if the domain of function below is \( D f=\mathbb{R}-\{-2\} \), what is \( a+b= \) ? \( f(x)=\frac{x^{2}-1}{x^{2}+a x+b} \)

To determine the values of \( a \) and \( b \) such that the function \[ f(x) = \frac{x^2 - 1}{x^2 + ax + b} \] has a domain of \( D_f = \mathbb{R} - \{-2\} \), we need to ensure that the denominator \( x^2 + ax + b \) is equal to zero at \( x = -2 \) and is non-zero for all other real numbers. First, we substitute \( x = -2 \) into the denominator: \[ (-2)^2 + a(-2) + b = 0 \] This simplifies to: \[ 4 - 2a + b = 0 \] Rearranging gives us: \[ b = 2a - 4 \quad \text{(1)} \] Next, we need to ensure that the quadratic \( x^2 + ax + b \) does not have any other real roots. The condition for a quadratic \( Ax^2 + Bx + C \) to have no real roots is that its discriminant must be less than zero. The discriminant \( D \) for our quadratic is given by: \[ D = a^2 - 4b \] Substituting \( b \) from equation (1) into the discriminant: \[ D = a^2 - 4(2a - 4) \] This simplifies to: \[ D = a^2 - 8a + 16 \] We want this discriminant to be less than zero: \[ a^2 - 8a + 16 < 0 \] Factoring the quadratic: \[ (a - 4)^2 < 0 \] The expression \( (a - 4)^2 \) is always non-negative and equals zero when \( a = 4 \). Therefore, the only solution is: \[ a = 4 \] Substituting \( a = 4 \) back into equation (1) to find \( b \): \[ b = 2(4) - 4 = 8 - 4 = 4 \] Thus, we have: \[ a = 4 \quad \text{and} \quad b = 4 \] Finally, we calculate \( a + b \): \[ a + b = 4 + 4 = 8 \] Therefore, the answer is: \[ \boxed{8} \]