How do you identify all asymptotes for #f(x)=(1+3x^2-x^3)/x^2#?

Question

Serenity Johnson · Accepted Answer

#x=0# and #x+y=3#

First of all one need to see, whether there areany holes or not, i.e. those values of #x# for which value of function exists but function is not defined. This happens when there are common factors between numerator and denominator. For example if #x-alpha# is a common factor both in numerator and denominator, though they cancel out, function may not be defined. Here, we have none.

Further, if denominator can be factorized, we have vertical asymptotes. If factors are #(x-alpha)(x-beta)#, then vertical asymptotes are #x=alpha#and #x=beta#. For example, here we have just #x# as factor in denominator (with repetition) and hence only vertical asymptote is #x=0#,.

Horizontal asymptotes appear when degree of numerator and denominator are same. It is not so. However as degree of numerator is just one more than that of denominator, there is a slant or oblique asymptote.

Observe that #(1+3x^2-x^3)/x^2=1/x^2+3-x#

and as #x->oo#, #f(x)->3-x#

Hence slanting asymptote is #y=3-x# or #x+y=3#

graph{(1+3x^2-x^3)/x^2 [-19.17, 20.83, -1.44, 18.56]}

Serenity Johnson · Answer

undefined To identify all asymptotes for the function $ f(x) = \frac{1 + 3x^2 - x^3}{x^2} $, we first need to consider the behavior of the function as $ x $ approaches certain values.

1. Vertical Asymptotes: Vertical asymptotes occur where the function approaches positive or negative infinity as $ x $ approaches a certain value. Vertical asymptotes occur when the denominator of the function approaches zero while the numerator does not. So, we find vertical asymptotes by setting the denominator equal to zero and solving for $ x $. In this case, the denominator is $ x^2 $, so the vertical asymptotes occur at $ x = 0 $.

2. Horizontal Asymptotes: Horizontal asymptotes occur when the function approaches a constant value as $ x $ approaches positive or negative infinity. To find horizontal asymptotes, we examine the behavior of the function as $ x $ approaches infinity. We can use the properties of limits to determine this behavior. For rational functions like this one, the degree of the numerator and the degree of the denominator will affect the horizontal asymptotes. In this case, the degree of the numerator is 3, and the degree of the denominator is 2. Since the degree of the numerator is greater, there is no horizontal asymptote. Instead, there is a slant asymptote.

3. Slant (Oblique) Asymptote: Slant asymptotes occur when the degree of the numerator is one more than the degree of the denominator. To find the slant asymptote, we perform polynomial long division or synthetic division to divide the numerator by the denominator. In this case, when we divide $ 1 + 3x^2 - x^3 $ by $ x^2 $, we get $ -x + 3 $ with a remainder of 0. So, the equation of the slant asymptote is $ y = -x + 3 $.

Therefore, the asymptotes for the function $ f(x) = \frac{1 + 3x^2 - x^3}{x^2} $ are:
- Vertical asymptote: $ x = 0 $
- Slant asymptote: $ y = -x + 3 $

Elijah Allen · Answer

undefined To identify the asymptotes of $ f(x) = \frac{1 + 3x^2 - x^3}{x^2} $, we need to consider both vertical and horizontal asymptotes.

**Vertical Asymptotes:**

Vertical asymptotes occur where the function approaches infinity or negative infinity as $ x $ approaches a particular value.

For $ f(x) $, vertical asymptotes can occur where the denominator $ x^2 $ equals zero:

$$ x^2 = 0 $$
$$ x = 0 $$

So, $ x = 0 $ is a vertical asymptote.

**Horizontal Asymptotes:**

Horizontal asymptotes describe the behavior of the function as $ x $ approaches positive or negative infinity.

To find the horizontal asymptote, we need to look at the degrees of the numerator and denominator:

The degree of the numerator is 3 (highest power of $ x $) and the degree of the denominator is 2.

When the degree of the numerator is greater than the degree of the denominator by 1, the horizontal asymptote is the line $ y = 0 $.

Therefore, the horizontal asymptote for $ f(x) $ is $ y = 0 $.

In summary, the function $ f(x) = \frac{1 + 3x^2 - x^3}{x^2} $ has a vertical asymptote at $ x = 0 $ and a horizontal asymptote at $ y = 0 $.