Identifying Asymptotes of a Hyperbola

Question

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?

Answer 1

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.

Answer 2

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

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 $x_v=-b$.

Analytically, we can prove this by using limits, as $x \to -b$ and $x \to \infty$.

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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.

Answer 3

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

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

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

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

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

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