\documentclass[12pt]{article}\usepackage{ammaths}\begin{document}

\partie{Homework \# 11}

{\bf An interesting formula}

In this homework, we will look at a formula that allows to find the formula of a polynomial function of the second
degree, knowing only the coordinates of its extremal point and two special characteristics of the curve.

\begin{multicols}{2}

In the picture on the right-hand side, $(x_0,y_0)$ are the coordinates of the extremal point, and $r$ and $d$ are the
horizontal and vertical distances between the two points. In the picture, $x_0$ and $y_0$ are positive, and they could
also be negative.\\ The number $r$ is called the \emph{radius} and $d$ the \emph{deviation}.

\begin{center}
\psset{xunit=0.8cm,yunit=0.7cm}
\begin{pspicture}(0,0)(8,6)
\psaxes[labels=none,ticks=none](0,0)(8,6) 
\psdots(4,1)(1,4)
\psplot{0.5}{7.5}{x 4 neg add x 4 neg add mul 3 div 1 add}
\psline[linestyle=dashed](4,0)(4,1)(0,1)
\uput[dr](4,0){$x_0$}
\uput[dl](0,1){$y_0$}
\psline[linestyle=dashed](1,0)(1,4)(0,4)
\psline[linewidth=1.25pt]{|-|}(-0.25,1)(-0.25,4)
\psline[linewidth=1.25pt]{|-|}(1,-0.25)(4,-0.25)
\uput[d](2.5,-0.25){$r$}
\uput[l](-0.25,2.5){$d$}
\end{pspicture}
\end{center}

\end{multicols}

\probpart{-- An example}

In this part, we will try to find the formula of a quadratic function whose extremal point is
$(3,7)$ with a deviation of $2$ for a radius of $4$.

\begin{enumerate}
\item Find the simplest linear function such that the image of $3$ is $0$.
\item Turn the previous function into a quadratic function such that the image of $3$ is $0$.
\item Deduce a quadratic function $g$ such that the image of $3$ is $7$.
\item \begin{enumerate}
	\item Use the radius and the deviation to find another point on the curve. 
	\item The ordinate of the new point should be the image of its abscissa under function $g$. Check that it is so
with the formula found previously.
	\item With only one multiplicative coefficient, turn the formula of function $g$ into the formula of a new
function $f$ that works for the two points.
	\end{enumerate}
\item By using the radius and deviation symmetrically, we get the coordinates of another point that should be on the
curve. Check that the formula works for this point too.
\item Expand the formula you found for $f$ in the previous question.
\item Draw the variations table of the function $f$, and then its sign table.
\end{enumerate}

\probpart{-- The general formula}

In this part, we don't know the coordinates $(x_0,y_0)$, the radius $r$  nor the deviation $d$. We will try to find a
general formula.

\begin{enumerate}
\item Find the simplest linear function such that the image of $x_0$ is $0$.
\item Turn the previous function into a quadratic function such that the image of $x_0$ is $0$.
\item Deduce a quadratic function $g$ such that the image of $x_0$ is $y_0$.
\item \begin{enumerate}
	\item Use the radius and the deviation to find another point on the curve. 
	\item The ordinate of the new point should be the image of its abscissa under function $g$. Check that it is so
with the formula found previously.
	\item Using the multiplicative coefficient $\frac{d}{r^2}$, turn the formula of function $g$ into the formula of
a new function $f$ that works for the two points.
	\end{enumerate}
\item By using the radius and deviation symmetrically, we get the coordinates of another point that should be on the
curve. Check that the formula works for this point too.
\end{enumerate}

\probpart{-- Applying the formula}

Use the formula to find the quadratic functions with the characteristics show below. In each case, give the initial
expression, then the expanded one and the variations table.

\begin{enumerate}
\item $x_0=-2$, $y_0=4$, $r=4$ and $d=-2$.
\item $x_0=7$, $y_0=-2$, $r=5$ and $d=3$.
\item $x_0=4$, $y_0=6$, $r=3$ and $d=-5$.
\end{enumerate}




\end{document}
