**Theorem: **Given a quadratic function with rational roots, the \(y\)-intercept is rational if and only if the stretch is rational.

**Proof:** If \(f\) is a quadratic function with rational roots \(m\) and \(n\) and vertical stretch \(a\), then \[f(x) = a(x – m)(x – n) \\ = a(x^2 – (m+n)x + mn) \\ = ax^2 – a(m+n)x + amn.\] The \(y\)-intercept is the value of \(f(x)\) when \(x = 0\), that is, \(f(0) = amn\).

A rational number times a rational number is always rational. Therefore, \(mn\) is rational, since \(m\) and \(n\) are rational. If \(a\) is rational, then \(amn\) is rational, and hence the \(y\)-intercept is rational.

A rational number times an irrational number is always irrational. Therefore, if \(a\) is irrational, \(amn\) is irrational, and hence the \(y\)-intercept is irrational. □

I’m offering this as an example of a simple proof. For students used to proofs from Geometry class, with their columns, I think it’s important to realize that two-column proofs, the sort that high school students so often despise, do not resemble what is generally considered a proof in mathematics beyond high school.

We can also add some interesting corollaries. Here’s one:

**Corollary: ** Given a quadratic function with integer roots, the y-intercept is an integer if (but not only if) the stretch is an integer.

**Proof:** As above, the \(y\)-intercept is \(amn\). The product of integers is always an integer, so if \(a\), \(m\), and \(n\) are integers, then so is the \(y\)-intercept.

However, it is possible for a non-integer times an integer to be an integer. For instance, if \(a = \frac{1}{3}\) and \(mn = 3\), then \(amn = 1\). Therefore, an integer \(y\)-intercept is not enough to prove that \(a\) is an integer. □

These are sometimes called “paragraph proofs”, and are the standard way of writing proofs in mathematics (including the square to indicate the end of the proof.