Question 1197629
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To determine the vertical asymptotes, write the function with numerator and denominator in factored form:<br>
{{{(x^2+x-30)/(x^2-2x-15)=((x+6)(x-5))/((x-5)(x+3))}}}<br>
The denominator of the fraction can't be 0, so the function value is not defined at x=5 or at x=-3; those two values of x are the potential vertical asymptotes.<br>
Since the factor (x-5) is in both numerator and denominator, the function is equivalent to<br>
{{{(x+6)/(x+3)}}}<br>
everywhere except at x=5.  So at x=5 there is a hole in the graph instead of a vertical asymptote.<br>
So there is a vertical asymptote and x=-3, but none at x=5.<br>
The horizontal asymptote of the function is the function value when x becomes very large positive or very large negative.  For very large positive or negative values of x, the linear and constant terms become insignificant, so the function value approaches {{{x^2/x^2=1}}}<br>
So the horizontal asymptote is y=1.<br>
ANSWERS: vertical asymptote x=-3; horizontal asymptote y=1.<br>
A graph, showing the given function and its horizontal asymptote...<br>
{{{graph(800,400,-20,20,-10,10,(x^2+x-30)/(x^2-2x-15),1)}}}<br>
Note the hole at x=5 won't show up on this graph, because the function is undefined only at that one point.  Graphing the given function on a very small interval around x=5 on a good graphing calculator will show the hole in the graph.<br>