How do you identify all asymptotes or holes for #f(x)=(x^3+3x^2+2x)/(3x^2+15x+12)#?

Question

Madelyn Anderson · Accepted Answer

There is a hole at #x=-1#
A vertical asymptote is #(x=-4)#

The slant asymptote is #y=x/3-2/3#
No horizontal asymptote

Let's simplify #f(x)# The numerator #=x^3+3x^2+2x=x(x^2+3x+2)# #=x(x+1)(x+2)# The denominator #=3x^2+15x+12=3(x^2+5x+4)# #=3(x+1)(x+4)# Therefore, #f(x)=(xcancel(x+1)(x+2))/(3cancel(x+1)(x+4))# There is a hole when #x=-1# #:. f(x)=(x(x+2))/(3(x+4))# As we cannot divide by #0#, #x!=-4# So, a vertical asymptote is #(x=-4)# As the degree of the numerator #># degree of the denominator, we expect a slant asymptote. Let #y=ax+b# be the slant asymptote #f(x)=(x^2+2x)/(3x+12)=(ax+b)+c/(3x+12)# #=((ax+b)(3x+12)+c)/(3x+12)# #:.x^2+2x=(ax+b)(3x+12)+c# Comparing the coefficients, of #x^2##=>#, #3a=1#, #a=1/3# of #x, ##=>#, #2=3b+12a#, #=># #b=-2/3# and #12b+c=0##=>#, #c=8# The slant asymptote is #y=x/3-2/3# #lim_(x->+-oo)f(x)=lim_(x->+-oo)x/3=+-oo# Therefore, no horizontal asymtote graph{(y-(x^2+2x)/(3x+12))(y-x/3+2/3)=0 [-10, 10, -5, 5]}

Xavier Alonso · Answer

undefined To identify all asymptotes or holes for the function $ f(x) = \frac{x^3 + 3x^2 + 2x}{3x^2 + 15x + 12} $, follow these steps:

1. **Factorize the function**: Factorize both the numerator and denominator if possible.

$$ f(x) = \frac{x(x^2 + 3x + 2)}{3(x^2 + 5x + 4)} $$

2. **Identify vertical asymptotes**: Set the denominator equal to zero and solve for $ x $. These are vertical asymptotes.

$$ 3x^2 + 15x + 12 = 0 $$
$$ x = -4 \text{ or } -1 $$

So, there are vertical asymptotes at $ x = -4 $ and $ x = -1 $.

3. **Identify holes**: Check for common factors between the numerator and denominator. If there are common factors that cancel out, there will be a hole at that point.

$$ f(x) = \frac{x(x^2 + 3x + 2)}{3(x^2 + 5x + 4)} = \frac{x(x + 1)(x + 2)}{3(x + 1)(x + 4)} $$

There's a common factor of $ x + 1 $, so there's a hole at $ x = -1 $.

4. **Identify horizontal asymptotes**: Compare the degrees of the numerator and denominator. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is at $ y = 0 $. If the degrees are equal, divide the leading coefficients to find the horizontal asymptote. If the degree of the numerator is greater, there is no horizontal asymptote.

In this case, the degrees of the numerator and denominator are both 3. Divide the leading coefficients:

$$ \text{Leading coefficient of numerator} = 1 $$
$$ \text{Leading coefficient of denominator} = 3 $$

So, the horizontal asymptote is given by $ y = \frac{1}{3}x $.

Therefore, the identified asymptotes and holes for the function $ f(x) = \frac{x^3 + 3x^2 + 2x}{3x^2 + 15x + 12} $ are:
- Vertical asymptotes at $ x = -4 $ and $ x = -1 $
- A hole at $ x = -1 $
- A horizontal asymptote at $ y = \frac{1}{3}x $.