How do you find the asymptotes for $h (x) = \frac{2 x + 3}{3 x + 1}$ $h(x) = (2x+3)/(3x+1 )$ ?

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Answer 1 · 2016-04-13T20:11:35

$Vertical asymptote at x = - \frac{1}{3}$ $color(blue)("Vertical asymptote at "x= -1/3)$

$Horizontal asymptote at y = \frac{2}{3} \to 0.66 ¯ 6$ $color(blue)("Horizontal asymptote at "y= 2/3 ->0.66bar6)$

The equation becomes undefined at $x = - \frac{1}{3}$ $x=-1/3$
The denominator is not 'allowed' to become 0

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As absolute $x$ $x$ becomes increasingly larger the +3 and +1 become insignificant.

So $lim_(xto+-oo) (2x+3)/(3x+1) -> 2/3xx x/x = 2/3$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$color(blue)("Vertical asymptote at "x= -1/3)$

$color(blue)("Horizontal asymptote at "y= 2/3 ->0.66bar6)$

Tony B

How do you find the asymptotes for h(x) = (2x+3)/(3x+1 )h(x)=2x+33x+1h(x) = (2x+3)/(3x+1 )?