Summer is giving me some time to tie up loose ends that inevitably get dropped during the busy-ness of the school year. Here’s one of those.

I hinted in a post from several months ago about a cool underlying pattern in the quadratic function family.  Most algebra students know how the coefficients a and c control the graph of $y=a\cdot x^2+b\cdot x+c$, but what does b do?

I wrote a Geogebra worksheet to allow exploration of a, b, and c.  On this page, there are sliders on the right side that allow users to vary the values of these three coefficients while the graph of $y=a\cdot x^2+b\cdot x+c$ changes live to reflect those values.

Over the past decade, I’ve become increasingly enamored with this approach to exploring the behavior of function families in advance of more formal analyses.  “Seeing” the effects of parameter can inform and guide the work you do later.  Students quickly recognize that c vertically changes the parabola’s position; closer inspection notes that c is the y-intercept.

Most also note that a changes the “width” of the parabola.  This is true enough, but (in my opinion) a clearer description is that a changes the quadratic’s height.  For any value of x, the y-values of $y=2x^2$ ($a=2$) are twice the y-values of $y=x^2$.  If you attempt to quantify the width, then $a=2$ means the corresponding points are $\displaystyle \frac{1}{\sqrt{2}}$ “wider”.  That just isn’t intuitive to anyone I know, and describing lead coefficients as vertical scale changes is an idea that applies to all functions. I eventually refocus these descriptions to vertical scale changes, but that’s not the point right now.

So what happens when you change b?  If you don’t already know the answer, I encourage you to explore the Geogebra worksheet before reading further.  Try to be precise.

SOLUTION ALERT!  Don’t read further if you want to solve this first.

Like c, varying b changes the position of the parabola, but not its shape.  The difference is that b moves the parabola both horizontally and vertically.   Closer observation suggests that the motion might be along a parabolic path.  Using a new Geogebra worksheet, I placed a trace on the vertex to record the “footprints” of the vertex as b changed.

That’s pretty compelling evidence.  The challenge students face at this point is defining an equation for the suspected parabola. Following the vertex of a parabola is a good proxy for following the entire parabola.

FAST FORWARD:  Through lots of trial-and-error, students eventually propose $y=-a\cdot x^2+c$.  That’s nice, but writing an equation isn’t a proof.  One of the most elegant proofs I’ve seen solves the system of equations defined by the original generic quadratic family and the proposed path of the vertex. A CAS is obviously an appropriate tool in this situation.

There are two solutions:  $(\frac{-b}{2\cdot a},\frac{4\cdot a\cdot c-b^2}{4\cdot a})$ and $(0,c)$ implying two graphical intersections, a fact verified by the vertex trace image above.  The proof lies in the first ordered pair–the generic form of the coordinates of the vertex of $y=a\cdot x^2+b\cdot x+c$–clearly establishing that the generic vertex always travels on the proposed path.  Nice.

What amazed me most about this problem is that I had been teaching quadratic equations for years and remembered from my time as a student what a and c did to the graph.  How is it that I had never explored b ?  How could such a pretty result have been overlooked?  No longer.  This is a project every time I teach an algebra class.