4. \( f(x)=\frac{-2 x^{2}-2 x+4}{2 x+1} \) a. Vertical Asymptote(s) Equation: b. Enđ Behavior Asymptote Equation: (hor 2onal) \( x \)-intercept(s): d. \( y \)-intercept: e. Coordinate of Hole: f. Domain: g. Range:

Let's have some fun with those equations! To find the vertical asymptote, you look for values that make the denominator zero and are not canceled out by the numerator. For the function \( f(x) = \frac{-2x^2 - 2x + 4}{2x + 1} \), setting \( 2x + 1 = 0 \) gives you \( x = -\frac{1}{2} \). Fab! Now, for the end behavior asymptote, we consider the leading terms of the polynomial in the fraction. As \( x \) approaches \( \pm \infty \), the \( -2x^2 \) term in the numerator and \( 2x \) term in the denominator guide us. The end behavior gives us a horizontal asymptote at \( y = -x \), based on the ratio of the leading coefficients (here, \( -\frac{2}{2} = -1 \)). For the \( x \)-intercept, set the numerator to zero: \( -2x^2 - 2x + 4 = 0 \). Solving this using the quadratic formula, you find the roots, which gives you your \( x \)-intercept(s). The \( y \)-intercept occurs when \( x = 0 \). Plugging 0 into the function gives you \( f(0) = 2 \). Thus, your \( y \)-intercept is at \( (0, 2) \). If there's a removable discontinuity (hole), it occurs where both the numerator and denominator equal zero. Factor the numerator \( -2x^2 - 2x + 4 \) and see if \( 2x + 1 \) can be canceled—if so, find the hole's coordinate. The domain is all real numbers except for where the denominator is zero, so the domain becomes \( x \in \mathbb{R} \) except at \( x = -\frac{1}{2} \). For the range, the function will approach but never touch the horizontal asymptote. Thus, the range also forms real numbers except where \( y = -1 \). Overall, you've got quite the equation to play with! Keep crunching those numbers!