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September 9th, 2009, 07:38 PM   #1
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Parabola and Line intersections?

I have this problem that I just cannot figure out, and was wondering if I could get any help here.

So basically we have to graph a generic parabola (y=x^2) and some lines (y=x+k) and figure out where it intersects once, twice, no times. That part is straightfoward enough, -1/4 is the single intersection, anything above is twice, anything below, none. What I'm having trouble with is to prove the conjecture that "When k>-1/4, there will be 2 intersections. There will be one intersection at k=-1/4, and no solutions when k<-1/4." I know that we're supposed to play around with the discrimminant and such, and I've gotten to the point where I suppose I've proved that there is one intersection at -1/4, using b^2-4ac with the equation x^2-x+1/4=k+1/4 (the previous two equations set equal to each other and modified by completing the square). What I'm really having a problem with is how to prove that numbers above this would have two intersections, and that numbers below would have none. Someone told me something about a "law of discriminants" but upon Googling this, I didn't really find anything. I get the feeling I'm missing something really obvious, but I don't know, and I can't figure this out at all. thanks to anybody who can help <3
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September 9th, 2009, 09:06 PM   #2
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Any intersections have a real x-coordinate that is a root of x = x + k.
That's a quadratic equation with roots 1/2 ?(k + 1/4), so there are two real values of x when k > -1/4 and no real roots when x < -1/4.
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