How can you show that a tilted parabola is not the graph of a function? Consider the parabola that is the graph of $f(x) = x^2$, and suppose we rotate this parabola by an angle $\theta$. Unless this $\theta$ is a multiple of $\pi$ radians, the result will not be the graph of some function. How can you show this? Or a slightly tougher question: in terms of a variable $\theta$ how do you exhibit two points on the rotated parabola with the same $x$-coordinate?
I can figure this out, but I want to know how other people would tackle this problem and the minimal machinery necessary to do so. This would make a good challenge question for a motivated student, and I would like to get ideas for how they might approach it or how I could guide them towards a fruitful approach. (I'll update the tags in response to answers)

The graph of $f$, rotated by $\theta$, would be the graph of a function if, and only if, it did not intersect any vertical line twice (the so-called "vertical line test").
A vertical line intersecting the graph of $f$, rotated by $\theta$, twice is the logically equivalent to a vertical line, rotated by $-\theta$, intersecting the graph of $f$ twice.
Let $m$ be the slope of a vertical line rotated by $-\theta$. We can take the slope of this line since $\theta$ is not a multiple of $\pi$ and the rotated line is therefore not vertical. Then consider the line $y = mx + 1$. I claim that this line intersects the given parabola twice.
Indeed, to find the x coordinates of an intersection, we must solve $x^2 = mx + 1$. That is, $x^2 - mx - 1 = 0$. Using the quadratic formula, we have $x = \frac{m \pm \sqrt{m^2 + 4}}{2}$. There are clearly 2 distinct solutions.
