How to find the asymptotes of #f(x)=2 - 3/x^2#?

Question

Elijah Allen · Accepted Answer

Vertical asymptote is #x =0# or #y#-axis

Horizontal asymptote is #y=2#

No oblique asymptote.

An ASYMPTOTE is a line that approches a curve, but NEVER meets it.

To find the vertical asymptote , put the denominator = 0 (because 0 cannot divide any number) and solve. This is where the function cannot exist.

Given below is the step-by-step walk through:

The curve can never touch #x =0#, or the #y#-axis , thus making it the vertical asymptote.

To find the horizontal asymptote , compare the degree of the expressions in the numerator and the denominator.

First, lets re-write the expression so we have one a common denominator.

Now we can compare the degrees of the numerator and the denominaotr.

The degree of the numerator = 2 and the degree of the denominator = 2.
Since the degrees are equal, the horizontal asymptote

#y = ("numerator's leading coefficient"/"denominator's leading coefficient")#

# y = 2/1 => y = 2#

The oblique asymptote is a line of the form y = mx + c.
Oblique asymptote exists when the degree of numerator = degree of denominator + 1
Here, the degree of the numerator = degree of the denominator = 2.
Therefore, the given function has no oblique asymptotes.

Chloe Alexander · Answer

undefined To find the asymptotes of the function $ f(x) = 2 - \frac{3}{x^2} $:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $. If any solutions are found, they represent vertical asymptotes.
   
2. Horizontal asymptotes: Examine the behavior of the function as $ x $ approaches positive or negative infinity. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is at $ y = 0 $. If the degree of the numerator equals the degree of the denominator, divide the leading coefficients to find the horizontal asymptote.

For the given function, there are no vertical asymptotes since the denominator $ x^2 $ is never zero. As $ x $ approaches positive or negative infinity, $ f(x) $ approaches $ y = 2 $. Therefore, there is a horizontal asymptote at $ y = 2 $.