How do you find the vertical, horizontal or slant asymptotes for $f (x) = (\frac{1}{x}) + 3$ $f(x) = (1/x) + 3$ ?

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Answer 1 · 2017-01-02T14:13:48

Horizontal: $\leftarrow y = 3 \to$ $larr y = 3 rarr$
Vertical : $↑ x = 0 ↓$ $uarr x = 0 darr$

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

Tn the quadratic form,

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

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

Likewise,

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

So,

the vertical asy,ptote is x = 0 and

the horizontal asymptote is y = 3.#.

Note

The equation $(y - m_{1} x - c_{1}) (y - m_{2} x - c_{2}) = k$ $(y-m_1 x-c_1)(y-m_2 x-c_2)=k$ represents a

hyperbola having the guiding asymptotes

$(y - m_{1} x - c_{1}) (y - m_{2} x - c_{2}) = 0$ $(y-m_1 x-c_1)(y-m_2 x-c_2)=0$ .

The hyperbola is rectangular, if $m_{1} m_{2} = - 1$ $m_1m_2=-1$ .

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