How do you graph #y=(x^3+27)/(3x^2+x)# using asymptotes, intercepts, end behavior?

1 Answer
Nov 29, 2017

Answer:

See below.

Explanation:

#y# axis intercepts occur when #x=0#

#y=((0)^3+27)/(x(0)^2+(0))=0#

(This is undefined, so no y axis intercept)

#x# axis intercepts occur when #y=0# i.e. when #x^3+27=0#

#x^3+27=0=>x=root(3)(-27)=-3#

(This will also have complex roots, but these will be of no help in graphing).

Point: #(-3 , 0) #

Vertical asymptotes occur where the function is undefined. i.e. #3x^2+x=0#

#3x^2+x=x(3x+1)=0=>x=0 and x=-1/3#

Asymptotes are the lines:

#color(blue)(x=0) and color(blue)(x=-1/3)#

Because the rational function is an improper fraction ( degree of numerator is higher than degree of denominator) an oblique asymptote occurs. The can be found by dividing numerator by denominator.

#(x^3+27)/(3x^2+x)=1/3x-1/9#

( we only need to divide enough to give us the terms of a line equation)

So the line:

#color(blue)(y=1/3x-1/9)# is an oblique asymptote:

End behaviour:

#x^3/(3x^2)=1/3x#

as #x->oo#,#color(white)(88)1/3x->oo#

as #x->-oo#,#color(white)(88)1/3x->-oo#

Graph:

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