Question 1186817
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(1) vertical asymptote at x=2: there is a factor of (x-2) in the denominator but not in the numerator<br>
(2) hole at x=-6: there are factors of (x+6) in both numerator and denominator<br>
(3) x-intercept (1,0): there is a factor of (x-1) in the numerator but not in the denominator<br>
At this point, the parts of the function we have are these:<br>
{{{f(x)=((x-1)(x+6))/((x-2)(x+6))}}}<br>
That function has a horizontal asymptote at y=1; we need a slant asymptote of y=x+7.  To get a slant asymptote, we need an additional factor (x-a) in the numerator such that<br>
{{{f(x)=((x-1)(x+6)(x-a))/((x-2)(x+6))}}}<br>
has quotient (x+7) (and we don't care about the remainder)<br>
We can determine the constant a using synthetic division of (x-1)(x-a) = x^2+(-a-1)x+a by x-2:<br><pre>

  2  |  1  -a-1  a
     |        2 ...
     +---------------
        1  -a+1 ...</pre>
Since we want the asymptote to be y=x+7, we need to have<br>
{{{-a+1=7}}}
{{{-a=6}}}
{{{a=-6}}}<br>
The additional factor we need in the numerator is (x-(-6)) = (x+6).<br>
ANSWER: {{{f(x)=((x-1)(x+6)(x+6))/((x-2)(x+6))}}}<br>
A graph, showing the slant asymptote...<br>
{{{graph(500,500,-20,10,-20,20,((x-1)(x+6)(x+6))/((x-2)(x+6)),x+7)}}}<br>