Question 1045870
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If there is a horizontal asymptote, then the degree of the numerator and denominator polynomials must be equal (which is true in this case) and the equation of the asymptote must be *[tex \Large y\ =\ \frac{p}{q}] where *[tex \Large p] is the lead coefficient of the numerator polynomial and *[tex \Large q] is the lead coefficient of the denominator polynomial.


So if your function is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{rx\ +\ s}{2x\ +\ t}]


and if the equation of the horizontal asymptote is


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ -4]


Then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{r}{2}\ =\ -4]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r\ =\ -8]


A vertical asymptote exists where the denominator polynomial is equal to zero, so if the vertical asymptote is


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 3]


then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2(3)\ -\ t\ =\ 0]


which is to say


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ 6]


In order for the function value to be zero, the numerator must be zero.  So, given an x-intercept of *[tex \Large (1,0)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -8(1)\ +\ s\ =\ 0]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ s\ =\ 8]


Putting it all together,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ -\frac{8x\ -\ 8}{2x\ -\ 6}]


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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