In high school mathematics, one common task is graphing a parabola, commonly given in the form
Searching Google on graphing parabolas found about 167,000 results, including Clayton State, and Regents Prep, and Sparknotes, and many many more. Quickly glancing through the results, the top few suggest making a data table, and then connecting points with a smooth curve. While this accomplishes the task, it doesn't give us any feel for shape. Graphing Calculator can easily plot parabolas for us. This Graphing Calculator file produced the following window. It's tough to see similarities when the equations are in this form, and we have to work to find the vertex.
There's a much better way to show the vertex, but to do so we need a different representation. We can complete the square! Unlike the name of this reference website, the math may not be fun, but it is very useful. To complete the square in the equation above:
We have two numbers to manipulate, one inside the square and one outside. It's a little awkward, because inside the square we have a minus and outside we have a plus, but mathematical equations aren't always beautiful and neat. To make things easier, let's call the value inside H and the value outside V. This makes our equation
Let's start by examining one graph in Graphing Calculator, y = \left[ x - 2 \right]^2 + 1.5 . The red dot indicates the vertex, which lies at x = 2 and y = 1.5, right on H and V.
You might be suspicious that I specifically chose 2 and 1.5, so let's look at other values of H and V. In all five cases shown, the vertex lies at the point (H, V).
If you like, you can experiment with various values of H and V through a GeoGebra Java applet. Click through to the handv.html page. Once the applet loads, you can click on the play button towards the bottom to watch an animation, which then becomes a pause button to stop. Alternatively, you can click on the H and V buttons on the bottom to drag them around the space.
Now that we've seen the relationship graphically, we can find the vertex mathematically. The value of Y sums the constant V and the value of \left[ x - H \right]^2 . We're working with real numbers, so the value of the square is never negative, and will be minimized when \left[ x - H \right] = 0 . This minimum occurs at x = H. At this point, we solve for the minimum value of Y:
Thus, our minimum value, the vertex, is at (H, V), as we've seen. If we need the original formula, we can solve B = -2 H and then substitute into V = C - \frac{B^2}{4} to find C = V + H^2 . Looking at both parabolic forms,