\documentclass[12pt]{exam}
\noprintanswers
\noaddpoints

\usepackage{amsmath}
\usepackage{graphicx,float}
\usepackage{wrapfig}
\usepackage{enumerate}

\usepackage{tikz}
\usepackage{pgfplots}


\begin{document}

\begin{questions}


\question Do the following for each curve below.
\begin{itemize}
\item Use implicit differentiation to compute $\frac{dy}{dx}$.
\item Find $\frac{dx}{dy}$.
\item Identify any points $(x,y)$ on the curve where your formulas for $\frac{dy}{dx}$ and $\frac{dx}{dy}$ both evaluate to $\frac{0}{0}$. Your formulas are invalid at these points.
\item For what values of $x$ and $y$ is the slope of the curve positive?
\item When is the tangent line horizontal?
\item When is the tangent line vertical?
\item Match your curve to one of the graphs at the bottom of the page. Check that the graph satisfies all the properties above.
\end{itemize}

\vspace{12pt}

\begin{parts}


\part $ \displaystyle y^{2}=x^{3}+x^{2}$
% \frac{dy}{dx} = \frac{ x(3x+2) }{ 2y }
% slope positive if y>0 and x>0 or if y<0 and x<-2/3
% tangent horizontal if x=-2/3 or if x=0 and y<>0 (never happens)
% tangent vertical if y=0 and x<>0 (never happens)
% alpha shape
\vspace{6pt}

\part $ \displaystyle x^4=x^{2}-y^{2}$
% \frac{dy}{dx} = \frac{ x( 1 + sqrt{2} x )( 1 - sqrt{2} x ) }{ y }
% slope positive if y>0 and x>1/sqrt{2} or -1/sqrt{2}<x<0  or  y<0 and x<-1/sqrt{2} or 0<x<1/sqrt{2}
% tangent horizontal if y<>0 and x=0 or 
% lemniscate
\vspace{6pt}

\part $ \displaystyle x^{2/3} + y^{2/3} = 1$
% \frac{dy}{dx} = - \sqrt[3]{ \frac{y}{x} } always valid
% slope positive if x and y have opposite signs
% tangent horizontal when y=0 and x<>0
% tangent vertical when x=0 and y<>0
% asteroid
\vspace{12pt}

\end{parts}

\begin{center}
\includegraphics[width=0.8\textwidth]{esergel33.png}
\end{center}

\end{questions}

\end{document}
