Free Response With Calculator (4) 4. Find the \( y \)-value(s) where \( x^{2}+y^{2}=-4 x-6 y-4 \) has vertical tangents.

To find the \( y \)-value(s) where the equation \( x^{2}+y^{2}=-4 x-6 y-4 \) has vertical tangents, we can first rewrite the equation in a more manageable form. Rearranging gives us: \[ x^2 + 4x + y^2 + 6y + 4 = 0 \] Next, we complete the square for both \( x \) and \( y \): \[ (x+2)^2 + (y+3)^2 = 9 \] This represents a circle centered at \( (-2, -3) \) with a radius of \( 3 \). For vertical tangents, we look for points where the slope of the tangent line is undefined, which occurs when \( \frac{dx}{dy} = 0 \). To find \( \frac{dx}{dy} \), we can use implicit differentiation on the circle equation. Implicitly differentiating gives: \[ 2x + 4 + 2y \frac{dy}{dx} + 6 \frac{dy}{dx} = 0 \] Simplifying, we get: \[ (2y + 6) \frac{dy}{dx} = - (2x + 4) \] Thus, \[ \frac{dy}{dx} = \frac{- (2x + 4)}{2y + 6} \] For vertical tangents, we require \( \frac{dx}{dy} = 0 \) or \( 2y + 6 = 0 \): \[ 2y + 6 = 0 \implies y = -3 \] Lastly, we check for \( x \): Plugging \( y = -3 \) back into the circle equation: \[ (x + 2)^2 + (3 - 3)^2 = 9 \implies (x + 2)^2 = 9 \implies x + 2 = \pm 3 \] This gives us \( x = 1 \) and \( x = -5 \). Therefore, the points with vertical tangents are \( (1, -3) \) and \( (-5, -3) \). Thus, the \( y \)-value where the equation has vertical tangents is: \[ \boxed{-3} \]