Question

For what values of a and m does f(x) have a horizontal asymptote at y = 2 and a vertical asymptote at x = 1?

f(x)=2x^m/x+a

which is the answer

a = –1, m = 0
a = 1, m = 0
a = –1, m = 1
a = 1, m = 1

Answer 1

$f(x)$ will have a horizontal asymptote of $y=2$ if

$\displaystyle\lim_{x\to\pm\infty}(f(x)-2)=0$

where at least one of these limits needs to be satisfied. Since we would like


$\dfrac{2x^m}{x+a}-2=\dfrac{2x^m-2(x+a)}{x+a}$

to converge to 0, we need the $x^m$ term to disappear in the numerator. The only way for that to happen is if $m=1$. Then

$\dfrac{2x-2(x+a)}{x+a}=-\dfrac{2a}{x+a}$

and this does indeed approach 0 as $x$ gets arbitrarily large.

Now, in order to have a vertical asymptote at $x=1$, all we need to do is set $a=-1$. Then

$\displaystyle\lim_{x\to-1}\frac{2x}{x+1}$

does not exist, as required.

Answer 2

Answer:

C

Step-by-step explanation:

a= -1, m=1