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Math Help - Finding intercepts?

  1. #1
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    Finding intercepts?

    Can anyone help me find the intercept of this

    x^2y-x^2+4y=0

    My second question is find the points of intersection of the graphs of the equations
    analytically.

    x^2+y^2=5

    x-y=1

    I know you have to set them to y so they can be equal to each other.

    So I did y=x-1

    But how would I do y^2=-x^2+5 ?
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  2. #2
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    Re: Finding intercepts?

    Quote Originally Posted by homeylova223 View Post
    Can anyone help me find the intercept of this

    x^2y-x^2+4y=0 <--- is this written correctly?

    My second question is find the points of intersection of the graphs of the equations
    analytically.

    x^2+y^2=5

    x-y=1

    I know you have to set them to y so they can be equal to each other.

    So I did y=x-1

    But how would I do y^2=-x^2+5 ?
    In this case I would replace the y of the first equation by x - 1. You'll get a quadratic equation in x:

    x^2 + (x-1)^2 = 5~\implies~2x^2-2x-4=0

    Solve for x and afterwards plug in the results into the equation y = x - 1 to get the y-values.
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  3. #3
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    Re: Finding intercepts?

    Quote Originally Posted by homeylova223 View Post
    Can anyone help me find the intercept of this

    x^2y-x^2+4y=0
    For the first one, the graph of this function touches the y-axis when x=0
    and touches the x-axis when y=0.

    Let one of these equal zero and discover the value of the other.
    This is how to find axis crossing-points.


    My second question is find the points of intersection of the graphs of the equations
    analytically.

    x^2+y^2=5

    x-y=1

    I know you have to set them to y so they can be equal to each other.

    So I did y=x-1

    But how would I do y^2=-x^2+5 ?
    I'd also recommend Earboth's solution.
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  4. #4
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    Re: Finding intercepts?

    Hmm I think I may have written it incorrectly

    here is what I mean

    (x^2)y-(x^2)+4y
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  5. #5
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    Re: Finding intercepts?

    Quote Originally Posted by homeylova223 View Post
    Hmm I think I may have written it incorrectly

    here is what I mean

    (x^2)y-(x^2)+4y
    Same thing if it all equals zero as before.
    Just place x=0 and find y
    or vice versa.
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  6. #6
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    Re: Finding intercepts?

    So If I place y=0 I get x^2(0)-x^2=0

    So Would my answer be (0,0)
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  7. #7
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    Re: Finding intercepts?

    Yes, that's all there is to it.
    The curve touches both axes at the origin.
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