Question 1102598
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Suppose a rational function is written with  numerator consisting of a constant coefficient and linear factors of the form (x-a), and with the denominator consisting of linear factors of the form (x-a).  Then...
(1) Each factor (x-a) that appears the same number of times in both numerator and denominator will produce a hole in the graph at x=a;
(2) Each factor (x-b) that appears only in the denominator will produce a vertical asymptote at x=b;
(3) Each factor (x-c) that appears only once (or any odd number of times) in the numerator will produce a root at x=c, with the graph crossing the x-axis at that point; and
(4) Each factor (x-d) that appears twice (or any even number of times) in the numerator will produce a root at x=d, with the graph just touching the x-axis at that point.<br>
(5a) The graph of the rational function will have a horizontal asymptote of y=0 if the degree of the numerator is less than the degree of the denominator;
(5b) The graph will have a horizontal asymptote of y=k if the degrees of the numerator and denominator are equal and the constant coefficient is k.<br>
Here is a rational function that shows most of these features, along with its graph (red) and the horizontal asymptote (green).<br>
{{{(4(x-2)(x+2)^2(x-1))/((x+1)(x-1)(x-3)(x+3))}}}<br>
{{{graph(600,400,-8,8,-20,20,4(x-2)(x+2)^2(x-1)/(x+1)(x-1)(x-3)(x+3),4)}}}<br>
The constant coefficient 4, along with the fact that both numerator and denominator are of degree 4, produces the horizontal asymptote y=4 (item 5b above).<br>
The identical factors of (x-1) in the numerator and denominator produce a hole in the graph at x=1 (item 1 above).  The hole is not apparent in the graph produced by the software used on this site.<br>
(If you graph this function on a good graphing calculator, with a small window either side of x=1, you should be able to see the hole.)<br>
The factors (x+3), (x+1), and (x-3) in the denominator produce the vertical asymptotes at x=-3, x=-1, and x=3 (item 2 above).<br>
The single factor (x-2) in the numerator produces a zero at x=2, with the graph crossing the x-axis at that point (item 3 above).<br>
And the double factor of (x+2) in the numerator produces a double root at x=-2, with the graph just touching the x-axis at that point (item 4 above).<br><br>
Use that example and the preceding discussion to determine the factors that are required for your example.<br>
Notice that the requirements in your problem require...
(1) a single root  (1 factor in the numerator);
(2) a double root  (2 factors in the numerator);
(3) two vertical asymptotes  (2 factors in the denominator); and
(4) a hole (1 factor each in numerator and denominator)<br>
With only factors causing those features, the numerator has 4 linear factors while the denominator has only 3.  If those were the only factors, then there would be no horizontal asymptote, as required.  To get the required horizontal asymptote y=2, you need another factor in the denominator (producing a third vertical asymptote) to make the degrees of the numerator and denominator the same; and you need a constant coefficient 2 in the numerator.<br>
That is why my example is similar to yours but has three vertical asymptotes instead of two.<br>