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\title[Shifting graphs; symmetries of graphs]
{Shifting graphs; symmetries of graphs}

\institute[College of the Holy Cross]
{
   MATH 133 -- Calculus with Fundamentals, section 1, Prof. Little\\
}

%\date[September 7, 2015]
%{September 7, 2015}

\begin{document}

\begin{frame}
  \titlepage
\end{frame}

\begin{frame}
{\bf Vertical shifts of graphs}

Shift the whole graph $y = f(x) = \sin(x)$ up (i.e. along the 
direction of the positive $y$-axis) by 2 units in the $xy$-plane:

\begin{figure}
\includegraphics[scale=.3]{VertShift.png}
\end{figure}

The new graph also satisfies the vertical line test (do you see why?)
So the shifted graph is also the graph of some function $y = g(x)$. 
{\it What is the formula for $g(x)$ to get the shifted graph?}

\end{frame}

\begin{frame}
{\bf Formula for a shift}

\begin{itemize}
\item<1-> More precisely, the shifting means that for each point $(x,y)$ on the graph 
$y = \sin(x)$, we want the point $(x,y+2)$ to be on the new shifted graph.
\item<2-> If $y = \sin(x)$ at a point on the original graph, then $(x, \sin(x) + 2)$
is a point on the shifted graph.  
\item<3->  This means the shifted graph comes from the function $y = g(x) = \sin(x) + 2$.
\item<4->  The general pattern is the same and we can ``work it both ways:''  If we know the graph
$y = f(x)$, then the graph $y = f(x) + c$ is a \emph{vertical shift}, up by $c$ units
along the $y$-direction if $c > 0$, down by $|c|$ units if $c < 0$.  
\end{itemize}


\end{frame}

\begin{frame}
{\bf A downward shift}

For instance, we can ask:  What does the graph $y = x^2 - 3$ look like?

\begin{itemize}
\item<1-> The graph $y = f(x) = x^2$ is the standard parabola with vertex at $(0,0)$, opening up.
\item<2-> Here we want $y = g(x) = x^2 - 3 = f(x) - 3$.  
\item<3-> This has the form $f(x) + c$ for $c = -3$.
\item<4-> Since $-3 < 0$, we are shifting the whole parabola \emph{down} by 3 units
\end{itemize}

\end{frame}

\begin{frame}
{\bf Shifted parabola}

\begin{figure}
\includegraphics[scale=.3]{VertShiftPara.png}
\end{figure}

(Because of the way the parabola is curving, a bit of an \emph{optical illusion} occurs here: 
it is not obvious that the vertical separation between the two graphs is always 3 units.  
But that can be seen from the points plotted as circles with the arrows.)
\end{frame}

\begin{frame}
{\bf Horizontal shifting}

Now, let's repeat the above sort of analysis, but shifting the graph $y = x^2$ by 2 units to the right.

\begin{figure}
\includegraphics[scale=.3]{HorizShiftPara.png}
\end{figure}

For instance, from $(0,0)$ on the original parabola, we get the corresponding point $(2,0)$ on
the shifted parabola.
\end{frame}

\begin{frame}
{\bf The equation of a horizontal shift}

\begin{itemize}
\item<1->  What is the equation of the shifted parabola?
\item<2->  For each point $(a,a^2)$ on the original parabola $y = x^2$, we get a corresponding
point $(a + 2, a^2)$ on the shifted parabola.
\item<3-> Writing $x = a + 2$, we get $a = x - 2$ so $y = a^2 = (x - 2)^2$.
\item<4-> Since the points on the shifted parabola have the form $(x,(x - 2)^2)$, that is the 
graph of $y = (x - 2)^2$. 
\item<5-> By the same reasoning, shifting \emph{left} by $1$ unit would give the graph 
$y = (x + 1)^2$:  If $(a,a^2)$ is on $y = x^2$, then $(a - 1,a^2)$ is on the graph of 
$y = (x + 1)^2$ since if $x = a - 1$, then $a = x+1$.  
\end{itemize}
\end{frame}

\begin{frame}
{\bf The general pattern}

\begin{itemize}
\item<1-> Shifting $y = f(x)$ horizontally gives the graph of $y = f(x + c)$ for some $c$, 
and vice versa, the graph $y = f(x + c)$ is a horizontal shift of $y = f(x)$. 
\item<2-> \emph{But, be careful -- this works differently from the vertical shifts}:  The $+c$ is
added to the $x$ ``\emph{inside}'' the function, \emph{and} the signs of $c$ work the 
\emph{opposite way from what you might expect}: if $c > 0$ the shift is to the \emph{left}, and if $c < 0$, the
shift is to the \emph{right}.
\item<3-> For example, $y = \sin(x - 1)$ is the graph $y = \sin(x)$ shifted to the \emph{right}
by one unit; while $y = \sin(x + \pi)$ is the sine graph shifted to the \emph{left} by 
$\pi$ units.  
\end{itemize}
\end{frame}

\begin{frame}
{\bf Even functions and $y$-axis symmetry of graphs}

\begin{itemize}
\item<1->  A function $f$ is said to be \emph{even} if $f(-x) = f(x)$ for all $x$.  
\item<2->  In other words, the $y$-coordinates on the points $(x,f(x))$ and $(-x,f(-x))$
are the same for all $x$.  
\item<3->  The function $f(x) = x^2$ is an example since $f(-x) = (-x)^2 = x^2 = f(x)$ for all $x$.
\item<4-> Saying $f$ is even is the same as saying the graph $y = f(x)$ has \emph{$y$-axis symmetry}.
\begin{figure}
\includegraphics[scale=.2]{Para.png}
\end{figure}
\end{itemize}
\end{frame}

\begin{frame}
{\bf Odd functions and their symmetry}

\begin{itemize}
\item<1->  A function $f$ is said to be \emph{odd} if $f(-x) = -f(x)$ for all $x$.
\item<2->  In other words, if $(x,f(x))$ is on the graph, then so is $(-x,f(-x)) = (-x,-f(x))$.
\item<3->  The function $f(x) = x^3$ is an example since $f(-x) = (-x)^3 = (-1)^3 x^3 = -x^3 = -f(x)$.
\item<4->  Our book says that $y = f(x)$ has \emph{symmetry about the origin} if $f$ is odd.
\item<5->  \emph{To visualize this}: If $(a,b)$ is a point in the plane, then $(-a,-b)$ can be obtained by \emph{rotating} the plane by $180$ degrees (or $\pi$ radians) about the origin.  
\item<6->  Rotating the whole plane this way, the graph of an odd function is \emph{taken to itself}
(but points in the first quadrant would get rotated around to the third quadrant, etc.).
\end{itemize}
\end{frame}

\begin{frame}
{\bf An odd function and symmetry about the origin}

\begin{columns}
\column{.5 \textwidth}
\begin{figure}
\includegraphics[scale=.27]{Cub.png}
\caption{$y = x^3$}
\end{figure}


\column{.5 \textwidth}
\begin{figure}
\includegraphics[scale=.27]{RotCub.png}
\caption{Rotations of $y = x^3$}
\end{figure}

\end{columns}
On the right, the dotted curves are rotations of $y = x^3$ by whole number multiples of $30$ degrees
counterclockwise.  The red arc gives the path followed by $(1.1, (1.1)^3)$ on $y = x^3$.
After a rotation through $180$ degrees, we're back to $y = x^3$(!)
\end{frame}


\end{document}

