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Math Help - Graph transformations...

  1. #1
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    Graph transformations...

    Hello! I am a little confused about graph transformations and I have some questions that I don't really know how to solve...Please help! Thank you!

    1)
    The transformations A,B and C are as follows:
    A: A translation of 1 unit in the negative y direction
    B: A stretching parallel to the y-axis (with x-axis invariant) with a scale factor of 2
    C: A reflection about the x-axis

    A curve undergoes in succession, the above transformations A,B and C and the equation of the resulting curve is y = -2(x^2) -6. Determine the equation of the curve before the transformations were effected. [Is it possible to view the equation of the resulting curve as y= -2 (x^2 -3)? But I think this will give a different original equation from if I use the given resulting equation??]

    2)
    The equation of a graph is y = (ax)/(bx-y), and y= c is the horizontal asymptote while x = d is the vertical asymtote of this graph. Given that c = d = 1/2, what are the values of a and b? [If I am not wrong, b should be 2, but I don't know how to find a....Is a = 1?]

    3)
    Consider the curve (5x + 10)^2 - (y-3)^2 = 25. Prove, using an algebraic method, that x cannot lie between two certain values (which are to be determined). [eartboth has solved this here but I don't understand the method...Can someone explain?]

    4)
    The curve of y = x^2 + (3/x) undergoes a reflection about the line y=4 and the point (2, 5.5) is mapped onto (2, 2.5). Hence find the equation of the new curve in the form y = f(x).

    Thank you very very much for helping me!
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  2. #2
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    Quote Originally Posted by Tangera View Post
    ...
    3)
    Consider the curve (5x + 10)^2 - (y-3)^2 = 25. Prove, using an algebraic method, that x cannot lie between two certain values (which are to be determined). ...
    ...
    I rewrote your equation into:

    \frac{(x+2)^2}{1^2} - \frac{(y-3)^2}{5^2} = 1

    which is the equation of a hyperbola with the center (-2, 3) and the semi-axes a = 1 and b = 5.

    Then I calculated the coordinates of the vertices and got the intervall in which x cannot lie.

    I've attached a drawing of the curve to show where you can find this interval.
    Attached Thumbnails Attached Thumbnails Graph transformations...-hyp_tangera.gif  
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  3. #3
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    Quote Originally Posted by Tangera View Post
    Hello! I am a little confused about graph transformations and I have some questions that I don't really know how to solve...Please help! Thank you!

    1)
    The transformations A,B and C are as follows:
    A: A translation of 1 unit in the negative y direction
    B: A stretching parallel to the y-axis (with x-axis invariant) with a scale factor of 2
    C: A reflection about the x-axis

    A curve undergoes in succession, the above transformations A,B and C and the equation of the resulting curve is y = -2(x^2) -6. Determine the equation of the curve before the transformations were effected. [Is it possible to view the equation of the resulting curve as y= -2 (x^2 -3)? But I think this will give a different original equation from if I use the given resulting equation??]

    ...
    You have:

    y = -2x^2-6 = -2(x^2+3)

    Reflection about the x-axis will yield: y = 2(x^2+3)

    The stretching parallel to the y-axis (with x-axis invariant) with a scale factor of \frac12 will yield: y = x^2+3

    The translation of 1 unit in the positive y direction will yield: \boxed{y = x^2+4}

    I've attached a sketch of the complete process. You start with the blue graph and you end with the red one.
    Attached Thumbnails Attached Thumbnails Graph transformations...-transformparab_tangera.jpg  
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