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{\Large {\bf Key 1}}

\begin{enumerate}

\item Easy

\item Find $m$ first which is $m=\frac{-14-4}{3-(-3)}=-3$.  So the equation
of the line is $y-(-14)=-3(x-3)$.  In slope intercept form, this is:
$y=-3x-5$

\item  Since $2 \geq 1$, we use the second formula to find
$f(2)=-(2^3)+2(2)+2=-2$

\item  This is not a function because it does not pass the vertical line
test.

\item  You should be able to graph this.  The zeros are: .532, .653, 2.879.
The relative max is:(0,1).  The relative min is:  (2,-3).  The function is
increasing on(x-values not y-values, not both):
$(-\infty,0)\bigcup(2,+\infty)$.  The function is decreasing on(again, only
x-values):(0,2).  The function is not constant on any interval.

\item  You can't have zero in the denominator \underline{and} you can't have
negatives under the squar root.  $x\neq5$ and $2x-3 \geq 0 \rightarrow
x\geq \frac{2}{3}$.  In interval notation:
$[\frac{2}{3},5)\bigcup(5,+\infty)$

\item  The function must be one to one.

\item  It's a line so it is one to one.  Solve $x=\frac{4-y}{6}$ for $y$ to
get $y=f^{-1}(x)=4-6x$

\item  $\frac{2+i}{2-i}=\frac{2+i}{2-i}\frac{2+i}{2+i} =
\frac{4+4i+i^2}{4-i^2}=\frac{3+4i}{5}=\frac{3}{5}+\frac{4}{5}i$

\item  You start at 1536 ft above the ground ($s_0=1536$).  You drop the
rock (you dont throw it, $v_0=0$).  Now solve for when it hits the ground
($s=0$):  $0=-16t^2+1536$  This gives $t=\pm \sqrt{96}$.  Throw away the
negative answer so $t=\sqrt{96}=4\sqrt{6}\approx 9.80$

\item  Now you start on the ground ($s_0=0$) and you shoot the rock up
($v_0=320$).  You want to find you when the rock is above the building
($s=1536$).  So you solve:  $-16t^2+320 \geq 1536$  Solving gives:
$(t-8)(t-12) \leq 0$  This give your critical numbers of 8 and 12.  You try
numbers into the equation for each of the three regions and find that the
solution is: $8 \leq t \leq 12$.

\item  Here you can read the critical number immediatly: 8 and 5.  You try
numbers into the equation in the three regions and find the solution is:
$\{x:\;x<5$ or $x \geq 8$\}.  In interval notation: $(-\infty,5)\bigcup[8,+\infty)$

\item  There are 4 points so $n=4$.  You must find $\sum x = 11$, $\sum y =
17$, $\sum xy = 58$, $\sum x^2 = 45$.  You put these into the formulas for
$a$ and $b$ to get:
$a=\frac{(4)(58)-(11)(17)}{(4)(45)-(11)^2}=\frac{45}{59} \approx .763$  and
$b=\frac{1}{4}(17-\frac{45}{59}(11))=\frac{127}{59} \approx 2.153$.  The
equation of the line is:  $y=\frac{45}{59}x+\frac{127}{59}$

\end{enumerate}


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