## Thomas' Calculus 13th Edition

The slope of the curve at $P_1$ is given as $m_1=-2$ and at $P_2$ is: $m_2=0$
1. From the graph, it can be seen that very near to $P_1$ is a point $Q(-1.6,-1)$. The slope $m_1$ of the curve at $P_1$ is given as $m_1=\frac{-1-(-1.4)}{-1.6-(-1.4)}=\dfrac{-1+1.4}{-1.6+1.4}=\frac{0.4}{-0.2}=-2$ 2. From the graph, it can be seen that very near to $P_2$ lies at the very top of a curve in the graph and the tangent line to the curve passing $P_2$ is parallel with the x-axis. Thus, this implies that the slope of that tangent line is $0$. $\implies m_2=0$