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%TCIDATA{<META NAME="Title" CONTENT="Graphs of Equations">}
%TCIDATA{Created=Mon Aug 19 14:52:24 1996}
%TCIDATA{LastRevised=Sun Mar 23 15:28:07 1997}
%TCIDATA{Language=American English}
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\begin{document}


\section{Graphs of Equations\label{Graphs of Equations}}

We say a point \textbf{satisfies} an equation in two variables $x$ and $y$
if the equation is true when the coordinates of the point are substituted
for the corresponding variables. For instance, $\left( 2,5\right) $
satisfies the equation $y^{2}-x^{2}=21$, since $5^{2}-2^{2}=21$.

\subsection{The Graph of an Equation\label%
{Definition of Graph of an Equation}}

The \textbf{graph} of an equation in $x$ and $y$ is the set of all points $%
\left( x,y\right) $ in the coordinate plane that satisfy the equation.

The graphs of most of the equations you will encounter are curves.

\subsection{Sketching a Graph by Plotting Points\label%
{Sketching a graph by plotting points}}

To sketch the graph of an expression by plotting points, first find some
number of points satisfying the expression (typically, the more points, the
better the sketch), and then join the points with curves or lines.
Generally, this method can only enable you to draw a good approximation to a
graph if you have additional information about the shape of the graph.

\begin{itemize}
\item[Example]  Sketch the graph of the equation 
\[
y=x^{2}-2
\]

\item[Solution]  To find some points satisfying the equation, it is
frequently useful to choose some values for one variable (most often $x$),
and then find corresponding values for $y$ by substituting these into the
equation. So, for instance, if you choose $x=0$, a $y$ value is obtained
from $y=0^{2}-2=-2$. Following this procedure, you can obtain the following
table: 
\[
\begin{tabular}{|l|l|l|}
\hline
$x$ & $y$ & $\left( x,y\right) $ \\ \hline
$-3$ & $7$ & $\left( -3,7\right) $ \\ 
$-2$ & $2$ & $\left( -2,2\right) $ \\ 
$-1$ & $-1$ & $\left( -1,-1\right) $ \\ 
$0$ & $-2$ & $\left( 0,-2\right) $ \\ 
$1$ & $-1$ & $\left( 1,-1\right) $ \\ 
$2$ & $2$ & $\left( 2,2\right) $ \\ 
$3$ & $7$ & $\left( 3,7\right) $ \\ \hline
\end{tabular}
\]
Now draw the points and connect them with a smooth curve\FRAME{dtbpFU}{%
2.898in}{1.9337in}{0pt}{\Qcb{$y=x^{2}-2$}}{}{quadplot1.bmp}{\special%
{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio
TRUE;display "PICT";valid_file "F";width 2.898in;height 1.9337in;depth
0pt;original-width 0pt;original-height 0pt;cropleft "0";croptop
"1";cropright "1";cropbottom "0";filename 'quadplot1.bmp';file-properties
"NPEU";}}\bigskip This type of curve is called a \textbf{parabola}. \bigskip 
\end{itemize}

\subsection{Definition of Intercepts\label{Definition of intercepts}}

You can improve your sketches of graphs somewhat by finding the points at
which the graphs cross the coordinate axes.

\begin{itemize}
\item  The \textbf{x-intercepts} of the graph of an equation are the
x-coordinates of points where the graph intersects the x-axis.

To find the x-intercepts, set $y=0$ and solve for $x$.

\item  The \textbf{y-intercepts} of the graph of an equation are the
y-coordinates of points where the graph intersects the y-axis.

To fnd the y-intercepts, set $x=0$ and solve for $y$.\bigskip 
\end{itemize}

\subsection{The Equation of a Circle\label{Equation of a Circle}\hfill}

You have seen some hints on how to sketch the graph of an equation in $x$
and $y$. Now consider the reverse problem: given a curve in the coordinate
plane, find an equation whose graph is exactly that curve; in other words,
find an equation satisfied exactly by the points on the curve.

Geometrically, a circle is the set of points all lying a given (fixed)
distance from another point (the center of the circle). Take as your problem
finding the equation of a circle of radius $r$ (where $r>0$) centered at the
point $\left( h,k\right) $.

You need an equation such that a given $x$ and $y$ satisfy the equation
exactly when the distance from $\left( x,y\right) $ to $\left( h,k\right) $
is $r$. According to the distance formula this translates to 
\[
\sqrt{\left( x-h\right) ^{2}+\left( y-k\right) ^{2}}=r
\]
Squaring both sides yields 
\[
\left( x-h\right) ^{2}+\left( y-k\right) ^{2}=r^{2}
\]
Furthermore, since both $\sqrt{\left( x-h\right) ^{2}+\left( y-k\right) ^{2}}
$ and $r$ are nonnegative, any $x$ and $y$ satisfying this equation must
also satisfy the first one, taking square roots of each side. Thus the two
equations are equivalent, and this is the desired equation. In particular,
if the center is the origin (so $h=k=0$), the equation is 
\[
x^{2}+y^{2}=r^{2}
\]

\subsection{Definition of Symmetry\label{Symmetries of Graphs in the Plane}}

Another piece of information you can use to improve your graphing is \textbf{%
symmetry}. Observe the graph of $y=x^{2}$\FRAME{dtbpF}{2.898in}{1.9337in}{0pt%
}{}{}{x_squared.bmp}{\special{language "Scientific Word";type
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'x_squared.bmp';file-properties "NPEU";}}Notice that the right half of the
graph (to the right of the y-axis) is the mirror image of the left half.
Knowing this about a graph in advance would be a help in sketching it.
Similarly, the top half of a graph may be the mirror image of the bottom
half. What types of such symmetries can you recognize before drawing a graph?

\begin{itemize}
\item  \label{Symmetry with respect to the x-axis}\textbf{Symmetry with
respect to the x-axis} means that the graph is unchanged when reflected
across the x-axis (the upper half is the mirror image of the lower half).

How can you determine this in advance? This symmetry says that a point $%
\left( x,y\right) $ lies on the graph if and only if its mirror point $%
\left( x,-y\right) $ does; so the condition you're looking for is that $%
\left( x,y\right) $ satisfies the equation of the graph exactly when $\left(
x,-y\right) $ does. In other words, replacing every $y$ value by its
opposite and leaving the $x$ values alone should leave the equation
unchanged!

$y^{2}-2y^{4}=x+1$ is an example, since $\left( -y\right) ^{2}-2\left(
-y\right) ^{4}=y^{2}-2y^{4}$. So this graph would have symmetry with respect
to the x-axis.\FRAME{dtbpF}{2.898in}{1.9337in}{0pt}{}{}{symmetry1.bmp}{%
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"NPEU";}}

\item  \label{Symmetry with respect to the y-axis}\textbf{Symmetry with
respect to the y-axis} means that the graph is unchanged when reflected
across the y-axis (the right half is the mirror image of the left half).

This symmetry says that a point $\left( x,y\right) $ satisfies the equation
of the graph exactly when $\left( -x,y\right) $ does. In other words,
replacing $x$ by $-x$ should leave the equation unchanged.

$y=x^{2}$ is an example, as you saw above; for $\left( -x\right) ^{2}=x^{2}$.

\item  \label{Symmetry with respect to the origin}\textbf{Symmetry with
respect to the origin} means that the graph is unchanged when rotated $180%
\UNICODE[m]{0xb0}$ around the origin, or in other words, when each point is
reflected across the origin.

This symmetry says that a point $\left( x,y\right) $ satisfies the equation
of the graph exactly when $\left( -x,-y\right) $ does. In other words,
replacing $x$ by $-x$ and $y$ by $-y$ should leave the equation unchanged.

An example is $x^{2}+xy+y^{2}=3$, for replacing $x$ by $-x$ and $y$ by $-y$
yields 
\begin{eqnarray*}
\left( -x\right) ^{2}+\left( -x\right) \left( -y\right) +\left( -y\right)
^{2} &=&3 \\
x^{2}+xy+y^{2} &=&3
\end{eqnarray*}
Its graph is\FRAME{dtbpF}{2.7665in}{2.0591in}{0pt}{}{}{symmetry3.bmp}{%
\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio
TRUE;display "PICT";valid_file "F";width 2.7665in;height 2.0591in;depth
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"1";cropright "1";cropbottom "0";filename 'symmetry3.bmp';file-properties
"NPEU";}}
\end{itemize}

\subsection{Using Symmetry and Intercepts to Sketch a Graph\label%
{Using symmetry and intercepts to sketch a graph}}

\begin{itemize}
\item[Example]  Find the x- and y-intercepts of the graph of the equation $%
y=x^{3}-9x$. Test the equation for symmetry and sketch its graph.

\item[Solution]  Find the y-intercept by setting $x=0$%
\[
y=0^{3}-9\left( 0\right) =0
\]
so the y-intercept is $0$. You can find the x-intercept by setting $y=0$%
\begin{eqnarray*}
0 &=&x^{3}-9x \\
0 &=&x\left( x^{2}-9\right)  \\
0 &=&x\left( x+3\right) \left( x-3\right) 
\end{eqnarray*}
so the x-intercepts are $0$, $-3$, and $3$.

Now test for symmetry. For symmetry with respect to the x-axis, replace $y$
by $-y$%
\[
-y=x^{3}-9x
\]
This is certainly not the same equation as $y=x^{3}-9x$, so the graph fails
to be symmetric about the x-axis. For symmetry with respect to the y-axis,
replace $x$ by $-x$%
\begin{eqnarray*}
y &=&\left( -x\right) ^{3}-9\left( -x\right)  \\
y &=&-x^{3}+9x
\end{eqnarray*}
This again is not the same as the original equation, so the graph fails to
be symmetric about the y-axis. For symmetry with respect to the origin, make
both replacements 
\begin{eqnarray*}
-y &=&\left( -x\right) ^{3}-9\left( -x\right)  \\
-y &=&-x^{3}+9x \\
y &=&x^{3}-9x\text{\quad (multiplying by }-1\text{ on each side)}
\end{eqnarray*}
So the graph \emph{does} have symmetry about the origin.

Finally, plot points for some positive values of $x$%
\[
\begin{tabular}{|l|l|l|}
\hline
$x$ & $y$ & $\left( x,y\right) $ \\ \hline
$0$ & $0$ & $\left( 0,0\right) $ \\ 
$1$ & $-8$ & $\left( 1,-8\right) $ \\ 
$1.5$ & $-10.125$ & $\left( 1.5,-10.125\right) $ \\ 
$2$ & $-10$ & $\left( 2,-10\right) $ \\ 
$2.5$ & $-6.875$ & $\left( 2.5,-6.875\right) $ \\ 
$3$ & $0$ & $\left( 3,0\right) $ \\ 
$4$ & $28$ & $\left( 4,28\right) $ \\ \hline
\end{tabular}
\]
Graph these points first, including the x- and y-intercepts found above (the
only one not accounted for in the table is $\left( -3,0\right) $)\FRAME{dtbpF%
}{2.898in}{1.9337in}{0pt}{}{}{plot_sym_int1.bmp}{\special{language
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\end{itemize}

\noindent \hrulefill

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\noindent \hrulefill

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