Assignment 17

course MTH 158

7/19/09 11:00 pm

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.

017. `* 17

Note that you can't use a calculator graph to document your solutions to these problems. You have to use the analytical methods as in the given solutions.

Documentation is required on tests, and while you may certainly use the calculator to see symmetry, intercepts etc., you have to support your solutions with the algebraic details of why the graph looks the way it does.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

Self-critique Rating:

*********************************************

Question: * 2.2.34 / 10 (was 2.2.6). Point symmetric to (-1, -1) wrt x axis, y axis, origin.

What point is symmetric to the given point with respect to each: x axis, y axis, the origin?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

symetric with the x axis (-1,1)

symetric with the y axis (1,-1)

symetric with the origin (1,1)

Confidence rating: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

* * There are three points:

The point symmetric to (-1, -1) with respect to the x axis is (-1 , 1).

The point symmetric to (-1, -1) with respect to the y axis is y axis (1, -1)

The point symmetric to (-1, -1) with respect to the origin is ( 1,1). **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

Self-critique Rating:

*********************************************

Question: * 2.2.43 / 19 (was 2.2.15). Parabola vertex origin opens to left. **** Give the intercepts of the graph and tell whether the graph is symmetric to the x axis, to the y axis and to the origin. Explain how you determined the answer to each question.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

I did not read anything in this chapter about the Parabola vertex

Confidence rating: 0

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

* * The graph intercepts both axes at the same point, (0,0)

The graph is symmetric to the x-axis, with every point above the x axis mirrored by its 'reflection' below the x axis. **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

I can see how this would look on a graph I just dont understand it

Self-critique Rating:

*********************************************

Question: * 2.2.48 / 24 (was 2.2.20). basic cubic poly arb vert stretch **** Give the intercepts of the graph and tell whether the graph is symmetric to the x axis, to the y axis and to the origin. Explain how you determined the answer to each question.

The graph s strictly increasing except perhaps at the origin where it might level off for just an instant, in which case the only intercept is at the origin (0, 0).

I dont understand this question either. I didnt see anything in this chapter about that

The graph is symmetric with respect to the origin, since for every x we have f(-x) = - f(x). For example, f(2) = 8 and f(-2) = -8. It looks like f(1) = 1 and f(-1) = -1. Whatever number you choose for x, f(-x) = - f(x).

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

Self-critique Rating:

*********************************************

Question: * 2.2.62 / 40 (was 2.2.36). 4x^2 + y^2 = 4 **** List the intercepts and explain how you made each test for symmetry, and the results of your tests.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

4x^2 + y^2 = 4

4x^2 = 4 Made the y (0) to find the x intercept

4x^2/4 = 4/4

x^2 = 1

sqrt x^2 = +/- sqrt(1)

x = +/- 1

x intercepts = (1,0) and (-1,0)

4x^2 + y^2 = 4

y^2 = 4 Made the x (0) to find the y intercept

sqrt (y^2) = sqrt (4)

y = +/- 2

y intercept = (0,2) and (0,-2)

all origins have symmetry with respect to x,y and origin

to test for x I set y = to -y to see if the equation was the same

to test for y I set x = -x to see if the original equation was the same

to test for the origin I set both x and y to -x and -y and the original equation was still the same

Confidence rating: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

* * Starting with 4x^2 +y^2 = 1 we find the x intercept by letting y = 0.

We get

4x^2 + 0 = 1 so

4x^2 = 1 and

x^2=1/4 . Therefore

x=1/2 or -1/2 and the x intercepts are

(1/2,0) and ( -1/2,0).

Starting with 4x^2 +y^2 = 1 we find the y intercept by letting x = 0.

We get

0 +y^2 = 1 so

y^2 = 1 and

y= 1 or -1, giving us y intercepts

(0,1) and (0,-1).

To test for symmetry about the y axis we substitute -x for x. If there's no change in the equation then the graph will be symmetric to about the y axis.

Substituting we get 4 (-x)^2 + y^2 = 1. SInce (-x)^2 = x^2 the result is 4 x^2 + y^2 = 1. This is identical to the original equation so we do have symmetry about the y axis.

To test for symmetry about the x axis we substitute -y for y. If there's no change in the equation then the graph will be symmetric to about the x axis.

Substituting we get 4 (x)^2 + (-y)^2 = 1. SInce (-y)^2 = y^2 the result is 4 x^2 + y^2 = 1. This is identical to the original equation so we do have symmetry about the x axis.

To test for symmetry about the origin we substitute -x for x and -y for y. If there's no change in the equation then the graph will be symmetric to about the origin.

Substituting we get 4 (-x)^2 + (-y)^2 = 1. SInce (-x)^2 = x^2 and (-y)^2 - y^2 the result is 4 x^2 + y^2 = 1. This is identical to the original equation so we do have symmetry about the origin. **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

Self-critique Rating:

*********************************************

Question: * 2.2.68 / 46 (was 2.2.42). y = (x^2-4)/(2x) **** List the intercepts and explain how you made each test for symmetry, and the results of your tests.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

the x intercept is (-2,0) and (2,0)

there are no y intercepts division by 0 is undefined so the line is vertical

we only have symmetry about the origin

Confidence rating: 2

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

* * We do not have symmetry about the x or the y axis, but we do have symmetry about the origin:

To test for symmetry about the y axis we substitute -x for x. If there's no change in the equation then the graph will be symmetric to about the y axis.

}Substituting we get y = ( (-x)^2 - 4) / (2 * (-x) ). SInce (-x)^2 = x^2 the result is y = -(x^2 - 4) / (2 x). This is not identical to the original equation so we do not have symmetry about the y axis.

To test for symmetry about the x axis we substitute -y for y. If there's no change in the equation then the graph will be symmetric to about the x axis.

Substituting we get -y = (x^2-4)/(2x) , or y = -(x^2-4)/(2x). This is not identical to the original equation so we do not have symmetry about the x axis.

To test for symmetry about the origin we substitute -x for x and -y for y. If there's no change in the equation then the graph will be symmetric to about the origin.

Substituting we get -y = ((-x)^2-4)/(2(-x)) SInce (-x)^2 = x^2 the result is -y = -(x^2-4)/(2x), or multiplying both sides by -1, our result is y = (x^2-4)/(2x). This is identical to the original equation so we do have symmetry about the origin. **