I am trying to find all of the asymptotes of a rational function, and the answer choices I am given are confusing. Because the degree of the function in the numerator was greater than the degree of the function in the denominator, I divided the numerator by the denominator using synthetic division to get the slant asymptote.
Rational Function $f(x) = \frac{x^2 +4x+3}{x-2}$
After dividing by x=2 (or (x-2)) $y=2x+8$
Answer Choices
A. $y=x+6, x=2$
B. $x=-x=-3$
C. $y=1, x=2$
D. $y=x+6, x=-2$
I am struggling to follow your options but let us simply discuss the asymptotes.
You have made a mistake in your synthetic division. For $ x \to ±\infty$ You should get this: $$ \textrm{lim}_{x \to ±\infty} f(x)= \textrm{lim}_{x \to ±\infty} \frac{(x+6)(x-2)+15}{x-2} =\textrm{lim}_{x \to ±\infty} \frac{(x+6)(x-2)}{x-2} + \textrm{lim}_{x \to ±\infty} \frac{15}{x-2} \\ = x+6 $$
So the slant asymptote is $y=x+6$.
Now we also have a horizontal asymptote when the denominator is $0$. This simply given by $x-2=0 \implies x=2$.