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Thread: Two lines meet. How do I find the x and y intercepts of this line on quadrant IV?

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    Two lines meet. How do I find the x and y intercepts of this line on quadrant IV?

    The two lines (-4x) and (-2x^2+120x) meet. As you can see, one is a quadratic function and the other is a linear function.
    I would like to know the x and y intercepts of quadrant IV.

    I tried to set both functions equal to each other and calculated both coordinates of x and y using the quadratic formula. The results are (0,0) and (62,0). But these answers do not provide the other intercept that -2x^2+120x creates when meeting -4x in quadrant IV.

    Can someone tell me how to find this said x and y coordinate?
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    Re: Two lines meet. How do I find the x and y intercepts of this line on quadrant IV?

    This post is a mess.

    You have a line and a parabola meeting. Not two lines. The term for a general line in space is called a curve.

    I would like to know the x and y intercepts of quadrant IV.
    What does this mean? Do you want the point $(x,y)$ of the intersection of these two curves?
    The results you posted seem to indicate this though you got the second one wrong.

    There are two points of intersection.

    $(0,0)$ is clearly one of them

    $-4x = -2x^2 + 120x$

    $2x^2 = 124x$

    if $x\neq 0$

    $2x = 124$

    $x = 62$

    and plugging this into the linear function we get $y=-4\cdot 62 = -248$

    So the point you're after is $(62, -248)$ which is indeed in the quadrant IV
    Thanks from topsquark and bossbasslol
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