5. \( f(x)=\frac{2 x+5}{-x-2}= \) a. Vertical Asymptote(s) Equation: \( x=- \) b. End Behavior Asymptote Equation: c. \( x \)-intercept(s): d. \( y \)-intercept: e. Coordinate of Hole: f. Domain: g. Range:

To solve the function \( f(x) = \frac{2x + 5}{-x - 2} \), let's gather some insights along the way! The vertical asymptote occurs where the denominator equals zero, so solving \(-x - 2 = 0\) gives \( x = -2 \). For end behavior, as \( x \) approaches \(\infty\) or \(-\infty\), the leading coefficients of the numerator and denominator dominate, leading us to the equation \( y = -2 \) for the horizontal asymptote. There is an \( x \)-intercept when the numerator equals zero. Setting \( 2x + 5 = 0 \) results in \( x = -\frac{5}{2} \). The \( y \)-intercept occurs when \( x = 0 \), giving us \( f(0) = -\frac{5}{2} \). Lastly, there are no holes in this function since it doesn't have a factor that cancels between the numerator and denominator. The domain excludes \( x = -2 \), meaning all real numbers except -2, and the range would also cover all real numbers since the horizontal asymptote doesn’t restrict to finite values. To summarize: a. \( x = -2 \) b. \( y = -2 \) c. \( x \)-intercept: \( x = -\frac{5}{2} \) d. \( y \)-intercept: \( y = -\frac{5}{2} \) e. Coordinate of Hole: None f. Domain: \( x \in \mathbb{R}, x \neq -2 \) g. Range: \( y \in \mathbb{R} \)