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\begin{document}


\title{On-line Math 21}


\email{david.johnson@lehigh.edu}


\urladdr{http://www.lehigh.edu/\textasciitilde{}dlj0}

\maketitle

\subsection{Curve Sketching}

\begin{example}
\( f(x)=x^{2}-2x+3 \). 
\end{example}

\subsubsection*{Solution}

Although it was only mentioned in passing, it is useful to first find the intercepts,
the places where the graph crosses an axis. This give a place to start drawing
the graph.


\subparagraph*{Intercepts}

The \( y \)-axis intercept is always easy to find. Just evaluate \( f(0) \).
In this case, \( f(0)=3 \), so \( (0,3) \) is the \( y \)-intercept. 

The \( x \)-intercepts, where \( f(x)=0 \), (there may be more than one) are
harder to find, since it requires solving an equation. Only try to solve these
by hand if it is straightforward. If the solution looks difficult, wait until
the rest of the graph is drawn to see where to look for solutions -- or use
Maple. In this case, we just have to solve the equation\begin{eqnarray*}
0 & = & x^{2}-2x+3.
\end{eqnarray*}
Wait a minute. Is that so easy to solve? Well, if you can't factor it with the
standard game of trying to write \[
x^{2}-2x+3=(x-?)(x-??),\]
which I can't get to work out, you then try the quadratic formula. What this
does is find the numbers \( r_{1} \) and \( r_{2} \) so that \[
x^{2}-2x+3=(x-r_{1})(x-r_{2}),\]
or, equivalently (and what we want), find the roots of the equation \( x^{2}-2x+3=0 \),
by a direct formula, in general, \( ax^{2}+bx+c=0 \) has roots \( r \) determined
by \[
r=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}.\]
In this case, \[
r=\frac{+2\pm \sqrt{4-4\cdot 3}}{2}.\]
But this is impossible. There are no (real number) solutions, since that square
root is of a negative number. So, there are no \( x \)-intercepts.


\subparagraph*{Increasing/decreasing, and critical points}

If we find the derivative of the function, \[
f'(x)=2x-2,\]
which clearly means that \( f'(x)=0 \) only when \( x=1 \), and that the number
line for \( f'(x) \) in this case is 

\vspace{0.3cm}
{\par\centering \includegraphics{ex1-num-line.ps} \par}
\vspace{0.3cm}

Thus \( x=1 \) is a local minimum (decreasing down to it, increasing away).
It is helpful to find the values at any critical point. Here, \( f(1)=1^{2}-2\cdot 1+3=2 \),
so \( (1,2) \) is the critical point.


\subparagraph*{Draw the graph}

Then use this information to draw a fair representation of the graph. 

\begin{itemize}
\item The first information you plot are the intercepts; {[}Link to ex1-1.gif{]}
\item Then plot the critical point (and value). I always mark it with a horizontal
line to remind myself that it is a critical point. {[}Link to ex1-2.gif{]}
\item Then fill in the graph, connecting the dots and making sure that the graph you
draw is increasing/decreasing where the number line indicates it should be.
{[}Link to ex1-3.gif{]}
\end{itemize}
{[}Each of these items should trigger the appearance of a new drawing with that
information added.{]}

\vspace{0.3cm}
{\par\centering \includegraphics{ex1-3.ps} \par}\vspace{0.3cm}

\end{document}
