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

1 Answer
Jan 2, 2017

Answer:

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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