\magnification=1200
\baselineskip=12pt
\nopagenumbers

\noindent
MATH 180\hfill FALL 1995

\noindent
10:00 SECTION\hfill FONG

\bigskip\noindent
\centerline{ASSIGNMENT \#7}

\bigskip\noindent
1.10, \#36.
A possible function is $y=-4\sin 2x$.

\bigskip\noindent
1.11, \#1.
The minimal degree is at least one more than the number of ``bends''
in the curve.  The sign of the leading coefficient is given
by the behavior of the curve as $x\rightarrow\pm\infty$.
$$\vbox{
\halign{\quad#\hfil&\quad#\hfil&\quad#\hfil&\quad#\hfil&
   \quad#\hfil&\quad#\hfil\cr
  Graph&i&ii&iii&iv&v\cr
  Degree&3&4&4&5&5\cr
  Leading Sign&$-$&$+$&$-$&$-$&$+$\cr}}$$

\bigskip\noindent
1.11, \#9.
The rational function $y=(x^2+2x+1)/(x^2-4)$
behaves like $x^2/x^2=1$ as $x\rightarrow\pm\infty$,
so the line $y=1$ is a horizontal
asymptote.  Vertical asymptotes arise when the denominator
$x^2-4=0$, {\it i.e.\/}, when $x=\pm2$.  To see whether
$y\rightarrow \infty$ or $y\rightarrow -\infty$,
we consider the sign of $y$ as $x$ gets close to
$-2$ from the left and as $x$ gets close to $-2$ from the
right.  This analysis has to be repeated for the line $x=2$.
The graph has $x$-intercept at $x=-1$ and $y$-intercept at
$y=-1/4$.

\smallskip\noindent
Comment:  A common mistake was drawing a graph
symmetric with respect to the $y$-axis with $x$-intercept
at $x=0$ rather than at $x=-1$.  This may have been suggested by
the graph on the calculator, but pictures at low resolutions
can mislead.  Be careful!

\bigskip\noindent
1.11, \#13.  $(1,1)$ lies on $ax^2+bx+c$, so that $a+b+c=1$.
The axis of symmetry $x=-b/2a$ passes through $(1,1)$, so that
$-b=2a$.  The $y$-intercept is 6, so that $c=6$.
This pins down the function: $y=5x^2-10x+6$.

\bigskip\noindent
1.11, \#17. The cubic $(x+3)x(x-4)$ has roots at
the given points $-3$, 0, 4.  Since $y\rightarrow-\infty$ as
$x\rightarrow\infty$, the leading coefficient of $y$ is negative.
So $y=-(x+3)x(x-4)=-x^3+x^2+12x$ does the trick.  The
graph is increasing on $(-1.5,2)$ and decreasing
on $(-\infty,-1.5)$ and $(2,\infty)$ (the points $-1.5$ and $2$ are
guesses from the picture).

\bigskip\noindent
1.11, \#23.  $(0,0)$ lies on the graph, so the numerator is
a quadratic with a root at $x=0$, as for example $x^2$.
The graph has no vertical asymptotes, so the denominator is a
quadratic with no real roots, as
for example $x^2+1$.  Since
the line $y=2$ is a horizontal asymptote, the ratio of the leading
coefficients of numerator and denominator is equal to 2.
Moreover, the graph lies between the lines $y=0$ and $y=2$.
The function $y=2x^2/(x^2+1)$ satisfies all of the preceding
requirements and indeed its graph has the indicated form.

%\bigskip\noindent
%Comments:

\end
