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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

User Cory Price
by
7.7k points

2 Answers

6 votes

Answer:

C

Explanation:

a= -1, m=1

User Purgoufr
by
8.4k points
5 votes

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



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


(2x-2(x+a))/(x+a)=-(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)(2x)/(x+1)

does not exist, as required.
User Mark Paspirgilis
by
8.0k points

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