Assignment 17

course Mth 158

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Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

11:35:47

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.

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11:37:54

query 2.2.10 (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?

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RESPONSE -->

Point (-1,1) symmetrical about the y axis is (1,-1)

Point (-1,1) symmetrical about the x axis is (-1,1)

Point (-1,1) symmetrical about the origin is (1,1)

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11:38:08

** 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). **

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11:40:00

**** query 2.1.19 (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.

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RESPONSE -->

The parabola intercepts at the origin (0,0)

The parabola is symmetrical about the x axis.

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11:40:13

** 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. **

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11:41:53

**** query 2.2.24 (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.

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I dont have a graphing calculator but it apears to me that the intercepts of the graph are: (-.5,0) (.5,0) (0,0)

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11:42:05

STUDENT SOLUTION: origin of the graph is (-.5,0) and (.5,0)

graph is symmetric to the origin

INSTRUCTOR COMMENT:

Check and see whether the graph passes thru the origin (0, 0), which according to my note it should (but my note could be wrong).

}If so, and if it is 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 believe the graph is symmetric with respect to the origin, and if so (0, 0) must be an intercept.

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11:46:23

**** query 2.2.40 (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.

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RESPONSE -->

4x^2 + y^2 = 4

4x^2 + 0 = 4

x^2 = 1

x = +-1

(1,0) (-1,0)

y^2 - 4 = 0

y^2 = 4

y = sqrt(4)

y = +-2

(0,2) (0,-2)

Intercepts are: (1,0) (-1,0) (0,2) (0,-2)

4-x^2 + -y^2 = 4 = 4x^2 + y^2 = 4

graph is symmetrical.

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11:47:13

** 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. **

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RESPONSE -->

Problem in book has the equation 4x^2 + y^2 = 4.

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11:51:45

**** query 2.2.46 (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.

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RESPONSE -->

y = (x^2 - 4) / 2 *x^4

y = (0^2 - 4)/2(0)^4

y = 0

(0,0)

-y = (x^2 - 4) / 2 *x^4 is not equivalent to y = (x^2 - 4) / 2 *x^4

y = (-x^2 - 4) / 2 *-x^4 is equivalent to y = (x^2 - 4) / 2 *x^4

-y = (-x^2 - 4) / 2 *-x^4 is not equivalent to origin

graph is symmetrical to the y axis.

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11:52:19

** 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. **

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RESPONSE -->

Problem in book has the equation y = (x^2 - 4) / 2 *x^4

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Assignment 17

Very good work. I'll have to double-check the correspondence with text problems; thanks for pointing out the discrepancies.