Question 1113031
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A rational function with a removable discontinuity at a point *[tex \Large a] has a factor of *[tex \Large x\ -\ a] in BOTH the numerator and denominator.  A rational function with an infinite discontinuity at a point *[tex \Large a] has a factor of *[tex \Large x\ -\ a] in the denominator ONLY.  A rational function with an *[tex \Large x]-intercept at *[tex \Large (a,\,0)] has a factor of *[tex \Large x\ -\ a] in the numerator only.


Hence, *[tex \Large x\ +\ 2] must be in both numerator and denominator, *[tex \Large x\ +\ 3] must be in the denominator only, and *[tex \Large x\ +\ 1] must be in the numerator only, thus:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r(x)\ =\ \frac{(x\ +\ 1)(x\ +\ 2)}{(x\ + \ 2)(x\ +\ 3)]


And we are almost there.  Note that *[tex \Large r(0)\ =\ \frac{1}{3}], but we want the *[tex \Large y]-intercept, that is the point *[tex \Large \(0,r(0)\)] to be *[tex \Large (0,6)].


Hence, we need a lead coefficient of 18 to complete all of the given specifications.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ \frac{18(x\ +\ 1)(x\ +\ 2)}{(x\ + \ 2)(x\ +\ 3)]


You don't specify the required form for your final answer so this form fits the requirements of the problem as you posted it.  If you need it multiplied out, I'll just leave that in your capable hands.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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