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 – 1/34y^{2} + 0.43y – π + 4.3y^{2} + 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.

It was 2-1/2 minutes for me; I made the problem up when I wrote the blog and did the graphing after I submitted it. Then I spent the next 1/2 hour confirming my graph and calculating all the detail numbers. I will post my results after some of you had a chance to do the problem and post your times.

Derek – just need to clarify before I pick up my pencil:

For these two “y” terms: 54/7y – 1/34y

^{2}is it(54/7)y – (1/34)y

^{2}Or

54/(7y) – 1/(34y

^{2}) I assume it's the former.Your first entry. Otherwise it would not be a parabola.

(54/7)y – (1/34)y

^{2}No graphing calculators, spreadsheets, or other tools — just the pencil and paper, folks.@Derek, thanks for the post. I really tried hard solving the function without using graphing calculators/spreadsheets but I couldnt. I hope you will be posting solution to this problem in future.

@Brad, I had similar doubt when i saw the equation. I read the equation as : ((54/7)y) – ((1/34)y

^{2}) .@Derek – good – just wanted to make sure that I hadn't taken leave of my senses.

In my view, in the most nonlinear equation, y = f(x), x is input and y is output. But when look this formula. Y is input and x is output. That kind expression is odd. Second, to understand for quick plot, more description may be required for this blog.

Do not think in the box. The equation presented is correct. This is basic math from high school and the plotting method described has several basic concepts. The final trick on quick-plotting is multiplying out to get the so-called slope of the parabola.

Plotting numerous parabolas in high school became tedious and I came up with this method on my own. There are YouTube videos that show how to set up the problem, but their final graphing method is tedious compared to my method. (I did the YouTube search after the article was written as a curiosity to see what is out on the internet.)

It would appear to me that changing the form of the equation to find the slope can be done using the quadratic formula. For some equations, this may be easy enough to do by inspection. You note you didn't go into that part; are you suggesting there are faster methods than the quadratic formula?

Regarding the counts, I note that in the general form the slope applies to the quadratic term. I noticed the following:

Y = m(x+c)

^{2}+ bY = m(Ω)

^{2}+ bdY/dΩ = 2m

count = m, 2m+m, 2m+2m+m, 2m+2m+2m+m, … = m, 3m, 5m, 7m, …

m= 1: 1, 3, 5, 7,…

m=2: 2, 6, 10, 14,…

m=3: 3, 9, 15, 21,…

It took me about 2.5 minutes to crunch the equation into a form I can plot and do the actual plot (last 30 or so seconds). My notes are below.

The spreadsheet (below) was used to make sure my plot matches up correctly – which when overlaid on the actual, I match pretty good (upper left). The plot on the far right was used to get the slope and help develop the coefficients into the equation I actually plotted.

Images appear to not come through well. One can pull a spreadsheet that has them at

http://www.dkoonce.com/B005_Quick-plot_parabola.xls

Derek's images aren't showing up entirely correctly, so you can see them properly by going here:

http://www.dkoonce.com/B005_Quick-plot_parabola.xls

Where you'll see his hand-calcs on one tab of the spreadsheet (as an image pasted into the spreadsheet); and, on another tab, the Excel plotted graphs.