<PRE><B>Question 1157588</B>
&lt;br&gt;
(1) Horizontal or oblique asymptotes&lt;br&gt;
The degrees of the numerator and denominator are the same, so there is a horizontal asymptote and no oblique asymptote.&lt;br&gt;
The ratio of the leading coefficients gives you the horizontal asymptote.&lt;br&gt;
ANSWER: Horizontal asymptote at y = 3/2.&lt;br&gt;
(2) Vertical asymptotes&lt;br&gt;
Vertical asymptotes will occur wherever there is a linear factor in the denominator that is not also in the numerator.&lt;br&gt;
Factor numerator and denominator:&lt;br&gt;
{{{((3x+1)(x-4))/((2x+1)(x-4))}}}&lt;br&gt;
There is a vertical asymptote were the factor (2x+1) in the denominator is equal to 0.&lt;br&gt;
ANSWER: There is a single vertical asymptote, at x = -1/2.&lt;br&gt;
What about the factors (x-4) in both numerator and denominator?&lt;br&gt;
For x=4, the denominator is 0 and so the function is undefined.  For all other values of x, the function is equivalent to {{{(3x+1)/(2x+1)}}}.  So the graph of the given function is the same as the graph of {{{3x+1)/(2x+1)}}} except that there is a hole in the graph at x=4.&lt;br&gt;
</PRE>