ap calculus bc mcq practice test b\n21. which of the following could be the slope - field for the…

Explanation:

Step1: Find equilibrium solutions

Set $\frac{dy}{dx}=y^{2}-1 = 0$. Then $(y - 1)(y + 1)=0$, so $y=1$ and $y = - 1$ are equilibrium solutions. At these values of $y$, the slope $\frac{dy}{dx}=0$, and the slope - field will have horizontal segments.

Step2: Analyze sign of $\frac{dy}{dx}$ in intervals

When $y>1$, $y^{2}-1>0$, so $\frac{dy}{dx}>0$ and the slopes are positive. When $-1<y<1$, $y^{2}-1<0$, so $\frac{dy}{dx}<0$ and the slopes are negative. When $y < - 1$, $y^{2}-1>0$, so $\frac{dy}{dx}>0$ and the slopes are positive.

Answer:

C