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Math Help - Finding an equation by using a graph...

  1. #1
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    Finding an equation by using a graph...

    I am given a graph and I am supposed to find the equations of these graphs and I am not sure how to go about doing that. Can someone help me? Thanks...

    P.S. Attached is the problem and ff you look at the top of the problem, thats the equation we are supposed to get. Thanks again...
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  2. #2
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    Quote Originally Posted by Godzilla View Post
    I am given a graph and I am supposed to find the equations of these graphs and I am not sure how to go about doing that. Can someone help me? Thanks...

    P.S. Attached is the problem and ff you look at the top of the problem, thats the equation we are supposed to get. Thanks again...

    You can see that the function is a transformation of the core function f(x)=x^2, which is a standard parabola with the vertex at the origin. The function y=-x^2 is identical, but flipped about the x-axis. Are you familar with the equation:

    y=a(x+h)+k

    ??
    In the graph we can see:

    a=-1, h=-2, k=4

    Do you understand this. The parabola is flipped around the x-axis, shifted two units to the right, and 4 units up.
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  3. #3
    Senior Member pacman's Avatar
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    vertex: (2,4)

    Is that 43) P(x) = -x^2 + 4x? Let's try manipulating it, P(x) = y = -x^2 + 4x,

    y = -x^2 + 4x, re-arranging it

    x^2 - 4x = -y or x^2 - 4x + y = 0 , completing the square, add 4 to both sides.

    x^2 - 4x + 4 = - y + 4,

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

    vertex: (2,4), it is confirmed that it is indeed the graph of P(x) = -x^2 +4x

    length of latus rectum = 4a = 1

    then a = 1/4. but i don't need it, i'm not in the graphing mood, the graph is there above already. You already have all the details in the graph . . . . .

    --------------------------------------------------------------------

    i) genearal form of a parabola: x^2 - 4x + y = 0

    ii) standard form of a parabola facing downward: (x - 2)^2 = -(y - 4).

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