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Math Help - I need this explained....

  1. #1
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    I need this explained....

    Hello,
    My notes have y=sqrt x^2-16
    Now it's -4 and 4 on the x axis, but just the top part of the y curve, as it starts with that. going out to each side.

    So the fixing up the equation above is y^2=x^2-16
    so y^2-x^2=-16 which turns into x^2-y^2=16.
    Now x^2=16 is x=4, But for y it's y = -4, but how does that work? I need x -4,+4.

    I don't understand how to come up with the -4 on x.
    Does this make sense? it's hard to write out without a graph.

    Graphing Calculator

    If you go to this graphing online you will see what I am talking about keying in the above equation in bold


    If someone can explain this would be great, thanks.
    Joanne

    just trying to understand it, because I am not getting it.
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  2. #2
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    Quote Originally Posted by bradycat View Post
    Hello,
    My notes have y=sqrt x^2-16
    Now it's -4 and 4 on the x axis, but just the top part of the y curve, as it starts with that. going out to each side.

    So the fixing up the equation above is y^2=x^2-16
    so y^2-x^2=-16 which turns into x^2-y^2=16.
    Now x^2=16 is x=4, But for y it's y = -4, but how does that work? I need x -4,+4.

    I don't understand how to come up with the -4 on x.
    Does this make sense? it's hard to write out without a graph.

    Graphing Calculator

    If you go to this graphing online you will see what I am talking about keying in the above equation in bold


    If someone can explain this would be great, thanks.
    Joanne

    just trying to understand it, because I am not getting it.
    Hi bradycat,

    y=\sqrt{x^2-16}

    means that "y" is the positive square root of x^2-16

    so \left(x^2-16\right)\ \ge\ 0

    \Rightarrow\ x^2\ \ge\ 16\Rightarrow\ x\ \ge\ 4 if x is positive

    or x\ \le\ -4 if x is negative.

    What exactly are you looking for ?
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  3. #3
    A riddle wrapped in an enigma
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    Quote Originally Posted by bradycat View Post
    Hello,
    My notes have y=sqrt x^2-16
    Now it's -4 and 4 on the x axis, but just the top part of the y curve, as it starts with that. going out to each side.

    So the fixing up the equation above is y^2=x^2-16
    so y^2-x^2=-16 which turns into x^2-y^2=16.
    Now x^2=16 is x=4, But for y it's y = -4, but how does that work? I need x -4,+4.

    I don't understand how to come up with the -4 on x.
    Does this make sense? it's hard to write out without a graph.

    Graphing Calculator

    If you go to this graphing online you will see what I am talking about keying in the above equation in bold


    If someone can explain this would be great, thanks.
    Joanne

    just trying to understand it, because I am not getting it.
    Hi brandycat,

    It seems to me that you are trying to find the x-intercepts of y=\sqrt{x^2-16}

    Set y = 0 and solve for x.

    r \:\: \sqrt{x^2-16}=0" alt="0=\sqrt{x^2-16}\:\r \:\: \sqrt{x^2-16}=0" />

    Square both sides now.

    x^2-16=0

    x^2=16

    x=\pm 4
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  4. #4
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    The graph of x^2- y^2= 16 is a hyperbola with two separated parts- one to the right of the y-axis, the other to the left.

    The graph of y= \sqrt{x^2- 16} is the upper half of that hyperbola and consists of two separate "wings". It meets the x-axis at x= 4 and x= -4 but does NOT cross the y- axis. In fact x is always either larger than or equal to 4 or less than or equal to -4.
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