If your solution to stated problem does not match the given
solution, you should self-critique per instructions at
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.
025. `* 25
Question:
* 3.5.12 (was
3.4.6). Upward parabola vertex (0, 2).
What equation matches this function?
The correct equation is y = x^2 + 2. How can you tell it isn't y = 2 x^2 +
2?. How can you tell it isn't y =
(x+2)^2.
GOOD STUDENT ANSWERS:
it isnt y = 2x^2 + 2 because that would make it a verical
stretch and it isnt vertical it is just a parabola.
it isnt y = (x+2)^2 because that would make it shift to the
left two units and it did not.
INSTRUCTOR NOTE:
Good answers. Here is
more detail:
The basic y = x^2 parabola has its vertex at (0, 0) and also
passes through (-1, 1) and (1, 1).
y = 2 x^2 would stretch the basic y = x^2 parabola
vertically by factor 2, which would move every point twice as far from the x
axis. This would leave the vertex at (0,
0) unchanged and would move the points (-1, 1) and (1, 1) to (-1, 2) and (1,
2).
Then y = 2x^2 + 2 would shift this function vertically 2
units, so that the points (0, 0), (-1, 2) and (1, 2) would shift 2 units upward
to (0, 2), (-1, 4) and (1, 4).
The resulting graph coincides with the given graph at the
vertex but because of the vertical stretch it is too steep to match the given
graph.
The function y = (x + 2)^2 shifts the basic y = x^2 parabola
2 units to the left so that the vertex would move to (-2, 0) and the points
(-1, 1) and (1, 1) to (-3, 1) and (-1, 1).
The resulting graph does not coincide with the given graph.
y = x^2 + 2 would
shift the basic y = x^2 parabola vertically 2 units, so that the points (0, 0),
(-1, 1) and (1, 1) would shift 2 units upward to (0, 2), (-1, 3) and (1,
3). These points do coincide with points
on the given graph, so it is the y = x^2 + 2 function that we are looking for. **
Self-critique (if necessary):
Self-critique Rating:
Question:
* 3.5.16 (was
3.4.10). Downward parabola.
The correct equation is y = -2 x^2. How can you tell it isn't y = 2 x^2?. How can you tell it isn't y = - x^2?
Your solution:
Confidence Assessment:
Given Solution:
* * The basic y = x^2 parabola has its vertex at (0, 0) and
also passes through (-1, 1) and (1, 1).
y = -x^2 would stretch the basic y = x^2 parabola vertically
by factor -1, which would move every point to another point which is the same
distance from the x axis but on the other side.
This would leave the vertex at (0, 0) unchanged and would move the
points (-1, 1) and (1, 1) to (-1, -1) and (1, -1).
y = -2 x^2 would stretch the basic y = x^2 parabola
vertically by factor -2, which would move every point to another point which is
twice the distance from the x axis but on the other side. This would leave the vertex at (0, 0)
unchanged and would move the points (-1, 1) and (1, 1) to (-1, -2) and (1,
-2). This graph coincides with the given
graph.
y = 2 x^2 would stretch the basic y = x^2 parabola
vertically by factor 2, which would move every point twice the distance from
the x axis and on the same side. This
would leave the vertex at (0, 0) unchanged and would move the points (-1, 1)
and (1, 1) to (-1, 2) and (1, 2). This
graph is on the wrong side of the x axis from the given graph. **
Self-critique (if necessary):
Self-critique Rating:
Question:
* 3.5.18 (was
3.4.12). V with vertex at origin.
What equation matches this function?
The correct equation is y = 2 | x | . How can you tell it isn't y = | x | ?.
How can you tell it isn't y = | x | + 2?
Your solution:
Confidence Assessment:
Given Solution:
* * The basic y = | x |
graph is in the shape of a V with a 'vertex' at (0, 0) and also passes
through (-1, 1) and (1, 1).
The given graph has the point of the V at the origin but
passes above (-1, 1) and (1, 1).
y = | x | + 2 would
shift the basic y = | x | graph vertically 2 units, so that the points (0, 0),
(-1, 1) and (1, 1) would shift 2 units upward to (0, 2), (-1, 3) and (1,
3). The point of the V would lie at (0,
2). None of these points coincide with
points on the given graph
y = 2 | x | would
stretch the basic y = | x | graph vertically by factor 2, which would move
every point twice the distance from the x axis and on the same side. This would leave the point of the 'V' at (0,
0) unchanged and would move the points (-1, 1) and (1, 1) to (-1, 2) and (1,
2). This graph coincides with the given
graph. **
Self-critique (if necessary):
Self-critique Rating:
Question:
* 3.5.30 (was
3.4.24). Transformations on y = sqrt(x).
What basic function did you start with and, in order, what
transformations were required to obtain the graph of the given function?
What is the function after you shift the graph up 2 units?
Your solution:
Confidence Assessment:
Given Solution:
* * ERRONEOUS STUDENT RESPONSE: y = x^2 + 2
INSTRUCTOR CORRECTION:
y = sqrt(x) is not the same as y = x^2.
y = sqrt(x) is the square root of x, not the square of x.
Shifting the graph of y = sqrt(x) + 2 up so units we would
obtain the graph y = sqrt(x) + 2. **
What is the function after you then reflect the graph about
the y axis?
** To reflect a graph about the y axis we replace x with -x.
It is the y = sqrt(x) + 2 function that is being reflected
so the function becomes
y = sqrt(-x) + 2. **
What is the function after you then fhist the graph left 3
units?
Your solution:
Confidence Assessment:
Given Solution:
* * To shift a graph 3 units to the left we replace x with x
+ 3.
It is the y = sqrt(-x) + 2 function that is being reflected
so the function becomes
y = sqrt( -(x+3) ) + 2. **
Self-critique (if necessary):
Self-critique Rating:
Question:
* 3.5.42 (was
3.4.36). f(x) = (x+2)^3 - 3.
What basic function did you start with and, in order, what
transformations were required to obtain the graph of the given function?
Describe your graph.
Give three points on your graph and tell to which basic points on the
graph of your basic function each corresponds.
Your solution:
Confidence Assessment:
Given Solution:
* * Starting with y = x^3 we replace x by x
+2 to shift the
graph 2 units left. We obtain
y = (x + 2)^3.
We then shift this graph -3 units vertically by adding
-3 to
the value of the function, obtaining
y = (x + 2 )^3 -
3.
The basic points of the y = x^3 graph are (-1, -1), (0, 0)
and (1, 1). Shifted 2 units left and
3
units down these points become
(-1 - 2, -1 -3
) = (-3, -4),
(0 - 2, 0 -3) = (-2,
-3) and
(1 - 2, 1 -3) = (-2,
-2).
The points you give might not be the same three points but
should be obtained by a similar process, and you should specify both the
original point and the point to which it is transformed. **
Self-critique (if necessary):
Self-critique Rating:
Question:
* 3.5.58 (was 3.4.40). h(x) = 4 / x + 2.
What basic function did you start with and, in order, what
transformations were required to obtain the graph of the given function?
Describe your graph.
Give three points on your graph and tell to which basic points on the
graph of your basic function each corresponds.
Your solution:
Confidence Assessment:
Given Solution:
* * Note on the function y = 1 / x.
Your text does not seem to have
included an explanation of the graph of y = 1 / x in this section.
Your author has good reasons for this (a very good discussion follows in the
next chapter when it discusses rational functions), but I think the
following explanation will be useful to many students.
A short table for y = 1 / x helps
explain the features of its graph:
| x |
y = 1 / x |
| 1/10 |
10 |
| 1/2 |
2 |
| 1 |
1 |
| 2 |
1/2 |
| 10 |
1/10 |
Before reading further you should hand-sketch a graph
corresponding to this table. You aren't likely to understand the rest
of this explanation without taking a couple of minutes to do so.
We see that as x increases from 1 to 2 to 10 the value
of 1/x decreases from 1 to 1/2 to 1/10. As x continues to increase, y
= 1/x will continue to decrease, but will still remain positive. Thus,
as we move to the right the graph descends toward the x axis, never reaching
it but becoming closer and closer, with no limit to how close it will
eventually become.
Now we move to the left of x = 1, moving through
decreasing x values which approach 0. As x decreases from 1 to 1/2 to
1/10, the value of y = 1 / x increases from 1 to 2 to 10. As the value
of x gets smaller and smaller, approaching 0 but never reaching 0, the value
of 1 / x increases more and more rapidly.
You should identify the points (1/10, 10), (1/2, 2),
(1, 1), (2, 1/2) and (10, 1/10) on the graph below, and compare this graph
to yours.
The positive x axis is a horizontal asymptote for the
graph: as we move to the right the graph approaches this line, with no
limit to how closely it will eventually approach.
The positive y axis is also an asymptote. The
closer x approaches 0, the greater the value of the function. There is
no limit to how closely x can approach 0, so there is no limit to how large
the values of 1 / x can become. There is no limit to how closely the
graph approaches the y axis, nor is there a limit to how large the y values
can become.
Now we consider the more abbreviated table below:
| x |
y = 1 / x |
| -2 |
-1/2 |
| -1 |
-1 |
| -1/2 |
-2 |
| 1/2 |
2 |
| 1 |
1 |
| 2 |
1/2 |
This table includes three negative values of x.
You should add the corresponding three points to your graph. If you do
so it should become clear to you how the entire graph of y = 1 / x is
constructed, and why.
Having plotted these points the trend of the graph
should be pretty clear to you. All negative values of x result in
negative values of y = 1 / x. As we move to the left from x = -1, the
graph moves closer to the x axis, and the negative x axis is a horizontal
asymptote. As we move from x = -1 toward x = 0, the graph approaches
the y axis as an asymptote.
The graph is depicted below. You should locate
the three new points (-2, -1/2), (-1, -1) and (-1/2, -2) on this graph.
Be sure you understand how the horizontal asymptotes on the positive and
negative x axes occur, as well as the vertical asymptotes on the positive
and negative y axes.
It is useful to also notice the symmetry behaviors of
the graph.
Now for an explanation of the problem:
We start with the basic reciprocal function y = 1 / x,
which has vertical asymptotes at the y axis and horizontal asymptotes along the x axis and passes through the points (-1, -1) and (1,
1).
To get y = 4 / x we must multiply y = 1 / x by 4. Multiplying a function by 4 results in a
vertical stretch by factor 4, moving every point 4 times as far from the x
axis.
This will not affect the location of the vertical or
horizontal asymptotes but for example the points (-1, -1) and (1, 1) will be
transformed to (-1, -4) and (1, 4). At
every point the new graph will at this point be 4 times as far from the x axis.
At this point we have the graph of y = 4 / x. The function we wish to obtain is y = 4 / x +
2.
Adding 2 in this manner increases the y value of each point
by 2. The point (-1, -4) will therefore
become (-1, -4 + 2) = (-1, -2). The
point (1, 4) will similarly be raised to (1, 6). Since every point is raised vertically by 2
units, with no horizontal shift, the y axis will remain an asymptote. As the y = 4 / x graph approaches the x axis,
where y = 0, the y = 4 / x + 2 graph will approach the line y = 0 + 2 = 2, so
the horizontal asymptotes to the right and left will consist of the line y = 2.
Our final graph will have asymptotes at the y axis and at
the line y = 2, with basic points (-1, -2) and (1, 6).
STUDENT QUESTION:
When you say multiplying a function
by 4, are you saying you would do this:</p>
1/x * 4/1 and if so, wouldn’t that give you 4/x then would you substitute the x
= 1 and x = -1 to get
the next set of points (-1, -4) and (1, 4)?</p>
INSTRUCTOR RESPONSE:
That is so.
However note that (1, 4) is 4 times as far from the x axis as the point (1, 1),
and (-1, -4) is 4 times as far from the x axis as (-1, -1).
Given a graph of y = 1 / x, we could have simply taken each point of the graph
and moved it 4 times as far from the x axis, without ever calculating y = 4 / x.
We might well want to calculate the values, but it's important to visualize,
without the point-by-point calculations, how the graphs transform.
For example the graph below depicts the
two functions. You should identify the points (-1, -1) and (1, 1), and
also the points (-1, -4) and (1, 4). Then look at the vertical blue line
and convince yourself that the point where this line intersects the y = 4 / x
graph is four times as far from the x axis as its intersection with the y = 1 /
x graph. Then visualize several other vertical lines, and observe how the
same statement applies no matter where you draw the vertical line (with the
exception of the y axis, where neither function is defined).
STUDENT QUESTION:
So, by adding the 2, the original points (-1, -4) and (1, 4)
were just used to show how 4/x would look, and then by adding the y = 4/x + 2 is
the “real” graph we are trying to obtain?
INSTRUCTOR RESPONSE:
That's about right.
When we add 2 to get the function y = 4 /
x + 2, we are increasing the y values by 2. This 'raises' the y = 4 / x
graph 2 units. The graph below depicts the previous graphs of y = 1 / x
and y = 4 / x, as well as the 'new' graph y = 4 / x + 2.
You can note that the point (1, 4) on the
y = 4 / x graph transforms to the point (1, 6), and the point (-1, -4) to the
point (-1, -2). The 'blue' vertical line intersects the 'new' graph at a
point 2 units higher than the y = 4 / x graph, and the same will be true of any
vertical line.
STUDENT QUESTION:
It says where y = 0, the y = 4/x + 2 graph
will approach the line y = 0 + 2 = 2……throws me off. I don’t understand this
part…..and the horizontal asymptotes to the right and left will consist of the
line y = 2…I tried
drawing the graph and looking at it, but it doesn’t appear that anything = 2.
INSTRUCTOR RESPONSE:
The x axis is the line y = 0.
To see this, pick several points on
the x axis. What is the y coordinate of each? Your answer will
always be that the y coordinate is 0.
If we raise each point of the line y = 0 by 2 units we will
get the line y = 2.
To see this, pick several points on the x axis and
'raise' each of them by 2 units in the y direction. Your new y
coordinate will always be 2. Your points will all lie on the line y =
2.
The y = 4 / x graph had an asymptote at the x axis.
That is, it was asymptotic to the line y = 0.
The y = 4 / x + 2 graph will therefore have an asymptote at y
= 2.
The graph below is the same as the previous graph, but the
line y = 2 has been added. You should observe how the graph of y = 4 / x +
2 approaches this line as a horizontal asymptote.
Here's a picture of the final graph of y = 4 / x + 2, along
with the horizontal asymptote y = 2.
Self-critique (if necessary):
Self-critique Rating:
Question:
* 3.5.58
/ 3.5.60 (was
3.4.54). f(x) = -4 sqrt(x-1).
What basic function did you start with and, in order, what
transformations were required to obtain the graph of the given function?
Describe your graph.
Give three points on your graph and tell to which basic points on the
graph of your basic function each corresponds.
Your solution:
Confidence Assessment:
Given Solution:
* * Starting with the basic function y = sqrt(x) we replace x
by x - 1, which shifts the graph right 1 unit, then we stretch the graph by
factor -4, which moves every point 4 times further from the x axis and to the
opposite side of the x axis.
The points (0, 0), (1, 1) and (4, 2) lie on the graph of the
original function y = sqrt(x).
Shifting each point 1 unit to the right we have the points
(1, 0), (2, 1) and (5, 2).
Then multiplying each y value by -4 we get the points
(1, 0), (2, -4) and (5, -8).
Note that each of these points is 4 times further from the x
axis than the point from which it came, and on the opposite side of the x axis.
**
STUDENT QUESTION
I see that you multiply the y values
by -4 but why do you multiply the y values and not the x values?
INSTRUCTOR RESPONSE
Think about the graph of y = 2 x^2
compared to the graph of y = x^2.
To make a brief table for either function you might use x values -1, 0 and 1.
The second function gives you points (-1, 1), (0, 0) and (1, 1).
The first function gives you points (-1, 2), (0, 0) and (1, 2).
If you multiply the y values of the first function by 2, you get the y values of
the second. The x values don't change.
In general if the function is y = f(x), multiplying f(x) by a 2 doubles the y
values, while adding, say, 5 would increase every y value by 5.
Self-critique (if necessary):
Self-critique Rating:
Question:
* 3.5.63
/ 3.5.66 (was
3.4.60). Piecewise linear (-4, -2) to
(-2, -2) to (2, 2) to (4, -2).
Describe your graphs of G(x) = f(x+2), H(x) = f(x+2) - 2 and g(x) = f(-x).
Give the four points on each of these graphs that correspond
to the four points labeled on the original graph.
Your solution:
Confidence Assessment:
Given Solution:
* * G(x) = f(x+2) shifts the points of the f(x) function 2
units to the left, so instead of going from (-4, -2) to (-2, -2) to (2, 2) to
(4, -2) the G(x) function goes from (-4-2, -2) to (-2-2, -2) to (2-2, 2) to
(4-2, -2), i.e., from (-6, -2) to (-4, -2) to (0, 2) to (2, -2).
H(x) = f(x+2) - 2 shifts the points of the f(x) function 1
unit to the left and 2 units down, so instead of going from (-4, -2) to (-2,
-2) to (2, 2) to (4, -2) the H(x) function goes from (-4-1, -2-2) to (-2-1,
-2-2) to (2-1, 2-2) to (4-1, -2-2), i.e., from (-5, -4) to (-3, 0) to (1, 0) to
(3, -4).
g(x) = f(-x) replaces x with -x, which shifts the graph
about the y axis. Instead of going from
(-4, -2) to (-2, -2) to (2, 2) to (4, -2)
the g(x) function goes from (4, -2) to (2, -2) to (-2, 2) to (-4, -2)
You should carefully sketch all these graphs so you can see
how the transformations affect the graphs.
**
Self-critique (if necessary):
Self-critique Rating:
Question:
* 3.5.70
/ 3.5.78 (was 3.4.72). Complete square and graph f(x) = x^2 + 4 x +
2.
Give the function in the designated form. Describe your graph this function.
Your solution:
Confidence Assessment:
Given Solution:
* * To complete the square on f(x) = x^2 + 4x + 2 we first
look at x^2 + 4x and note that to complete the square on this expression we
must add (4/2)^2 = 4.
Going back to our original expression we write f(x) = x^2 +
4x + 2 as
f(x) = x^2 + 4x + 4 - 4 + 2 the group and simplify to get
f(x) = (x^2 + 4x + 4) - 2.
Since the term in parentheses is a perfect square we write this as
f(x) = (x+2)^2 - 2.
This shifts the graph of the basic y = x^2 function 2 units
to the left and 2 units down, so the basic points (-1, 1), (0, 0) and (1, 1) of
the y = x^2 graph shift to (-3, -1), (-2, -2) and (-1, -1). **
STUDENT QUESTION
I don’t understand how the function went from
f(x) = x^2 + 4x + 4 - 4 + 2 to
f(x) = (x^2 + 4x + 4) – 2.
How did you get the +4 when the
previous step had a positive and a negative 4? Wouldn’t that cancel each other
out?
INSTRUCTOR RESPONSE
The object was to get a perfect square out of x^2 + 4 x.
That's why you squared 4/2, giving you 4. x^2 + 4 x + 4 is a perfect square.
Of course we can't just add 4 to the original expression--that would change its
value.
However we can add and subtract 4, which doesn't change the value
of the expression.
This is how we got the expression x^2 + 4 x + 4 - 4 + 2.
Grouping the perfect square our expression becomes
(x^2 + 4 x + 4) - 4 + 2
To simplify this expression we need only find -4 + 2, which is -2. Our
expression becomes
(x^2 + 4 x + 4) - 2, which we can write as
(x + 2)^2 - 2.
Self-critique (if necessary):
Self-critique Rating: