Identify the asymptotes. \[ t(x)=\frac{2-x}{x^{2}+x-6} \]

To find the asymptotes of the function \( t(x) = \frac{2-x}{x^2+x-6} \), we need to consider both vertical and horizontal asymptotes. First, let's determine the vertical asymptotes by finding the values of \( x \) that make the denominator equal to zero. We start by factoring the denominator: \[ x^2 + x - 6 = (x-2)(x+3) \] Setting the denominator to zero gives us: \[ (x-2)(x+3) = 0 \] From this, we find the values: \[ x = 2 \quad \text{and} \quad x = -3 \] Thus, the vertical asymptotes are at \( x = 2 \) and \( x = -3 \). Next, we find the horizontal asymptote by analyzing the degrees of the numerator and denominator. The degree of the numerator (which is 1) is less than the degree of the denominator (which is 2). This means that as \( x \) approaches infinity or negative infinity, \( t(x) \) approaches 0. Therefore, the horizontal asymptote is \( y = 0 \). To summarize, the function has vertical asymptotes at \( x = 2 \) and \( x = -3 \), and a horizontal asymptote at \( y = 0 \).