Dear Uncle Colin,

I want to find the oblique asymptote of $f(x) = \frac{2x^2 + 3x -9}{x+2}$, but I don’t want to learn polynomial division. What do you recommend?

Genuine Reason: It’s Difficult

Hi, GRID, and thanks for your message!

I would recommend learning to divide polynomials but since you’ve ruled that out, I’ll give some alternatives.

You can adjust the fraction by cleverly adding 0 until you have something that’s easy to divide.

For example, if the $3x$ were a $4x$, it’d be easy to farm off the first couple of terms, right? So let’s add an $x$ and subtract an $x$:

$f(x) = \frac{2x^2 + 4x - x - 9}{x+2} = 2x + \frac{-x-9}{x+2}$

And that woud be easy if the -9 were a -2, so let’s write it as -2-7:

$f(x) = 2x + \frac{-x-2-7}{x+2} = 2x - 1 - \frac{7}{x+2}$

And as $x$ gets large, that approaches $2x-1$, which is the asymptote

### Remainder theorem

The remainder from dividing the top of the fraction, $2x^2 + 3x - 9$, by the bottom, $x+2$, is the value of the top when $x = -2$ – which is -7.

We can write the top as $(2x^2 - 3x -2) - 7$, and factorise the first bracket as $(2x-1)(x+2)$. Then $f(x) = \frac{(2x-1)(x+2) - 7}{x+2}$, or $2x-1 - \frac{7}{x+2}$. As before, the asymptote leaps out.

### Yarr!

Hello, Mathematical Pirate! It’s a while since I’ve seen you. Don’t let the Ninja catch you with that calculator!

Arr. I shall put A MILLION in as $x$. And I get 1,999,998.999993

I’ll keep a lookout, shall I?

That’s two million, minus a tiny bit more than 1. So the asymptote is $2x-1$. Full sail for the next port!

Yikes. I’m not sure I recommend piratical methods, but they seem to work here!

I hope that helps!

- Uncle Colin