To prove this in the general case, that there are no quadratics that meet this condition, could you use the general form of the equation for each of these, do a lot of algebra, and show that the solution set is empty for the parameters of the general equations? The k in the general form would be 0 since the parabola has to intercept the x axis, which makes the algebra slightly easier. Would that be enough? Proving that incorrect may answer the question you're asking. Of course, that might result in an infinite solution set, as opposed to an empty one:
Going back to mmm's suggestion about plotting, in this form of the equation,
\(\displaystyle $f(x) = a x^2 + b x + c$\)
you would also have to test for a < 0. Try plotting \(\displaystyle $g(x) = -x^2$\) and its reciprocal, for example. If this is a yes/no question, the answer is 'yes,' as shown by these two examples. Keep in mind that all you have to do is come up with ONE case that meets the condition. Proving that it can't happen is a lot harder. To prove that it does happen, though, all you have to do is show one example.
You might want to try finding the conditions under which this is true or not true. That is, for what values of the a, b, and c parameters (or a, h, and k if you use that form) does the quadratic intercept its reciprocal and intercept the x axis only once. That might be a little more interesting.