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

1 Answer
Nov 17, 2016

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

Explanation:

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]}