As I was nearing the end of my article yesterday, something creeped into my head and lingered in the shadows for a while. This morning, it came into the light.
The function that allows us to find the input that corresponds to the output of another function has a name: It’s the inverse function.
One thing I’ve struggled with as a teacher is distinguishing a Quadratic FUNCTION from “the Quadratic FORMULA”, especially since I otherwise tend to see functions as formulas.
So, let’s forget about “the Quadratic Formula” entirely for a moment and ask ourselves instead: If \(f(x)\) is a quadratic function, what is its inverse?
The standard process for finding a function’s inverse is to reverse \(x\) and \(y\) and solve for the new \(y\), but I’m going to skip that step and instead solve for \(x\): \[y=a(x-h)^2+k \\ y-k=a(x-h)^2 \\ \frac{y-k}{a}=(x-h)^2 \\ \pm\sqrt{\frac{y-k}{a}}=x-h \\ h\pm\sqrt{\frac{y-k}{a}}=x \\ x=h\pm\sqrt{\frac{y-k}{a}} \]
That’s it. If \[f(x) = a(x-h)^2 + k\] then \[f^{-1}(y) = h\pm\sqrt{\frac{y-k}{a}}\]
All the mysticism that is “THE QUADRATIC FORMULA” (writ large and in Impact), reduced to a lowly “just an inverse”.
We can’t do this directly using the standard form of the quadratic because \(x\) and \(x^2\) get in our way: To find the inverse, we’d need to use the Quadratic Formula, so it’s circular to use it to find the Quadratic Formula.
But in the vertex form, where \(x\) only appears once, it’s the triviality shown above.
Rewriting this form to the Quadratic Formula is a bit more work, but we can rely on the mapping between the vertex and standard forms: \[y=a(x-h)^2+k \\ = ax^2 – 2ahx + ah^2 + k \\ y = ax^2 + bx + c \\ b = -2ah \implies h = -\frac{b}{2a} \\ c = ah^2 + k \implies k = \frac{4ac – b^2}{4a} \\ \therefore h\pm\sqrt{\frac{y-k}{a}} = -\frac{b}{2a}\pm\sqrt{\frac{y – \frac{4ac – b^2}{4a}}{a}} \\ = -\frac{b}{2a}\pm\sqrt{\frac{4ay – 4ac + b^2}{4a^2}} \\ = -\frac{b\pm\sqrt{b^2 + 4a(y – c)}}{2a}\]
Hence the inverse of \(f(x) = ax^2 + bx + c\) is \[f^{-1}(y) = -\frac{b\pm\sqrt{b^2 + 4a(y – c)}}{2a}\]
The so-called “Quadratic Formula” is simply this when \(y\) = 0. It’s not any special formula, it’s a special case of the inverse. But by skipping the Vertex Form version and just jumping directly to it, as we usually do, we disguise all of this.
The key to understanding the mysticism that is “THE QUADRATIC FORMULA” (writ large and in Impact) is that there is no “quadratic formula”. There is no spoon.