Using the concept of vertical and horizontal asymptotes, it is found that the function will have a horizontal asymptote at y = 2 and a vertical asymptote at x = 1 for a = -1 and m = 1.
In this problem, the function is:
[tex]f(x) = \frac{2x^m}{x + a}[/tex]
It has a vertical asymptote at x = 1, hence:
x + a = 0
1 + a = 0
a = -1
It has a horizontal asymptote at y = 2, hence:
[tex]\lim_{x \rightarrow \infty} f(x) = 2[/tex]
[tex]\lim_{x \rightarrow \infty} \frac{2x^m}{x + a} = 2[/tex]
[tex]\lim_{x \rightarrow \infty} \frac{2x^m}{x} = 2[/tex]
[tex]2^{m-1} = 2[/tex]
Then, since we want to simplify, the exponents at the numerator and the denominator have to be equal, hence m = 1.
More can be learned about asymptotes and end behavior at https://brainly.com/question/28037814
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