I’m teaching Algebra 2 this summer for my school. In a recent test on quadratic functions, I gave a question I thought would be a little different, but still reachable for those willing to make connections or exert a little creativity.

Write a system of quadratic functions that has exactly one solution: (1,1).

Their handheld graphing calculators were allowed. Some students definitely had difficulty with the challenge, some gave a version of the answer I expected, and one adopted a form I knew was possible, but doubted anyone would actually find during a test situation.

I show my students’ solutions below. But before you read on, can you give your own solution?

**SOLUTION ALERT! **Don’t read further if you want to find your own solution.

WHAT I EXPECTED

We’ve had many discussions in class about the power of the Rule of 4–that math ideas can be expressed numerically, graphically, algebraically, and verbally. When you get stumped in one representation, being able to shift to a different form is often helpful. That could mean a different algebraic representation, or a different Rule of 4 representation altogether.

The question is phrased verbally asking for an algebraic answer. But it asks about a solution to a system of equations. I hoped my students would recall that the graphical version of a system solution is equivalent to the point(s) where the graphs of the equations intersected. In my mind, the easiest way to do this is to write quadratic functions with coincident vertices. And this is most easily done in vertex form. The cleanest answer I ever got to this question was

A graphical representation verifies the solution.

Another student recognized that if two parabolas shared a vertex, but had different “slopes”, their only possible point of intersection was exactly the one the question required. Here’s a graphical version of her answer.

From these two, you can see that there is actually an infinite number of correct solutions. And I was asking them for just one of these! 🙂

WHAT I KNEW, BUT DIDN’T EXPECT

Another way to solve this question makes use of the geometry of quadratic graphs. If two quadratics have the same leading coefficients, they are the same graph, intersect exactly once, or never intersect. This is a very non-trivial idea for most students. While I’m not convinced the author of the following solution had this in mind when he answered the question, his solution works because of that fact. Here’s what J wrote on last week’s test and its graph.

J used more equations than he needed, but had he restricted himself to just two equations, I’m not sure the lovely pattern would have been so obvious.

This is a very different (and super cool) answer than what I expected my students to produce. **Lesson re-learned: **Challenge your students, give them room to express creativity and individuality, and be prepared to be amazed by them.

NEXT STEPS

J’s answer actually opens the door to other avenues of exploration.

- Can you generalize the form of all of J’s equations, essentially defining a family of quadratics? Can you prove that all members of your generalization satisfy the question posed and that no other answers are possible?
- Can you find forms of other generalized families of quadratic functions whose only solution is (1,1)?
- Notice that there were two types of solutions above: A) those with coincident vertices and different lead coefficients and B) those with identical lead coefficients and different vertices. Are these the only types of quadratics that can answer this question? That is, is there a system of quadratics with (1,1) as the only solution that have identical vertices and lead coefficients? Could both be different and (1,1) be the only solution?
- If I relax the requirement that the quadratics be
*functions*, what other types of quadratics are possible? [This could be a very nice calculus question!]

For my part, I’m returning to some of these questions this week to stretch and explore my student’s creativity and problem-solving.

I’d love to hear what you or your students discover.