for what values of a and m does f(x) have a horizontal asymptote at y = 2 and a vertical asymptote at x =…

Explanation:

Step1: Find vertical - asymptote condition

For a rational function $y = \frac{f(x)}{g(x)}$, vertical asymptotes occur at the values of $x$ that make the denominator $g(x)=0$. Given $f(x)=\frac{2x^{m}}{x + a}$, setting the denominator $x + a=0$, we get $x=-a$. Since the vertical asymptote is at $x = 1$, then $-a=1$, so $a=-1$.

Step2: Find horizontal - asymptote condition

Case 1: If $m = 0$, then $f(x)=\frac{2}{x - 1}$, and $\lim_{x\rightarrow\pm\infty}\frac{2}{x - 1}=0$. Case 2: If $m = 1$, then $f(x)=\frac{2x}{x - 1}$. We can rewrite $\frac{2x}{x - 1}$ as $\frac{2x-2 + 2}{x - 1}=2+\frac{2}{x - 1}$. Then $\lim_{x\rightarrow\pm\infty}\frac{2x}{x - 1}=\lim_{x\rightarrow\pm\infty}(2+\frac{2}{x - 1}) = 2$.

Answer:

C. $a=-1,m = 1$