## University Calculus: Early Transcendentals (3rd Edition)

The estimated slope of the curve at $P_1$ is $2.5$. The estimated slope of the curve at $P_2$ is $-2.5$.
To estimate the slope $m$ of the curve at point $P(a,b)$ without the equation of the function, we find another point $Q(c, d)$ as near to $P$ as possible in the graph, then the slope of the curve will approximately be equal with the slope of $PQ$: $$m=\frac{d-b}{c-a}$$ 1) Find the slope $m_1$ of the curve at $P_1(-1.6,1.5)$, or $P_1(-8/5,3/2)$ In the graph, we see that very near to $P_1$, there is this point $Q(-5/3,4/3)$. By calculating the slope of $P_1Q$, we will have the estimated slope $m_1$ of the curve at $P_1$: $$m_1=\frac{\frac{4}{3}-\frac{3}{2}}{-\frac{5}{3}-(-\frac{8}{5})}=\frac{-\frac{1}{6}}{\frac{8}{5}-\frac{5}{3}}=\frac{-\frac{1}{6}}{-\frac{1}{15}}=\frac{15}{6}=2.5$$ 2) Find the slope $m_2$ of the curve at $P_2(1.6,1.5)$, or $P_2(8/5,3/2)$ In the graph, we see that very near to $P_2$, there is this point $R(5/3,4/3)$. By calculating the slope of $P_2R$, we will have the estimated slope $m_2$ of the curve at $P_2$: $$m_2=\frac{\frac{4}{3}-\frac{3}{2}}{\frac{5}{3}-\frac{8}{5}}=\frac{-\frac{1}{6}}{\frac{1}{15}}=-\frac{15}{6}=-2.5$$ In the graph, we see that very near to $P_2$, there is this point $R(1.7,1.5)$. By calculating the slope of $P_2R$, we will have the estimated slope $m_2$ of the curve at $P_2$: $$m_2=\frac{1.5-1.3}{1.7-1.9}=\frac{0.2}{-0.2}=-1$$