College Algebra selected end-of-assignment questions and solutions for Assignment 17

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.

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?

Your solution:

Confidence Assessment:

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.

Your solution:

Confidence Assessment:

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):

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).

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.

Your solution:

Confidence Assessment:

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.

STUDENT QUESTION

I was able to get the x and y axis numbers but I still can’t figure out how to substitute them in to the equations to find the symmetry and the solution confused me more. Got any suggestions???

INSTRUCTOR RESPONSE

See if you can answer the following:

If you replace x in the equation

4 x^2 + y^2 = 1

with -x, what is your equation?

What do you get when you simplify the resulting equation?

Is the resulting equation the same as the original equation, or different?

If you replace y in the equation

4 x^2 + y^2 = 1

with -y, what is your equation?

What do you get when you simplify the resulting equation?

Is the resulting equation the same as the original equation, or different?

If you replace x in the equation

4 x^2 + y^2 = 1

with -x, and replace y with -y, what is your equation?

What do you get when you simplify the resulting equation?

Is the resulting equation the same as the original equation, or different?

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.

Your solution:

Confidence Assessment:

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. **

PARTIAL STUDENT SOLUTION

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

INSTRUCTOR RESPONSE

You have substituted -x for x, which is the first step in checking for symmetry about the y axis.

Now simplify this expression. Is the simplified expression equal to the original expression?

If so you have symmetry about the y axis.

To find the x intercept, note that a point is on the x axis if and only if its y coordinate is zero. So you substitute y = 0 and solve for x.

To find the y intercept, note that a point is on the y axis if and only if its x coordinate is zero. So you substitute x = 0 and solve for y.

Self-critique (if necessary):

Self-critique Rating: