the curves x = -y² - 8y - 12 and x = y + 6 are graphed. which expression represents the area bounded by the…

Explanation:

Step1: Find intersection - points

Set $-y^{2}-8y - 12=y + 6$. Rearrange to get $y^{2}+9y + 18 = 0$. Factor: $(y + 6)(y+3)=0$. So $y=-6$ and $y=-3$.

Step2: Determine the right - hand and left - hand functions

The right - hand function is $x_{1}=y + 6$ and the left - hand function is $x_{2}=-y^{2}-8y - 12$. The area $A$ between two curves $x = x_{1}(y)$ and $x = x_{2}(y)$ from $y = a$ to $y = b$ is given by $A=\int_{a}^{b}(x_{1}(y)-x_{2}(y))dy$.

Step3: Calculate the integrand

$x_{1}(y)-x_{2}(y)=(y + 6)-(-y^{2}-8y - 12)=y + 6 + y^{2}+8y + 12=y^{2}+9y + 18$.

Answer:

C. $\int_{-6}^{-3}(-y^{2}-9y - 18)dy$