How do you identify all asymptotes or holes for #f(x)=(x^3-4x^2+3x)/(3x^2-15x+18)#?

1 Answer
Dec 29, 2017

Answer:

See below.

Explanation:

First try to factor numerator and denominator:

#(x^3-4x^2+3x)/(3x^2-15x+18)=(x(x-1)(x-3))/(3(x-2)(x-3))=(x(x-1))/(3(x-2)#

Notice we can cancel the factor #(x-3)#. This shows that #x=3# is a removable discontinuity. This is sometimes called a point discontinuity or a singularity.

So there is a hole at #x=3#

Vertical asymptotes occur where the function is undefined. This occurs when #x=2# ( zero denominator )

So the line #x=2# is a vertical asymptote.

#(x(x-1))/(3(x-2))=(x^2-x)/(3x-6)#

The degree of the numerator is greater than the degree of the denominator, so there will be an oblique asymptote. Will can find the equation of this by polynomial division.

#(x^2-x)/(3x-6)=1/3x+1/3#

We only need to divide until we have the equation of a line, so the remainder of 2 in this case is irrelevant to us.

So the line #y= 1/3x +1/3# is an oblique asymptote.

For end behaviour, we only need concentrate on:

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

as #x->oo# , #x/3->oo#

as #x->-oo# , #x/3->-oo#

GRAPH:

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