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Question 113821: 6. Determine the behavior of 
Domain :
x – intercept(s) :
Hole(s) :
y – intercept :
Vertical Asymptote(s) :
Oblique Asymptote :
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! Domain:
Start with the given function
 Set the denominator equal to zero. Remember, dividing by 0 is undefined. So if we find values of x that make the
denominator zero, then we must exclude them from the domain.
 Add 3 to both sides
 Combine like terms on the right side
Since makes the denominator equal to zero, this means we must exclude from our domain
Answer:
So our domain is: 
which in plain English reads: x is the set of all real numbers except 
So our domain looks like this in interval notation
\cup\left(3,\infty \right))
x-intercept(s):
Start with the given function
 To find the x-intercepts, set f(x) (which is y) equal to zero
 This means the numerator is equal to zero. Remember the denominator cannot equal zero
 Factor the left side (note: if you need help with factoring, check out this
solver)
Now set each factor equal to zero:
 or 
 or  Now solve for x in each case
Answer:
So our solutions are or
This means the x-intercepts are (-3,0) and (2,0)
Hole(s):
Start with the given function
 Factor
Answer:
Since the expression does not simplify further, there are no exceptions to make about the two expressions. So in this case there are no holes in this function
y-intercept(s):
Start with the given function
 To find the y-intercept, plug in x=0
 Simplify
 Divide
Answer:
So the y intercept is (0,2)
Vertical Asymptote(s):
Since the value x=3 is excluded from the domain, and there are no holes, there is one vertical asymptote at x=3
Answer:
So the vertical asymptote is
Oblique Asymptote:
To find the oblique asymptote, divide  using synthetic division
Start with the given expression 
First lets find our test zero:
Set the denominator  equal to zero
Solve for x.
so our test zero is 3
Now set up the synthetic division table by placing the test zero in the upper left corner and placing the coefficients of the numerator to the right of the test zero.
Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 1)
Multiply 3 by 1 and place the product (which is 3) right underneath the second coefficient (which is 1)
Add 3 and 1 to get 4. Place the sum right underneath 3.
Multiply 3 by 4 and place the product (which is 12) right underneath the third coefficient (which is -6)
Add 12 and -6 to get 6. Place the sum right underneath 12.
Since the last column adds to 6, we have a remainder of 6. This means is not a factor of 
Now lets look at the bottom row of coefficients:
The first 2 coefficients (1,4) form the quotient
So the oblique asymptote is the line 
Notice if we graph and it's asymptotes, we can visually verify our answers:
 Graph of with the oblique asymptote (green) and the vertical asymptote (blue)
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