Question 987330
<pre>
Instead of doing your problem for you, I'll do one exactly in every detail
like yours so that you can do yours yourself using this as a model.  You
won't learn it as well if I do it for you. I'll do this one instead:
</pre>
State the vertical, horizontal asymptotes and zeros of the rational function, 
{{{f(x) =  (2x^2 + 9x + 6)/(x^2 + 7x + 12)}}}
Why is there no zero at x = –3?
<pre>
First we'll get the horizontal asymptote's equation:

Since the numerator and denominator have the same degree, the horizontal
asymptote's equation is 
{{{y}}}{{{""=""}}}{{{matrix(1,5,Leading,coefficient,in,the,numerator)/matrix(1,5,Leading,coefficient,in,the,denominator)}}}   
That's
{{{y}}}{{{""=""}}}{{{2/1}}}
{{{y}}}{{{""=""}}}{{{2}}}

Now let's answer the question "Why is there no zero at x = –3?"
Let's substitute -2 for x and see:

{{{"f(-3)" =  (2(-3)^2 + 9(-3) + 6)/((-3)^2 + 7(-3) + 12)}}}
{{{"f(-2)" =  (2(9)-27 + 6)/(9 - 21 + 12)}}}
{{{"f(-2)" =  (18-27 + 6)/(9 - 21 + 12)}}}
{{{cross("f(-2)" =  0/0)}}}

One of the impossible things in mathematics is division
by zero.  So that's why I crossed this out, and it's why f(-3)
is not a zero.

Now we'll go back and find the vertical asymptotes

{{{f(x) =  (2x^2 + 9x + 6)/(x^2 + 7x + 12)}}}

We factor numerator and denominator:

{{{f(x) =  ((x+3)(2x+3))/((x+3)(x+4))}}}

Now since the (x+3)'s will cancel, that will give us another function g(x)
which is exactly like f(x) except it WILL have a zero at x=-3 whereas f(x)
does not have a zero there:

So we get

{{{g(x) =  (2x+3)/(x+4)}}}

And we find the vertical asymptote's equation by setting the denominator = 0:

x+4 = 0

So the vertical asymptote's equation is x = -4

Now we graph g(x) by graphing the horizontal and vertical asymptotes (in green)
and making a table of values:

 x   y
-14  2.5
-9   3
-5   7
-3  -3
 0   1
 6  1.5    
  
{{{drawing(400,4000/11,-15,7,-10,10,
green(line(-4,20,-4,-20),line(-20,2,20,2)),
graph(400,4000/11,-15,7,-10,10,(2x+3)/(x+4)))}}}

That's the graph of g(x) but the graph of f(x) must
have a hole at (-3,-3) because that's not part of the
graph:

{{{drawing(400,4000/11,-15,7,-10,10,
circle(-3,-3,.5),
green(line(-4,20,-4,-20),line(-20,2,20,2)),
graph(400,4000/11,-15,7,-10,10,(2x+3)/(x+4)))}}}

Edwin</pre>