Question 1153489
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I don't understand what you did for the double root at x=1.<br>
And the solution from the other tutor has a single root at x=1 -- not a double root.<br>
This problem is greatly complicated by requiring a horizontal asymptote at a particular value and a hole in the graph at a particular value.<br>
The requirements are....
(a) a hole at (1,3)
(b) a vertical asymptote at x=4
(c) a horizontal asymptote at y=-2
(d) a double root at x=-1<br>
Note your work shows a horizontal asymptote at y=-4 instead of y=-2.  I will assume the y=-2 is the actual requirement.<br>
We'll take care of the vertical asymptote and the double root first -- those are straightforward.<br>
For the vertical asymptote at x=4 we need a factor of (x-4) in the denominator; for a double root at x=-1, we need two factors of (x+1) in the numerator:<br>
{{{((x+1)^2)/(x-4)}}}<br>
Next, for the hole at (1,3), we need factors of (x-1) in both numerator and denominator:<br>
{{{((x+1)^2(x-1))/((x-4)(x-1))}}}<br>
At this point, the degree of the numerator is greater than the degree of the denominator; there will not be a horizontal asymptote.  We need an additional factor in the denominator to make the degrees the same.<br>
Furthermore, the additional factor has to be such that the hole in the graph is at (1,3).  We need to determine that additional factor.
We first insert a constant factor -2 to get the required horizontal asymptote:<br>
{{{(-2(x+1)^2(x-1))/((x-4)(x-1)(x-a))}}}<br>
Now we add the additional unknown factor (x-a) in the denominator and evaluate the function at x=1, ignoring the factors of (x-1) in numerator and denominator, to determine the unknown factor, knowing that the y value at the hole is 3:<br>
{{{(-2(2^2))/((-3)(1-a)) = 3}}}
{{{8/(3(1-a)) = 3}}}
{{{8 = 9-9a}}}
{{{9a = 1}}}
{{{a = 1/9}}}<br>
The required additional factor in the denominator is (x-1/9); the function is now complete:<br>
{{{(-2(x+1)^2(x-1))/((x-4)(x-1)(x-1/9))}}}<br>
A graph with window [-4,4,-2,5] showing the double root at -1:<br>
{{{graph(400,400,-4,4,-2,5,(-2(x+1)^2(x-1))/((x-4)(x-1)(x-1/9)))}}}<br>
Note the graphing utility on this forum will not show the hole in the graph; but the graph passes through (1,3).<br>
A graph with window [-40,40,-10,10] showing the horizontal asymptote at y=-2:<br>
{{{graph(600,400,-40,40,-10,10,(-2(x+1)^2(x-1))/((x-4)(x-1)(x-1/9)))}}}<br>