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 the vertical - asymptote

The vertical asymptote of a rational function $y = \frac{f(x)}{g(x)}$ occurs at the values of $x$ that make the denominator $g(x)=0$. For the function $f(x)=\frac{2x^{m}}{x + a}$, setting the denominator equal to zero gives $x=-a$. Since the vertical asymptote is at $x = 1$, then $-a = 1$, so $a=-1$.

Step2: Find the horizontal - asymptote

Case 1: If $m>1$, as $x\to\pm\infty$, $y=\frac{2x^{m}}{x + a}\to\pm\infty$. Case 2: If $m = 1$, $y=\frac{2x}{x + a}=\frac{2x}{x(1+\frac{a}{x})}=\frac{2}{1+\frac{a}{x}}$. As $x\to\pm\infty$, $y\to2$. Case 3: If $m = 0$, $y=\frac{2}{x + a}$, and as $x\to\pm\infty$, $y\to0$. So, when $m = 1$ and $a=-1$, the function has a horizontal asymptote at $y = 2$ and a vertical asymptote at $x = 1$.

Answer:

a=-1, m = 1