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Math Help - Find the set for "m" so that the graphs of y=mx and y=x^2+m intersect a single poi

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    Find the set for "m" so that the graphs of y=mx and y=x^2+m intersect a single poi

    please help!!

    find the set for "m" so that the graphs of y=mx and y=(x^2)+m intersect in a single point.
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    First, we gotta set f(x)=mx and g(x)=x^2+m, now set f(x)=g(x)

    x^2+m-mx=0\implies x^2-mx+m=0

    The discriminant of this must be zero (such that f(x) and g(x) intersect in a single point), so

    m^2-4m=0\implies m=0,4
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    Quote Originally Posted by ginax7 View Post
    please help!!

    find the set for "m" so that the graphs of y=mx and y=(x^2)+m intersect in a single point.
    ginax7, please consider using a different colored font. You know, black really is an acceptable color for typing. Really!

    Find the set for "m" so that the graphs of y = mx and y = (x^2)+m intersect in a single point.
    Let
    mx = x^2 + m

    x^2 - mx + m = 0

    This must have only one solution, so the discriminant must be 0:
    (-m)^2 + 4m = 0

    m^2 - 4m = 0

    m(m - 4) = 0

    So either m = 0 or m = 4.

    -Dan
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    thanks

    thanks guys! sorry bout the pinkness! It's the first day of school and I havent even learned about discriminants I dunno why hes giving us problems like these
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  5. #5
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    Quote Originally Posted by ginax7 View Post
    thanks guys! sorry bout the pinkness! It's the first day of school and I havent even learned about discriminants I dunno why hes giving us problems like these
    Suppose you have the following quadratic equation ax^2+bx+c=0,\,\forall a,b,c\in\mathbb R,\,a\ne0

    The following formula solves this equation

    x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

    The expression \Delta=b^2-4ac it's called discriminant, which says who rules here!

    If \Delta>0, then the equation has two real and different roots.

    If \Delta=0, the the equation has two real and equal roots

    If \Delta<0, the equation has two complex roots.

    Hope it helps
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