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Math Help - What steps must I take to graph equations such as y=y/2+1

  1. #1
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    What steps must I take to graph equations such as y=y/2+1

    So I was wondering how do I solve equations where for example it is in the form y=y/2+x where instead of being in the standard form y=mx+b we have something where y = my + nx + b (where in the example I gave m = 1/2 n = 1 and b = 0). What steps should I take to figure out how to graph this? According the graphing application I am using (Grapher) the slope appears to be 2 but, how do I figure this stuff out?

    Also, to further this what if I wanted to graph the equation:
    y=ay^2 + ax^2 + my + nx + b?

    What should I look into to do this?
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  2. #2
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    Quote Originally Posted by thyrgle View Post
    So I was wondering how do I solve equations where for example it is in the form y=y/2+x where instead of being in the standard form y=mx+b we have something where y = my + nx + b (where in the example I gave m = 1/2 n = 1 and b = 0). What steps should I take to figure out how to graph this? According the graphing application I am using (Grapher) the slope appears to be 2 but, how do I figure this stuff out?
    Solve the equation
    y=\frac12 y + x for y:

    y=\frac12 y + x~\implies~\frac12 y = x~\implies~y = 2x

    Also, to further this what if I wanted to graph the equation:
    y=ay^2 + ax^2 + my + nx + b?

    What should I look into to do this?
    The graph of the 2nd equation is a circle. To draw a circle you need to know the coordinates of the center and the length of the radius:

    y=ay^2 + ax^2 + my + nx + b

    0=ay^2 + ax^2 + nx + my - y  + b

    0=y^2 + x^2 + \frac na x+ \frac{m-1}{a}y  + \frac ba

    Now complete the squares:

    y^2 + \frac{m-1}{a}y +\left(\frac{m-1}{2a} \right)^2 + x^2 + \frac na x +\left(\frac{n}{2a} \right)^2= - \frac ba+\left(\frac{m-1}{2a} \right)^2 +\left(\frac{n}{2a} \right)^2

    \left(y +\left(\frac{m-1}{2a}\right) \right)^2 + \left(x +\left(\frac{n}{2a} \right) \right)^2= - \frac ba+\left(\frac{m-1}{2a} \right)^2 +\left(\frac{n}{2a} \right)^2

    So the center is at M\left(\left(-\frac{n}{2a} \right)\ ,\  \left(-\frac{m-1}{2a}\right) \right) and r^2= - \frac ba+\left(\frac{m-1}{2a} \right)^2 +\left(\frac{n}{2a} \right)^2
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