Pre calc FInd the domain and asymptotes/holes y= 3(x-2)/(x-3)(x-2) I hope i wrote the above correctly if anything its a fraction 3(x-2) over (x-3)(x-2) Please show the steps and explain and if you can also graph it I need to check if I did this right and I am understanding it

College Algebra
1st Edition
ISBN:9781938168383
Author:Jay Abramson
Publisher:Jay Abramson
Chapter5: Polynomial And Rational Functions
Section5.6: Rational Functions
Problem 5TI: Find the vertical asymptotes and removable discontinuitiesof the graph of f(x)=x225x36x2+5x.
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Pre calc

FInd the domain and asymptotes/holes

y= 3(x-2)/(x-3)(x-2)

I hope i wrote the above correctly if anything its a fraction 3(x-2) over (x-3)(x-2)

Please show the steps and explain and if you can also graph it

I need to check if I did this right and I am understanding it

  

Expert Solution
Step 1

Given function is:

 

Domain of a function is defined as all the values for which the function is defined.

A rational function is not defined when the denominator is 0.

To determine the domain, the function is simplified.

Cancelling (x – 2) from numerator and denominator:

The above function is not defined when x – 3 is 0.

Hence,

Therefore, the domain of the function is R – {3}.

3 (x-2)
(x-3)x-2)
3
y =
(x-3)
x 3 0
x 3 3 03
Step 2

Asymptotes are the lines which the curve approaches but never crosses. Asymptotes can be horizontal, vertical or oblique.

 

Here the degree of the numerator is less than the degree of the denominator. Hence the horizontal asymptote is at y = 0.

 

The vertical asymptote can be obtained by equating the denominator (in simplified form) to 0.

Here, the denominator is x – 3.

Hence, the vertical asymptote is:

 

A hole is observed for a graph, if there is a common factor in numerator and denominator.

Here, the rational function has x – 2 in both the numerator and denominator. Hence, the hole for the function is given by:

 

r-3 0
x3 3 03
x - 2 0
x 2 2 02
x 2
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