\documentclass[12pt]{article}
\usepackage{graphicx}
\usepackage{amsmath}

\begin{document}
\begin{figure}
\begin{center}
\includegraphics[width=.8\linewidth]{SF1.png}
\end{center}
\caption{The slope field for $\frac{dy}{dx} = \frac{1}{4}y(5 - y)$.}
\end{figure}

\begin{figure}
\begin{center}
\includegraphics[width=.8\linewidth]{SF2.png}
\end{center}
\caption{The slope field for $\frac{dy}{dx} = \frac{1}{4}(x^2 + y^2)$.}
\end{figure}

\begin{figure}
\begin{center}
\includegraphics[width=.8\linewidth]{SF3.png}
\end{center}
\caption{The slope field for $\frac{dy}{dx} = \frac{x - y}{x + y}$.}
\end{figure}

\begin{figure}
\begin{center}
\includegraphics[width=.8\linewidth]{SF1S.png}
\end{center}
\caption{Solutions of the equation $\frac{dy}{dx} = \frac{1}{4}y(5 - y)$
with $y(0) = .5$, $y(0) = 3$, $y(0) = 7$.}
\end{figure}

\begin{figure}
\begin{center}
\includegraphics[width=.8\linewidth]{SF2S.png}
\end{center}
\caption{Solutions of $\frac{dy}{dx} = \frac{1}{4}(x^2 + y^2)$ with $y(-3) = -3$ and $y(-3) = -1$.}
\end{figure}

\begin{figure}
\begin{center}
\includegraphics[width=.8\linewidth]{SF3S.png}
\end{center}
\caption{Solutions of $\frac{dy}{dx} = \frac{x - y}{x + y}$ with $y(-3) = -2$, $y(0) = 2$.}
\end{figure}


\end{document}

\begin{figure}
\begin{center}
\includegraphics[width=.8\linewidth]{needleregion.png}
\end{center}
\caption{The region for problem 4.}
\end{figure}

\end{document}



