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What is an example of a function that has a vertical asymptote at $x = - 1$ $x = -1$ and a horizontal asymptote at $y = 2$ $y = 2$ ?

1 Answer

Mar 5, 2018

$f (x) (= y) = 2 + \frac{1}{x + 1}$ $f(x) (=y) = 2+1/(x+1)$

Explanation:

We know that the relation $ˆ y = \frac{1}{ˆ x}$ $haty=1/hatx$ has a vertical asymptote at $ˆ x = 0$ $hatx=0$ and a horizontal asymptote at $ˆ y = 0$ $haty=0$

If we wish to shift all points left one unit (i.e. the vertical asymptote to $x = - 1$ $x=-1$ ) then we need to replace $ˆ x$ $hatx$ with $x + 1$ $x+1$ (so when $x = - 1$ $x=-1$ then $ˆ x = 0$ $hatx=0$ , the vertical asymptote).

Similarly to shift all points up two units (i.e. the horizontal asymptote to $y = 2$ $y=2$ ) we need to replace $ˆ y$ $haty$ with $y - 2$ $y-2$

Therefore the required relation would become
$XXX y - 2 = \frac{1}{x + 1}$ $color(white)("XXX")y-2=1/(x+1)$

Converting this into function form (with $y = f (x)$ $y=f(x)$ )
we have
$XXX y = 2 + \frac{1}{x + 1}$ $color(white)("XXX")y=2+1/(x+1)$

enter image source here

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