Question 202900
<font face="Garamond" size="+2">

1) Has a horizontal asymptote at *[tex \Large y = 2] because the numerator and denominator are of the same degree, hence there is a horizontal asymptote at the value of the lead coefficient of the numerator divided by the lead coefficient of the denominator.  This one is false.


2) The graph of a rational function can have *[tex \Large x] intercepts.  The graph will intercept the *[tex \Large x]-axis at any zero of the numerator polynomial that is not also a zero of the denominator polynomial.  Consider *[tex \Large \frac{x^2-3x+2}{4 - x}].  This one is false.


3) As in #1, the horizontal asymptote is at *[tex \Large y = a] where *[tex \Large a] is the value of the lead coefficient of the numerator divided by the lead coefficient of the denominator.  Hence, the horizontal asymptote is at *[tex \Large y = 4] not *[tex \Large y = -3].  This one is false.


4) This one looks false on the face of it, because you would expect a vertical asymptote at any zero of the denominator polynomial.  However, note that -2 is a zero of the denominator, but it is also a zero of the numerator.  Hence, while there is a discontinuity at -2, there is no asymptote.  In order to prove it, you need a little trick from the Calculus called L'Hôpital's Rule.  It says that if:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \lim_{x\rightarrow c}\ f(x) = 0\text{ or }\pm\infty]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \lim_{x\rightarrow c}\ g(x)= 0\text{ or }\pm\infty]


and if


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \lim_{x\rightarrow c}\ \frac{f'(x)}{g'(x)}] exists, then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \lim_{x\rightarrow c}\ \frac{f(x)}{g(x)} = \lim_{x\rightarrow c}\ \frac{f'(x)}{g'(x)}]


So for the function given in #4,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ f'(x) = 2] and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ g'(x) = 2x] and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \lim_{x\rightarrow -2}\ \frac{f'(x)}{g'(x)} = \frac{2}{2(-2)}=-\frac{1}{2}]


not *[tex \Large \pm\infty] as would be the case if there were an asymptote at that point.


Therefore #4 is the true statement.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
</font>