It’s been 3 years, so I’m a liiiittle fuzzy, but I think I remember how it went down. Never mind I actually remember it perfectly lol
So if you’re graphing a function which is a quotient of polynomials (like a “cubed” function divided by a “squared” function, like you can see in the snap) there are usually asymptotes associated with the graph. Can be lines, can be parabolas. You find it by performing long division on the two polynomials–it’ll be their quotient. From the snap, it looks like they’re graphing 2x^3/(x^2-x-6), and the found asymptote is 2x+2.
The follow-up question that needs to be answered is “Does the graph ever intersect the asymptote at any point?” Sometimes it does, and you need to be able to represent it on the graph. The old textbook way involved taking your solved-for asymptote (the 2x+2 here) and setting it equal to the graphed equation ( 2x^3/(x^2-x-6)), solving out for x, and seeing if it equals a value. This probably requires some quadratic formula tinkering in most scenarios.
I noticed something though. You could answer the “does it cross?” question far, far more simply by looking at the remainder from the original long division. See that “14x+12″ which is left over on the board? If that can be set equal to 0 and solved out for x, then a) yes it crosses, and b) it crosses at the value of your solved-for x. You just need to take that x and plug it into your line to find y.
So in short, instead of kids have to solve 2x+2 = 2x^3/(x^2-x-6) for x, I realized you only have to solve 14x+12 = 0. In which x= -12/14, which simplifies to -6/7. And then the y comes from plugging in that x to the solved-for line y=2x+2.
Glancing at it now, I also know what the proof is for this, and why it works infallibly:
My method hones in on the x-value which will make the remainder 0. If the remainder is zero, that means the division between the numerator and denominator polynomial is exactly equal to the value of the asymptote line for this x. It’s almost as simple as saying “Because 6/3 = 2 with no remainder, that means 6/3 REALLY DOES equal 2. As opposed to, for example, 7/3 = 2 R1, where 7/3 doesn’t really equal 2.)
So by virtue of these two Y VALUES being exactly equal to each other at a specified X VALUE, it follows that my method has identified a coordinate (X, Y) which denotes a point of intersection between the graphed function and its asymptote.
Quod Erat Demonstrandum bitches
titsay
theartofmadeline
yousharknotpass
jbbartram-illu
germansingergirl
kaledo-art
