\( 5^{\circ} \) Justify that as \( M \). Consider the function \( f \) defined for all \( x \neq-1 \) as \( : f(x)=-\frac{x^{2}+2 x+5}{2 x+2} \). Let ( \( C \) ) be its curve in the plane of an orthonormal system with unit 1 cm . \( \mathbf{1}^{\circ} \) Study the limits of \( f \) at the open bounds of its domain . Deduce the existance of an asymptote to be specified . \( 2^{\circ} \) Justify that the curve \( (C) \) has an oblique asymptote \( (D) \) with equation \( y=-\frac{1}{2} x-\frac{1}{2} \).

To study the limits of the function \( f(x) = -\frac{x^{2}+2x+5}{2x+2} \) at the open bounds of its domain, we first analyze the limits as \( x \) approaches the excluded point \( -1 \). Evaluating the limit as \( x \to -1 \): \[ \lim_{x \to -1} f(x) = \lim_{x \to -1} -\frac{x^{2} + 2x + 5}{2x + 2} = -\frac{1 + 2 - 5}{-2 + 2} = \text{undefined} \] Now, examine the limits as \( x \to \infty \) and \( x \to -\infty \): \[ \lim_{x \to \infty} f(x) = \lim_{x \to \infty} -\frac{x^{2} + 2x + 5}{2x + 2} \approx -\frac{x^{2}}{2x} = -\frac{1}{2} x \rightarrow -\infty \] \[ \lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} -\frac{x^{2}}{2x} \approx \frac{1}{2} x \rightarrow +\infty \] Thus, as \( x \to -1 \) the limit does not exist (vertical asymptote), while as \( x \to \pm\infty \), \( f(x) \) implies the presence of horizontal and/or oblique asymptotes. Considering \( f(x) \) for the oblique asymptote, we perform synthetic division of the polynomial \( -x^2 - 2x - 5 \) by \( 2x+2 \): 1. Divide: \(-\frac{1}{2} x - \frac{1}{2}\) This confirms \( (C) \) has an oblique asymptote given by: \[ y = -\frac{1}{2}x - \frac{1}{2} \] Conclusively, setting \( D \) as the equation of said asymptote leads us to understand the curve's behavior at infinity.