# Identifying Asymptotes of a Hyperbola

basic hyperbolic functionHow would I find the vertical and horizontal asymptotes of a $y = \frac{1}{x}$ function algebraically? For example, $y = -\frac{2}{x+3}-1$ (as you would type into a calculator). Simply, how do I find the x and y values by looking at this equation? In other words, the middle point where the asymptotes in the picture has moved and the whole graph has been vertically stretched, where are the asymptotes now?

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Pic if you dont want to use calculus. – Gastón Burrull May 15 '12 at 0:32
@Marvis, don't you think your edit has changed the meaning of the question? – Gerry Myerson May 15 '12 at 1:13
@GerryMyerson Am sorry. I understand now. I have now reverted it to the old state. – user17762 May 15 '12 at 1:17
Caroline, Marvis took the trouble to edit your question into decent shape. Why did you muck it up? "as you would type into a calculator" is no help at all, as different calculators use different versions of the syntax. Do you, or don't you, want $y={-2\over{x+3}}-1$? If not, and if you can't figure out how to do TeX, can you write out what you mean in words? – Gerry Myerson May 15 '12 at 23:48
Caroline, what don't you understand about the answers Peter and I have posted? You can edit your question over and over again, but at some point you have to engage with the people who are trying to help you, and let them know what you do, and what you don't, understand. – Gerry Myerson May 16 '12 at 0:11

I assume that's $y=-{2\over x+3}-1$. There's a vertical asymptote where the function is undefined, and the function is undefined where it involves division by zero. For the horizontal asymptote, you have to ask yourself, what happens when $x$ grows without bound? when $x$ gets very large negative?

EDIT: Reading OP's comments and revised question, I think maybe this approach will be more what's wanted.

I take it you know that for $y=1/x$ the horizontal asymptote is $y=0$ and the vertical asymptote is $x=0$, and they meet at the point $(0,0)$. $y=1/(x+3)$ is the same graph but shifted 3 units to the left; that doesn't change the horizontal asymptote, but it shifts the vertical asymptote to $x=-3$. $y=-1/(x+3)$ flips (or, reflects) the graph in the $x$-axis, but has no effect on the asymptotes, which remain $y=0$ and $x=-3$. $y=-2/(x+3)$ stretches the graph, but again has no effect on the asymptotes. Finally, $y={-2\over x+3}-1$ lowers the graph by 1, so it changes the horizontal asymptote to $y=-1$, while not affecting the vertical asymptote.

So, the new asymptotes are $y=-1$ and $x=-3$, which meet at the point $(-3,-1)$, which is what you are calling the middle point.

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Ok, but if it is parametriced? asymptotes not implies indefinition – Gastón Burrull May 15 '12 at 0:34
@GastónBurrull, I'm not sure what you are getting at. Can you illustrate with an example? – Gerry Myerson May 15 '12 at 0:35
You must do a certain rotation before. Asymptotes can be in any direction. – Gastón Burrull May 15 '12 at 0:40
@GastónBurrull, the question asked about "a $y=1/x$ function". If you have something else in mind, please spell it out. Enough with the cryptic comments. – Gerry Myerson May 15 '12 at 0:52
No, title says "a Hyperbolic Function", he just shows an example. – Gastón Burrull May 15 '12 at 1:02

Not all that different from Gerry Myerson's answer, but expanding the explanation of the transformations and including graphs.

basic (parent) graph, $y=\dfrac{1}{x}$; asymptotes: $y=0$, $x=0$

scale by 2 in the $y$ direction ($y\to2y$), $y=\dfrac{2}{x}$; asymptotes: $y=0$, $x=0$

reflect over the $x$-axis ($y\to-y$), $y=-\dfrac{2}{x}$; asymptotes: $y=0$, $x=0$

translate left 3 ($x\to x-3$), $y=-\dfrac{2}{x+3}$; asymptotes: $y=0$, $x=-3$

translate down 1 ($y\to y-1$), $y=-\dfrac{2}{x+3}-1$; asymptotes: $y=-1$, $x=-3$

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Your pictures are worth my thousand words. – Gerry Myerson May 17 '12 at 23:48

In general, an homographic function $$y= \frac{ax+b}{cx+d}$$

where $c \neq 0$, has two asymptotes.

1. Horizontal asypmptote: $$y_h=\frac{a}{c}$$

2. Vertical asypmptote: $$x_v=-\frac{d}{c}$$

The canonic form of an homographic function is

$$y=\frac{a}{x+b}+c$$

where $a\neq 0$. Put this way, the asymptotes are $y_h=c$ and $x_v=-b$.

Analitically, one would prove this by using limits, as $x \to -b$ and $x \to \infty$.

If one is to generalize to any hyperbola, we use the defining equation:

$$\frac{(x-k)^2}{b^2}-\frac{(y-h)^2}{a^2}=1$$

Then the asymptotes are precisely

$$y_1 = \frac{a}{b}(x-k)+h$$ $$y_2 = -\frac{a}{b}(x-k)+h$$

If it is the case the hyperbola is the conjugate

$$\frac{(y-h)^2}{a^2}\frac{(x-k)^2}{b^2}=1$$

the asymptotes are (clearly) the same.

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I'd write the vertical asymptote as $x_v$ rather than $y_v$. – Gerry Myerson May 15 '12 at 1:09
@GerryMyerson True. Fixing. – Pedro Tamaroff May 15 '12 at 1:10