 'black'
graph, increasingly so as x (and hence the 'blue' value) increases. This will result in a 'peak' of the quotient graph
somewhere around x = 7.5, a bit to the left of the peak of the 'black' graph. **
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Self-critique (if necessary):
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The quotient will be further from the x axis than the 'black' graph wherever the 'blue' graph is within 1 unit of the origin
@&
Almost. Using your words, except for the last:
The quotient will be further from the x axis than the 'black' graph wherever the 'blue' graph is within 1 unit of the x axis.
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Self-critique rating:
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Question: `qQuery problems 7-8
Sketch the graph of y = x^2 - 2 x^4 by first sketching the graphs of y = x^2 and y = -2 x^4.
How does the result compare to the graph of y = x^2 - x^4, and how do you explain the difference?
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Your solution:
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From y=x^2 to y = -2 x^4 the graph goes from a general parabola to flipping about the x axis and then bumping up to touch the x axis at (-.5,0) and (.5,0) before going back down
y = x^2 - x^4 must have a double root (or multiplicity at zero ) because while the graph does touch the origin it doesn’t cross the x axis again until (2.5,0) (-2.5,0)
All graphs pass through the origin
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confidence rating #$&*:
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Given Solution:
`a** At x = 0, 1/2, 1 and 2 we have x^2 values 0, 1/4, 1 and 4, while -x^4 takes values 0, -1/16, -1 and -16, and -2
x^4 takes values 0, -1/8, -2 and -32.
All graphs clearly pass through the origin.
The graphs of y = x^2 - x^4 and y = x^4 - 2 x^4 are both increasingly negative at far right and far left.
Graphical addition will show that y = x^2 - x^4 takes value 0 and hence passes thru the x axis when the graphs have
equal but opposite y values, which occurs at x = 1 and x = -1. To the left of x = -1 and to the right of x = 1 the negative
values of -x^4 overwhelm the positive values of x^2 and the sum graph will be increasingly negative, with values
dominated by -x^4. Near x = 0 the graph of y = -x^4 is 'flatter' than that of y = x^2 and the x^2 values win out, making
the sum graph positive.
y = x^2 - 2 x^4 will take value 0 where the graphs are equal and opposite in value; this occurs somewhere between x =
.8 and x = .9, and also between x = -.9 and x = -.8, which places the zeros closer to the y axis than those of the graph of
y = x^2 - x^4. The graph of y = -2 x^4 is still flatter near x = 0 than the graph of y = x^2, but not as flat as the graph of
y = -x^4, so while the sum graph will be positive between the zeros the values won't be as great. Outside the zeros the
sum graph will be increasingly negative, with values dominated by -2x^4. **
The graphs you constructed, based on the basic points and behaviors of the functions y = x^2, y = -x^4 and y = -2 x^4, should have had the same characteristics and basic properties of those in the series of figures below:
The figure below depicts the graphs of y = x^2 and y = -x^4, with x and y gridlines at unit intervals.
The graph of y = x^2 - x^4 is superimposed below:
The figure below depicts the graphs of y = x^2, y = -2 x^4 and y = x^2 - 2 x^4:
The graphs of y = x^2 - x^4 and y = x^2 - 2 x^4 are depicted below:
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Self-critique (if necessary):
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Our graphs did match.
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@&
Not bad, but you could do more.
You didn't describe how you constructed the graphs, and your descriptions of the graphs themselves addressed fewer of the important characteristics than did the given solution.
Recommendation: Study the given solutions and be sure you understand all the details and and why they are important, as well as how to describe them in words.
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Self-critique rating:
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Question: `qHow does the shape of the graph change when you add x to get y = -2 x^4 + x^2 + x, and how
do you explain this change?
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Your solution:
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The graph still passes through origin
The reason the left hand side is much lower than the right is the graph Is adjusting to get the numbers larger
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confidence rating #$&*: 2
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Given Solution:
`a** At x = 0 there is no change in the y value, so the graph still passes through (0, 0).
As x increases through positive numbers we will have to increase the y values of y = x^2 -2 x^4 by greater and greater
amounts. So it will take a little longer for the negative values of -2 x^4 to 'overwhelm' the positive values of x^2 + x than
to overcome the positive values of x^2 and the x intercept will shift a bit to the right.
As we move away from x = 0 through negative values of x we will find that the positive effect of y = x^2 is immediately
overcome by the negative values of y = x, so there is no x intercept to the left of x = 0.
The graph in fact stays fairly close to the graph of y = x near (0, 0), gradually moving away from that graph as the values
of x^2 and -2 x^4 become more and more significant. **
The graphs you constructed, based on the basic points and behaviors of the functions y = x^2, y = -x^4 and y = -2 x^4, should have had the same characteristics and basic properties of those in the series of figures below:
The figure below depicts the graphs of y = x^2 and y = -x^4, with x and y gridlines at unit intervals.
The graph of y = x^2 - x^4 is superimposed below:
The figure below depicts the graphs of y = x^2, y = -2 x^4 and y = x^2 - 2 x^4:
The graphs of y = x^2 - x^4 and y = x^2 - 2 x^4 are depicted below:
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Self-critique (if necessary):
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My new graph has points at (-.3,-.2) (0,0) and maximum at (.6,.7)
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If you understand the assignment and were able to solve the previously given problems from your worksheets, you should be able to complete most of the following problems quickly and easily. If you experience difficulty with some of these problems, you will be given notes and we will work to resolve difficulties.
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Question: The graphs of two linear functions are depicted below. The first function f(x) has x-intercept (-8/3, 0) and y-intercept (0, 4), while the second function g(x) has y-intercept (0, 1) and slope -1/2.
You could easily find the formulas for these two functions and answer subsequent questions using these formulas, but don't do that. The task here, as you have seen, is to be able to construct graphs of combinations of functions. The purpose is to give you a deeper understanding of such functions and enhance your ability to think about and visualize complex trends.
Construct an approximate graph of the sum of these two functions.
At what approximate x coordinate do you estimate your graph of the sum intercepts the x axis?
Construct an approximate graph of the difference function f(x) - g(x).
At what approximate x coordinate do you estimate your graph of the difference function intercepts the x axis?
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Your solution:
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Construct an approximate graph of the sum of these two functions.
5=5
. At what approximate x coordinate do you estimate your graph of the sum intercepts the x axis?
-1.5
@&
At x = -1.5 both graphs are positive. The graph of the sum would intercept the x axis only if the y value was zero. You can't get zero by adding two positive values.
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Construct an approximate graph of the difference function f(x) - g(x).
-3=-3
At what approximate x coordinate do you estimate your graph of the difference function intercepts the x axis?
-2.7
@&
At x = -2.7 the first graph is very close to its zero at (-8/3, 0) and the second is positive with a value greater than 2. The difference of the two y values cannot be zero.
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confidence rating #$&*: 1
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Question: For the graphs of the preceding, sketch the graph of the product function f(x) * g(x).
On what interval(s) is the graph of the product function positive, and on what intervals negative?
How could you have determined these intervals, based on the given graphs, without actually having sketched the graph of the product function?
On what interval(s) is the graph of the product function increasing and on what intervals decreasing?
For large negative x, is the product function positive or negative, and how could you tell by just looking at the graphs of f(x) and g(x)?
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Your solution:
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the product function f(x) * g(x)
4=4
5
@&
For x > 5 one graph is below the x axis and the other above. The product is therefore negative for all values of x > 5.
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(0,5)
@&
(0, 5) is a point on the coordinate plane, not a number. So the statement
(0, 5) < x < infinity
is not properly formed. (0, 5) cannot be less than any single number.
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Multiplying the two functions
For the large x the product would be neg because you can see when you multiply a positive number by a neg you get negative
@&
I think your explanation is to the point. Just to be sure, make sure you either meant the following or that you understand why all the following is necesary to answer this question:
For large positive x the first graph is positive and the second negative, so the product is indeed negative.
For large negative x the first graph is negative and the second positive, so the product would again be negative.
*@
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Self-critique rating:
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Question: For the same functions f(x) and g(x), sketch the graph of the quotient function f(x) / g(x).
Describe all vertical and horizontal asymptotes of your graph.
On what intervals is your graph positive, and on what intervals negative?
How could you have answered this question by just looking at the graphs of the two functions?
Given just the graphs of the two functions, how would you determine whether and where the given quotient function has one or more vertical asymptotes?
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Your solution:
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(1=1)/(4=4)
My graph is constant so anything above y=1/4 or below would be an asymptote
1/4
Negative infinity
Given just the graphs of the two functions, how would you determine whether and where the given quotient function has one or more vertical asymptotes?
Black graph
(0,1.5) (0,.5)
Blue graph
(0,4.5) (0,-3.5)
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@&
Before further consideration of this problem, see if you can answer the following:
On what intervals is f(x) positive, and on what intervals is it negative?
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On what intervals is g(x) positive, and on what intervals is it negative?
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What has to be true of the values of two numbers a and b for the value of a/b to be negative?
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What has to be true of the values of f(x) and g(x) for f(x) / g(x) to have a negative value?
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On what intervals of the x axis will this condition hold?
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Where is f(x) equal to zero?
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Where is g(x) equal to zero?
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If f(x) is zero what can you say about the value of f(x) / g(x)?
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If g(x) is zero what can you say about the value of f(x) / g(x)?
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How do the vertical asymptotes of a quotient function happen?
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confidence rating #$&*: 1
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Question: In the graph below the linear function is f(x), the nonlinear function g(x).
Think about the graphs of the functions f(x) + g(x), f(x) * g(x) and f(x) / g(x).
What will be the x intercepts of each of these functions?
On what intervals will each function be positive?
What will be the sign of each of these functions for large positive x, and for large negative x?
What will be the horizontal and vertical asymptotes of the quotient function?
Describe the behavior of the quotient function near each of its vertical asymptotes.
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Your solution:
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What will be the x intercepts of each of these functions?
5,-3,1/4
On what intervals will each function be positive?
5
1/4
What will be the sign of each of these functions for large positive x, and for large negative x?
5,-3,1/4
I am not really sure what you mean by “large positive x, and for large negative x”???
What will be the horizontal and vertical asymptotes of the quotient function?
(3,0) (0,-1) (0,-2) (4,0)
Describe the behavior of the quotient function near each of its vertical asymptotes.
The graph curves right up before it can touch the asymptote
Then on another one it curves down
The next will curve up just to get close then curve back down
The next curves down and keeps going
@&
The following questions all have to do with the answers to the posed questions. See if you can answer these questions:
One what intervals is f(x) positive, and on what intervals is it negative?
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On what intervals is g(x) positive, and on what intervals is it negative?
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For what x values is f(x) zero?
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For what x values is g(x) zero?
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At what value or values of x do the functions have equal and opposite y values?
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If the graphs follow the apparent trend, then if x is a large positive number, is f(x) positive or negative, and is g(x) positive or negative, and would f(x) or g(x) have the greater magnitude?
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If the graphs follow the apparent trend, then if x is a large negative number, is f(x) positive or negative, and is g(x) positive or negative, and would f(x) or g(x) have the greater magnitude?
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What has to be true of the values of f(x) and g(x) in order for a vertical asymptote to occur at some value of x?
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#$&*
*@
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confidence rating #$&*: 1
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Self-critique rating:
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"
@&
You're making progress on this, but it's clear that you don't completely understand it.
Check my notes and see if you can revise your work on the last four questions, and answer the questions I've posed there.
I would love to see you ace this course.
*@