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) a=-1,m=0 a=1,m=0 a=-1,m=1 a=1,m=1

To determine the values of and for which the function \(f(x) = \frac{2x^m}{x + a}\) has a horizontal at and a vertical asymptote at , we need to analyze the behavior of the function as approaches infinity and as approaches 1.1. **Horizontal Asymptote at :** - For a rational function \(\frac{P(x)}{Q(x)}\), the horizontal asymptote is determined by the degrees of the polynomials \(P(x)\) and \(Q(x)\). - If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is . - If the degree of the numerator is equal to the degree of the denominator, the horizontal asymptote is . - If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote (there may be an oblique asymptote instead). In this case, we want the horizontal asymptote to be . This implies that the degrees of the numerator and the denominator must be equal, and the leading coefficient of the numerator divided by the leading coefficient of the denominator must be 2. Therefore, (since the degree of the numerator is 1) and the leading coefficient of the numerator is 2. The leading coefficient of the denominator is 1 (since the degree of the denominator is also 1). So, the horizontal asymptote condition is satisfied when .2. **Vertical Asymptote at :** - A vertical asymptote occurs where the denominator of the rational function is zero, provided the numerator is not zero at the same point. - For \(f(x) = \frac{2x^m}{x + a}\), the denominator is zero when , which gives . We want the vertical asymptote to be at , so we set , which gives .Combining both conditions, we find that and .Therefore, the correct answer is: