Question

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

Answer 1

Horizontal: `larr y = 3 rarr`
Vertical : `uarr x = 0 darr`

Explanation:

graph{x(y-3)(x(y-3)-1)=0x^2 [-20, 20, -10, 10]} `y=f(x)=1/x-3`,

Tn the quadratic form,

`x(y-3)=1` .....(1)

As `x to 0, y-3 to +-oo` and this gives `y to +-oo`.

Likewise,

as`y-3 to 0, x to +-oo`

So,

the vertical asy,ptote is x = 0 and

the horizontal asymptote is y = 3.#.

Note

The equation `(y-m_1 x-c_1)(y-m_2 x-c_2)=k` represents a

hyperbola having the guiding asymptotes

`(y-m_1 x-c_1)(y-m_2 x-c_2)=0`.

The hyperbola is rectangular, if `m_1m_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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