How do you find all the asymptotes for function #y=(x^2-4)/(x)#?

Question

Sophie Adams · Accepted Answer

The vertical asymptote is #x=0# and oblique asymptote
#y=x# A line #x=a# is a vertical asymptote of a function f(x) if #lim_(x->a)f(x)=+-oo# A line #y = b# is a horizontal asymptote of a function f(x) if #lim_(x->+-oo)=b# An oblique asymptote for a function f(x) has the formula #y=cx+d# where #c=lim_(x->oo)f(x)/x# and #d=lim_(x->oo)(f(x)-cx)# Hence we have that #f(x)->+-oo , x->0# and #c=limf(x)/x=lim(x^2-4)/x^2=1# #d=lim((x^2-4)/x-x)=0# Hence #x=0# and #y=x#

Serenity Johnson · Answer

undefined To find the asymptotes of the function $ y = \frac{x^2 - 4}{x} $, we need to consider both vertical and horizontal asymptotes.

Vertical asymptotes occur where the denominator of the function becomes zero, but the numerator does not. So, to find the vertical asymptotes, we set the denominator equal to zero and solve for x:
$$ x = 0 $$

For horizontal asymptotes, we analyze the behavior of the function as $ x $ approaches positive or negative infinity. To do this, we compare the degrees of the numerator and denominator. In this case, both the numerator and the denominator have the same degree (both are degree 1), so we look at the leading coefficients:
$$ \lim_{x \to \pm \infty} \frac{x^2 - 4}{x} = \lim_{x \to \pm \infty} \frac{x^2}{x} = \lim_{x \to \pm \infty} x = \pm \infty $$

Therefore, there are no horizontal asymptotes.

In summary:
- The vertical asymptote is $ x = 0 $.
- There are no horizontal asymptotes.