SOLUTION: What are the zeros and asymptotes of {{{y = (x^2-3x)/(x+4)}}}

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Question 303223: What are the zeros and asymptotes of y+=+%28x%5E2-3x%29%2F%28x%2B4%29


Answer by Edwin McCravy(20065) About Me  (Show Source):
You can put this solution on YOUR website!

To find the zeros, we set the rational function equal to 0:

%28x%5E2-3x%29%2F%28x%2B4%29=0
x%5E2-3x=0
x%28x-3%29=0


So the graph crosses the x-axis at 0 and 3:

To find the vertical asymptote(s), we set the denominator
equal to zero:

x-4=0
x=4

That is the equation of a vertical line through 4 on the x-axis:

Since the degree of the numerator x%5E2-3x is 2 and the degree
of the denominator is 1, the numerator has a greater degree so this
rational function cannot have a horizontal asymptote.  However since
the degree of the numerator is exactly 1 degree greater than the 
denominator, there is a slanted or oblique asymptote which we find 
by long division:

             x -  7  
x + 4)x² -  3x +  0
      x² + 4x
           -7x +  0
           -7x - 28
                 28

We ignore the remainder and the slanted or oblique asymptote
is the line whose equation is y=x-7. Drawing in the
two asymptotes in green and the graph in red, we have:


   
Edwin