Question 631227
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WARNING! Danger, Will Robinson.  Terminology Error!  *[tex \LARGE y\ =\ 2x\ -\ 4] is NOT a horizontal asymptote.  Horizontal asymptotes are horizontal, hence the name.  Linear functions with a non-zero (2 is not zero) slope are NOT horizontal.  Asymptotes that are slanted or oblique are called Slant Asymptotes or Oblique Asymptotes.


The fact that the function has zeros at -2 and 3 tells us that the factors of the numerator polynomial are *[tex \LARGE (x\ +\ 2)], *[tex \LARGE (x\ -\ 3)], and some constant *[tex \LARGE k] (because all polynomial equations *[tex \LARGE k\left(p(x)\right)\ =\ 0] where *[tex \LARGE p(x)] is a polynomial with degree *[tex \LARGE n] and *[tex \LARGE k\ \neq\ 0]have identical solution sets)


Hence, the numerator is *[tex \LARGE kx^2\ -\ kx\ -\ 6k]


The fact that the function has a vertical asymptote of *[tex \LARGE -1] means that the denominator polynomial has a zero at *[tex \LARGE -1], therefore the denominator polynomial must be *[tex \LARGE x\ +\ 1].


If a rational function has a numerator that is one degree greater than the degree of the denominator, then the function will have a slant asymptote equal to the quotient of a polynomial long division of the numerator by the denominator.


Perform the polynomial long division of *[tex \LARGE kx^2\ -\ kx\ -\ 6k\ \div\ x\ +\ 1].


Your quotient will have a factor of *[tex \LARGE k] in it, but if you set the quotient equal to the given slant asymptote *[tex \LARGE 2x\ -\ 4], you will very quickly see the value of *[tex \LARGE k].


Then it is simply a matter of constructing your function from the derived numerator and denominator.


Go to <a href="http://www.purplemath.com/modules/polydiv2.htm">Purple Math Polynomial Long Division</a> if you need a refresher on polynomial long division.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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