# How do you find the vertical, horizontal or slant asymptotes for f(x) = (1/x) + 3?

Jan 2, 2017

Horizontal: $\leftarrow y = 3 \rightarrow$
Vertical : $\uparrow x = 0 \downarrow$

#### Explanation:

graph{x(y-3)(x(y-3)-1)=0x^2 [-20, 20, -10, 10]} $y = f \left(x\right) = \frac{1}{x} - 3$,

$x \left(y - 3\right) = 1$ .....(1)

As $x \to 0 , y - 3 \to \pm \infty$ and this gives $y \to \pm \infty$.

Likewise,

as$y - 3 \to 0 , x \to \pm \infty$

So,

the vertical asy,ptote is x = 0 and

the horizontal asymptote is y = 3.#.

Note

The equation $\left(y - {m}_{1} x - {c}_{1}\right) \left(y - {m}_{2} x - {c}_{2}\right) = k$ represents a

hyperbola having the guiding asymptotes

$\left(y - {m}_{1} x - {c}_{1}\right) \left(y - {m}_{2} x - {c}_{2}\right) = 0$.

The hyperbola is rectangular, if ${m}_{1} {m}_{2} = - 1$.

Here, from (1), it is immediate that

x(y-3)=0 gives the pair of perpendicular asymptotes.

See the Socratic graph.

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