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Math Help - Questions on Graph transformations

  1. #1
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    Questions on 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). [I know how the given equation can be expressed to become the standard equation of a hyperbola, but I am not sure what is the "algebraic method" the question is asking for...]

    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 know how the given equation can be expressed to become the standard equation of a hyperbola, but I am not sure what is the "algebraic method" the question is asking for...]
    From

    (5x+10)^2-(y-3)^2=25~\iff~25(x+2)^2-(y-3)^2=25~\iff~ \frac{(x+2)^2}{1^2} - \frac{(y-3)^2}{5^2} = 1

    you know:

    The center of the hyperbola is C(-2, 3)

    The vertices of the hyperbola are V_1(-3, 3) , V_2(-1,3)

    Therefore x \notin (-3, -1)

    Second attempt

    Solve the equation for x:

    x = -2\pm\frac15 \cdot \sqrt{25+(y-3)^2}

    The smallest value the radical can be is 25 if (y-3) = 0

    The value of x lies between:

    -2-\frac15 \cdot 5 < x < -2+\frac15 \cdot 5

    -3 < x < -1
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  3. #3
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    Quote Originally Posted by earboth View Post

    Second attempt

    Solve the equation for x:

    x = -2\pm\frac15 \cdot \sqrt{25+(y-3)^2}

    The smallest value the radical can be is 25 if (y-3) = 0

    The value of x lies between:

    -2-\frac15 \cdot 5 < x < -2+\frac15 \cdot 5

    -3 < x < -1
    Ermm...sorry could you explain why do you want the value in the radical to be smallest? Thank you!
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