Results 1 to 8 of 8

Math Help - Test Review

  1. #1
    Newbie
    Joined
    Mar 2012
    From
    Canada
    Posts
    11

    Test Review

    Have a couple more questions I'm struggling to understand for tomorrows test, if anyone can reply and explain to me how I reach my answer exactly or provide me a link where I can learn how to conquer the equation that would be greatly appreciated.
    Question 1:
    Consider the circle given X^2+Y^2=1. Determine the equation of the transformed function if the graph undergoes a horizontal stretch by a factor of 1/2, vertical stretch by a factor of 3, horizontal translation of 1 unit right and a vertical translation of 4 units down. (I essentially don't know where to start here)


    Question 2:
    Consider the conic section given by (x-2)^2 + (y+3)^2. Convert to general form and state the coordinates of the center and vertices.
    36 16

    Question 3:
    AngleƟ has a terminal arm in the third quadrant. If cotƟ= 12, the values of the remaining five primary and reciprocal trig ratios.
    5

    Any help is greatly appreciate!
    Thanks,

    Shane
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,517
    Thanks
    771

    Re: Test Review

    Quote Originally Posted by Prentz View Post
    Question 1:
    Consider the circle given X^2+Y^2=1. Determine the equation of the transformed function if the graph undergoes a horizontal stretch by a factor of 1/2, vertical stretch by a factor of 3, horizontal translation of 1 unit right and a vertical translation of 4 units down.
    Consider a horizontal stretch by a factor of 1/2. Then each point (x, y) on the plot becomes (x/2, y). Let's call this new point (x', y'); then x' = x/2 and y' = y, or x = 2x' and y = y'. Substituting x(x') and y(y') into the original equation, we get (2x')^2 + (y')^2 = 1, which is the new equation. The remaining steps can be done similarly.

    Quote Originally Posted by Prentz View Post
    Question 2:
    Consider the conic section given by (x-2)^2 + (y+3)^2. Convert to general form and state the coordinates of the center and vertices.
    This is not an equation. However, for any R, (x-2)^2 + (y+3)^2 = R^2 is a circle with center (2, -3) and radius R.

    Quote Originally Posted by Prentz View Post
    Question 3:
    AngleƟ has a terminal arm in the third quadrant. If cotƟ= 12, the values of the remaining five primary and reciprocal trig ratios.
    Since a terminal arm is in the third quadrant, sin Ɵ and cos Ɵ are negative. Denoting x = -cos Ɵ and y = -sin Ɵ, you can write a relation between x and y and then use the Pythagorean theorem to find x and y.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2012
    From
    Canada
    Posts
    11

    Re: Test Review

    Thank you Emakarov, to clear it up and if this helps the first question 2 was suppose to be:
    [U](x-2)^2/36 + [U](y+3)^2/16
    for some reason it didn't allow the number to stay under the equation for the division.
    Also Question 3 is: cotƟ=12/5
    again the equation didn't have the 5 under it.
    If this changes anything please let me know!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,517
    Thanks
    771

    Re: Test Review

    [QUOTE=Prentz;711136]the first question 2 was suppose to be:
    [U](x-2)^2/36 + [U](y+3)^2/16[quote]My guess is this means \frac{(x-2)^2}{36}+\frac{(y+3)^2}{16}. This is still not an equation, but \left(\frac{x-2}{6}\right)^2+\left(\frac{y+3}{4}\right)^2=1 corresponds to an ellipse with the center at (2, -3).

    Quote Originally Posted by Prentz View Post
    Convert to general form and state the coordinates of the center and vertices.
    By "vertices" do you mean foci (plural of "focus")? They are on the major axes at the distance \sqrt{a^2-b^2} from the center (a and b are semi-major axis and semi-minor axis, respectively).

    Quote Originally Posted by Prentz View Post
    Also Question 3 is: cotƟ=12/5
    This does not change the solution method.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Mar 2012
    From
    Canada
    Posts
    11

    Re: Test Review

    Thank you for the help I have sorta out question 1 and 3 now and as far as question 2 yes I mean find the center and foci need to be found.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,517
    Thanks
    771

    Re: Test Review

    Quote Originally Posted by Prentz View Post
    as far as question 2 yes I mean find the center and foci need to be found.
    I am not sure if this presents a difficulty to you in view of post #4.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,412
    Thanks
    1328

    Re: Test Review

    [QUOTE=emakarov;711139][QUOTE=Prentz;711136]the first question 2 was suppose to be:
    [U](x-2)^2/36 + [U](y+3)^2/16
    My guess is this means \frac{(x-2)^2}{36}+\frac{(y+3)^2}{16}. This is still not an equation, but \left(\frac{x-2}{6}\right)^2+\left(\frac{y+3}{4}\right)^2=1 corresponds to an ellipse with the center at (2, -3).

    By "vertices" do you mean foci (plural of "focus")? They are on the major axes at the distance \sqrt{a^2-b^2} from the center (a and b are semi-major axis and semi-minor axis, respectively).
    The vertices of an ellipse are not the foci, they are the points on the ellipse where the curvature is maximum or minimum. They are the points at the ends of the axes. For an ellipse of the form \frac{(x- x_0)^2}{a^2}+ \frac{(y- y_0)^2}{b^2}= 1 they are at (x_0, y_0+ b), (x_0, y_0- b), (x_0+ a, y_0)], and [tex]x_0- a, y_0)[]/tex].

    This does not change the solution method.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Mar 2012
    From
    Canada
    Posts
    11

    Re: Test Review

    Thank you halls I appreciate you clearing that up, Is it possible to have you apply that to the equation I posted above? Just to give me a better understand of how I complete the question?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Pre-Cal Test Review Questions
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: November 7th 2011, 08:56 AM
  2. review for assessment test
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: January 19th 2009, 11:04 PM
  3. Algebra Review for Test
    Posted in the Algebra Forum
    Replies: 1
    Last Post: October 5th 2008, 07:30 AM
  4. Test Review
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: February 17th 2008, 07:48 AM
  5. ..a little help in my review sheet for test
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: November 13th 2006, 04:50 AM

Search Tags


/mathhelpforum @mathhelpforum