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)
 A: 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.
A: To show not a function:
One idea could be to show that the line $x = 0$ will intersect the rotated graph of $y = x^2$. To do this you could notice that this line would have had equation $y = \tan(\frac{\pi}{2}-\theta) \cdot x = \cot(\theta) \cdot x$ prior to the rotation. We can solve $\cot(\theta) \cdot x = x^2$. We get that $x = 0$ and $x = \cot(\theta)$ give two intersection points for the parabola and the line prior to the rotation.
In the "bad" case that $\cot(\theta) = 0$ we know that the graph of the rotated parabola is symmetric about the $x$-axis.
A: To show the equivalent notion that a non-vertical line meets an "upright" parabola in two points, consider the polar equation of such a parabola with its focus at the origin and with semi-latus rectum $p$:
$$r = \frac{p}{1-\sin\theta}$$
For $|\sin\theta|\neq1$ (that is, for non-vertical directions), angles $\theta$ and $\theta+\pi$ yield (finite) positive values for $r$, corresponding to points on the parabola in "opposite directions" from the focus. $\square$

Alternatively, but less concisely, the parabola rotated by angle $\phi$ has polar equation
$$r = \frac{p}{1-\sin(\theta-\phi)}$$
Points in the vertical directions correspond to $\theta=\pm\pi/2$, so we have
$$r = \frac{p}{1\pm\cos\phi}$$
For $|\cos\phi|\neq 1$, we get the two values of $r$, and the two points on the parabola on a vertical line through the the focus. $\square$

The switch to the polar context is a little "advanced", but the dead-simple way it identifies the points of intersection is tough to beat.
