<PRE><B>Question 113821</B>
&lt;b&gt;Domain:&lt;/b&gt;



{{{f(x)=(x^2+x-6)/(x-3)}}} Start with the given function



{{{x-3=0}}} 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.




{{{x=0+3}}}Add 3 to both sides



{{{x=3}}} Combine like terms on the right side






Since {{{x=3}}} makes the denominator equal to zero, this means we must exclude {{{x=3}}} from our domain



Answer:

So our domain is:  *[Tex \LARGE \textrm{\left{x|x\in\mathbb{R} x\neq3\right}}]


which in plain English reads: x is the set of all real numbers except {{{x&lt;&gt;3}}}


So our domain looks like this in interval notation

*[Tex \Large \left(-\infty, 3\right)\cup\left(3,\infty \right)]


&lt;hr&gt;



&lt;b&gt;x-intercept(s):&lt;/b&gt;


{{{f(x)=(x^2+x-6)/(x-3)}}} Start with the given function



{{{0=(x^2+x-6)/(x-3)}}} To find the x-intercepts, set f(x) (which is y) equal to zero



{{{x^2+x-6=0}}} This means the numerator is equal to zero. Remember the denominator cannot equal zero




{{{(x+3)(x-2)=0}}} Factor the left side (note: if you need help with factoring, check out this 
&lt;a href=http://www.algebra.com/algebra/homework/playground/change-this-name4450.solver&gt;solver&lt;/a&gt;)




Now set each factor equal to zero:


{{{x+3=0}}} or {{{x-2=0}}}


{{{x=-3}}} or {{{x=2}}}  Now solve for x in each case



Answer:

So our solutions are {{{x=-3}}} or {{{x=2}}}

This means the x-intercepts are (-3,0) and (2,0)


&lt;hr&gt;


&lt;b&gt;Hole(s):&lt;/b&gt;


{{{f(x)=(x^2+x-6)/(x-3)}}} Start with the given function



{{{f(x)=(x+3)(x-2)/(x-3)}}} 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



&lt;hr&gt;


&lt;b&gt;y-intercept(s):&lt;/b&gt;



{{{f(x)=(x^2+x-6)/(x-3)}}} Start with the given function



{{{f(0)=(0^2+0-6)/(0-3)}}} To find the y-intercept, plug in x=0



{{{f(0)=(-6)/(-3)}}} Simplify



{{{f(0)=2}}} Divide



Answer:


So the y intercept is (0,2)



&lt;hr&gt;




&lt;b&gt;Vertical Asymptote(s):&lt;/b&gt;



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 {{{x=3}}}


&lt;hr&gt;



&lt;b&gt;Oblique Asymptote:&lt;/b&gt;


To find the oblique asymptote, divide {{{(x^2+x-6)/(x-3)}}} using synthetic division



Start with the given expression {{{(x^2 + x - 6)/(x-3)}}}


First lets find our test zero:


{{{x-3=0}}} Set the denominator {{{x-3}}} equal to zero


{{{x=3}}} 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.&lt;TABLE cellpadding=10&gt;&lt;TR&gt;&lt;TD&gt;3&lt;/TD&gt;&lt;TD&gt;|&lt;/TD&gt;&lt;TD&gt;1&lt;/TD&gt;&lt;TD&gt;1&lt;/TD&gt;&lt;TD&gt;-6&lt;/TD&gt;&lt;/TR&gt;&lt;TR&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;|&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;/TR&gt;&lt;TR&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;/TR&gt;&lt;/TABLE&gt;

Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 1)

&lt;TABLE cellpadding=10&gt;&lt;TR&gt;&lt;TD&gt;3&lt;/TD&gt;&lt;TD&gt;|&lt;/TD&gt;&lt;TD&gt;1&lt;/TD&gt;&lt;TD&gt;1&lt;/TD&gt;&lt;TD&gt;-6&lt;/TD&gt;&lt;/TR&gt;&lt;TR&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;|&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;/TR&gt;&lt;TR&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;1&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;/TR&gt;&lt;/TABLE&gt;

    Multiply 3 by 1 and place the product (which is 3)  right underneath the second  coefficient (which is 1)

    &lt;TABLE cellpadding=10&gt;&lt;TR&gt;&lt;TD&gt;3&lt;/TD&gt;&lt;TD&gt;|&lt;/TD&gt;&lt;TD&gt;1&lt;/TD&gt;&lt;TD&gt;1&lt;/TD&gt;&lt;TD&gt;-6&lt;/TD&gt;&lt;/TR&gt;&lt;TR&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;|&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;3&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;/TR&gt;&lt;TR&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;1&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;/TR&gt;&lt;/TABLE&gt;

    Add 3 and 1 to get 4. Place the sum right underneath 3.

    &lt;TABLE cellpadding=10&gt;&lt;TR&gt;&lt;TD&gt;3&lt;/TD&gt;&lt;TD&gt;|&lt;/TD&gt;&lt;TD&gt;1&lt;/TD&gt;&lt;TD&gt;1&lt;/TD&gt;&lt;TD&gt;-6&lt;/TD&gt;&lt;/TR&gt;&lt;TR&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;|&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;3&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;/TR&gt;&lt;TR&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;1&lt;/TD&gt;&lt;TD&gt;4&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;/TR&gt;&lt;/TABLE&gt;

    Multiply 3 by 4 and place the product (which is 12)  right underneath the third  coefficient (which is -6)

    &lt;TABLE cellpadding=10&gt;&lt;TR&gt;&lt;TD&gt;3&lt;/TD&gt;&lt;TD&gt;|&lt;/TD&gt;&lt;TD&gt;1&lt;/TD&gt;&lt;TD&gt;1&lt;/TD&gt;&lt;TD&gt;-6&lt;/TD&gt;&lt;/TR&gt;&lt;TR&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;|&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;3&lt;/TD&gt;&lt;TD&gt;12&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;/TR&gt;&lt;TR&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;1&lt;/TD&gt;&lt;TD&gt;4&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;/TR&gt;&lt;/TABLE&gt;

    Add 12 and -6 to get 6. Place the sum right underneath 12.

    &lt;TABLE cellpadding=10&gt;&lt;TR&gt;&lt;TD&gt;3&lt;/TD&gt;&lt;TD&gt;|&lt;/TD&gt;&lt;TD&gt;1&lt;/TD&gt;&lt;TD&gt;1&lt;/TD&gt;&lt;TD&gt;-6&lt;/TD&gt;&lt;/TR&gt;&lt;TR&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;|&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;3&lt;/TD&gt;&lt;TD&gt;12&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;/TR&gt;&lt;TR&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;&lt;/TD&gt;&lt;TD&gt;1&lt;/TD&gt;&lt;TD&gt;4&lt;/TD&gt;&lt;TD&gt;6&lt;/TD&gt;&lt;/TR&gt;&lt;/TABLE&gt;

Since the last column adds to 6, we have a remainder of 6. This means {{{x-3}}} is &lt;b&gt;not&lt;/b&gt; a factor of  {{{x^2 + x - 6}}}

Now lets look at the bottom row of coefficients:


The first 2 coefficients (1,4) form the quotient


{{{x + 4}}}





So the oblique asymptote is the line {{{y=x+4}}}




Notice if we graph {{{f(x)=(x^2+x-6)/(x-3)}}} and it&#39;s asymptotes, we can visually verify our answers:



{{{ graph( 500, 500, -10, 15, -10, 15, (x^2+x-6)/(x-3),x+4, 1000(x-3)) }}} Graph of {{{f(x)=(x^2+x-6)/(x-3)}}} with the oblique asymptote {{{y=x+4}}} (green) and the vertical asymptote {{{x=3}}} (blue)


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