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\problemgrouptitle{Secant and Tangent Lines}
\item Consider the curve described by the function $y=x^3+x^2-2x+3$~.
\begin{enumerate}[label=(\alph*)]
\item Show that the points $(-1,5)$ and $(0,3)$ lie on the curve.
\item Determine the slope of the secant line passing through the points $(0,3)$ and $(-1,5)$.	
\item\label{itm:secant} Let $Q$ be the arbitrary point $(x,x^3+x^2-2x+3)$ on the curve. Find the slope of the secant line passing through $Q$ and $(-1,5)$.
\item Use your answer in \ref*{itm:secant} to determine the slope of the tangent line to the curve at the point $(-1,5)$.
\end{enumerate}
\answer{
\begin{enumerate}
\item $(-1)^3+(-1)^2-2(-1)+3=5$, $(0)^3+(0)^2-2(0)+3=3$
\item $\displaystyle\frac{3-5}{0-(-1)}=-2$
\item $\displaystyle\frac{x^3+x^2-2x+3-5}{x-(-1)}=x^2-2\ \ (x\neq -1)$
\item $\displaystyle\lim_{x\to-1}\left(x^2-2\right)=-1$
\end{enumerate}
}
