Mathematics: 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

Ithinks it’s a but not sure...

For what values of a and m does f(x) have a horizontal

Ответ:

raizagisselle7772

13.03.2022

Mathematics

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

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.

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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

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