assignment 17

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course Mth 158

10/16 5

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?

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Your solution:

points symmetric to (-1,1) with respect to:

x axis= (-1,1)

y axis= (1, -1)

origin = (1,1)

confidence rating #$&*: 3

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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). **

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Self-critique (if necessary):ok

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Self-critique Rating:

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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.

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Your solution:

Looking at the graph, the intercepts of both axis are right on (0,0) As for symmetry, if you were to fold it on the x-axis then it would be symmetrical, but it is not on the y-axis.

confidence rating #$&*: 3

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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. **

 

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Self-critique (if necessary):

ok

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Self-critique Rating:ok

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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.

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Your solution:

It's harder to tell what the intercepts are, but although the line curves to the right, it seems to level off right on top of (0,0) before continuing to increase with a curve to the left. So i'd say the intercept for the x and y axis is (0,0)

It is not symmetrical on either the x or the y axis

confidence rating #$&*: 2

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Given Solution:

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).

 

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).

 

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Self-critique (if necessary):

to be honest I don't really get how to tell whether the origin is symmetric.

@& The question isn't whether the origin is symmetric, but whether the graph is symmetric with respect to the origin.

This will be the case if for every point (x, y) there is a point on the graph for which the origin is the midpoint between than point and (x, y).

In other words, it's so if every point (x, y) has an opposite point (-x, -y) on the graph. Be sure you see that the origin would be the midpoint between two such points.

If y = f(x), the two points are (x, f(x)) and (-x, -f(x) ).

It follows that the condition for symmetry with the origin is f(-x) = - f(x).

Since f(-1) = - f(1), and f(-2) = - f(2), it appears that the graph might be symmetric with respect to the origin.

This condition is equivalent to the following:

If you draw a line segment from any point to the origin, then continue in the same direction for an equal distance, you will be at another point of the graph.*@

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Self-critique Rating:ok

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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.

 

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Your solution:

4x^2 +y^2 = 4 to find the x-int you let y=0

4x^2+0^2=4

4x^2=4

x^2=1

x=(+-)1 so the coordinates for the x int are (-1,0),(1,0)

to find y-int let x=0

4(0)^2 +y^2 = 4

y^2=4

y=2 so the coordinates for the y-int would be (0,2)

To test for symmetry in x-axis you would substitute -y for y and see if there's a change in the result of the equation. 4x^2 +y^2 = 4 would become 4x^2 -y^2 = 4 but -y^2 would just become y^2 anyway so the x-axis has symmetry

do the same for the y-axis except substitute -x for x. 4(-x)^2 -y^2 = 4 there still isn't a change, so apparently the y-axis has symmetry as well somehow

to find whether the origin has symmetry you substitute x and y with -x and -y but you find that the equation still doesn't change, so the origin is also symmetrical.

confidence rating #$&*: 3

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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. **

 

STUDENT COMMENT

 

Ok, I see we worked the problem very closely but I’m confused on where the 4x^2 + y^2 = 4 went.

INSTRUCTOR RESPONSE:

 

If you let y = 0 the equation becomes 4 x^2 + 0^2 = 1. As you see we then solve for x to obtain the x intercepts.

If you let x = 0 the equation becomes 4 * 0^2 + y^2 = 1. As you see we then solve for y to obtain the y intercepts.

Symmetry about the y axis means that you find the same y values at -x as you do at x.

Symmetry about the x axis means that you find the same x values at -y as you do at y.

Symmetry at the origin says that if (x, y) is a point on the graph, so is (-x, -y).

The given solution applies these tests.

 

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Self-critique (if necessary):

ok... why did you start with 4x^2 +y^2 = 1 instead of 4x^2 + y^2 = 4 ?

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Self-critique Rating: ok

@& That's probably a typo; possibly the problem was changed in the text and I failed to modify the solution when I modified the question.

In any case if you use 4 instead of 1, the same reasoning works with only minor modifications in some of the numbers.*@

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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.

 

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Your solution:

y = (x^2-4)/(2x) let y = 0 to find x-int

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

0= (x^2-4)

0=(x+2)(x-2)

so coordinates for x-intercepts are (-2,0),(2,0)

now let x=0 to find y-int

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

y=-4/0 So is it undefined?

X-axis symmetrical?

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

no

y-axis symmetrical?

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

no

origin symmetrical?

y = (x^2-4)/(2x) is -y = (-x^2-4)/(2(-x)) = -y = -(x^2-4)/(2x) multiply both sides by -1 and get

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

yes

confidence rating #$&*: 3?

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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. **

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Self-critique (if necessary):ok

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Self-critique Rating:ok"

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&#Good responses. See my notes and let me know if you have questions. &#