Many, many years ago, during my Algebra 2 class, we had the classic assignment of graphing parabolas. After a while of “select X, calculate Y,” I came to the conclusion that there had to be a simpler way, especially since I started to see a pattern. Yes, there is the basic pattern of the parabola being open to the top, bottom, or either side, but I started to see something a bit more unique.
The first thing I started to see was the slope of the parabola. The second thing I noticed was the vertex. I thought, “If there is a standard graphing form for a straight line, there must be one for a parabola.” This lead to the following similarity:
- Straight line: y = m ⋅ x +b
- Parabola: y = m (x + c)2 + b
It is understood that for the straight line, m = slope and b = an intercept on the y-axis. But for the parabola, the b value is the intercept on the value of -c.
Getting to this form requires completing the square for the parabolic equation. I will not go into this with this blog; the intent here is how to plot quickly. I only have to figure out one point and the rest is a simple count method. For the next steps, I will assume the parabola opens upwards.
With the above form, one would know that there is a point at (-c,b) on the graph. From this point, all that is needed is a counting rule away from this point — simple trick No. 1. For a slope of 1, the count is “over 1, up 1… over 1, up 3… over 1, up 5…” etc. Basically, it's just using every odd number. Once the graph of one side is done, the other side is just mirrored.
The real trick comes in factoring in the slope. Simply multiply the “up” count by the slope. Thus, for a slope of 3, the count would go like this:
over 1, up 3; over 1, up 9; over 1, up 15… etc.
I presented this concept to my math teacher to see how she would respond. With the simplicity in mind, she provided me with a problem to graph. A few quick steps, and I finished in less than a minute. She was impressed, and gave me one that was a bit more difficult. Again, the graph was completed in less than a minute. She continued on until I was eventually stumped. I went home and thought about my approach until I was able to really figure out all the details. The next day, I approached the teacher with the simple solution. She then upped the difficulty to something on the extreme level, such as this:
x = 54/7y – 1/34y2 + 0.43y – π + 4.3y2 + 5.34
With my newly discovered thought process, this became a cinch. Care to see how fast you can plot it? No graphing calculators, spreadsheets, or other tools — just the pencil and paper, folks.