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Suppose we have a second-order curve in general form:

(1) $a_{11}x^{2}+2a_{12}xy+a_{22}y^{2}+2a_{13}x+2a_{23}y+a_{33}=0$.

I'd like to know if there is a simple condition that ensures that the curve has at least one point on on or below the $x$ axis, i.e. that the left-hand side of (1) is nonpositive.

In the trivial case that the curve is a parabola, the discriminant being nonnegative is just such a condition. But what happens in the general case?

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Solve for $y$ in the form $y= A(x) \pm \sqrt{B(x)}$ and estimate. More abstract versions are just variant of this.

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Since my parameters are themselves complicated functions, I was hoping to avoid this... – Felix Goldberg Feb 11 '13 at 16:03
Perhaps you should provide this --- and any other relevant information --- in the body of the question. – Gerry Myerson Feb 11 '13 at 22:30

We may regard the left-hand side of the equation of thecurve as a quadratic polynomial in $x$. If $D(y)$ is its discriminant (with respect to $x$), then $D(y)\ge 0$ iff there exists a point with the second coordinate $y$ on the curve. Solve this inequality for $y$ and check whether its minimal solution is negative:)))

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I am not sure I understand the question. A point ON the $x$ axis is given by $y=0,$ so $a_{11} x^2 + 2 a_{13} x + a_{33}=0.$ That IS a parabola, and so we know what the solutions are (if any). If if the discriminant is positive, you are golden. If the discriminant is 0, pick a random value of $x$ ($0$ is easiest, but that might be the solution to your quadric), and check if $y$ is positive. If the discriminant is negative, take $x = 0$ and see what the solutions are to the resulting quadric in $y.$

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