How do you find the asymptotes for #f(x)= (2x+1)/(x-1)#?

Question

Serenity Jones · Accepted Answer

The vertical asymptote is #x=1#
The horizontal asymptote is #y=2#
No slant asymptote

As you cannot divide by #0#, #x!=1# The vertical asymptote is #x=1# As the degree of the numerator #=# the degree of the denominator, there is no slant asymptote. #lim_(x->+-oo)f(x)=lim_(x->+-oo)(2x)/x=2# The horizontal asymptote is #y=2# graph{(y-(2x+1)/(x-1))(y-2)(y-100x+100)=0 [-18.02, 18.03, -9.01, 9.01]}

Anna Allen · Answer

undefined To find the asymptotes of the function $ f(x) = \frac{2x+1}{x-1} $:

1. Determine if there are any vertical asymptotes by identifying any values of $ x $ that make the denominator equal to zero. In this case, the denominator $ (x-1) $ equals zero when $ x = 1 $. Therefore, there is a vertical asymptote at $ x = 1 $.

2. Determine if there are any horizontal asymptotes by examining the behavior of the function as $ x $ approaches positive or negative infinity. Use the concept of limits:

$$ \lim_{x \to \infty} \frac{2x+1}{x-1} $$
$$ \lim_{x \to -\infty} \frac{2x+1}{x-1} $$

To evaluate these limits, divide both the numerator and denominator by the highest power of $ x $:

$$ \lim_{x \to \infty} \frac{2 + \frac{1}{x}}{1 - \frac{1}{x}} $$
$$ \lim_{x \to -\infty} \frac{2 + \frac{1}{x}}{1 - \frac{1}{x}} $$

As $ x $ approaches positive or negative infinity, the terms involving $ \frac{1}{x} $ become negligible compared to other terms. Thus, the limit simplifies to:

$$ \lim_{x \to \infty} \frac{2}{1} = 2 $$
$$ \lim_{x \to -\infty} \frac{2}{1} = 2 $$

Hence, there is a horizontal asymptote at $ y = 2 $.

In summary, the function $ f(x) = \frac{2x+1}{x-1} $ has a vertical asymptote at $ x = 1 $ and a horizontal asymptote at $ y = 2 $.