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\begin{center}\huge MATH 1112 - EXAM 3 - FALL 2016\\
SOLUTION
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\begin{flushleft}\vspace{0.5in}
28 October 2016 \\
Instructor: Tom Cuchta
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\textbf{Instructions:}
\begin{itemize}
	\item  Show all work, clearly and in order, if you want to get full
	credit. If you claim something is true \textbf{you must show work backing up your claim}. I reserve the right to take off points if I cannot see how you arrived at your answer (even if your final answer is correct).
	\item Justify your answers algebraically whenever possible to ensure full credit. 	
	\item  Circle or otherwise indicate your final answers.
	\item  Please keep your written answers brief; be clear and to the point.
	\item  Good luck!
\end{itemize}
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\newcommand{\soln}{\underline{\hspace*{5cm}}}

\begin{questions}
		\question[21] Write in the form $a+bi$, where $a$ and $b$ are real numbers.
		\begin{parts}
			\part[3] $\sqrt{-41}$ \\
			\textit{Solution}: $\sqrt{41}i$
			\part[6] $\left(2+3i\right)-\left(1-4i\right)$\\
			\textit{Solution:} Calculate
			$$(2+3i)-(1-4i) = (2-1) + (3-(-4))i = 1+7i.$$
			\part[6] $\left(3-i\right)\left(4+2i\right)$ \\
			\textit{Solution:} Calculate
			$$\begin{array}{ll}
			(3-i)(4+2i) &= 12+6i-4i-2i^2 \\
			&= 12 + 2i+2 \\
			&= 14+2i.
			\end{array}$$
			\part[6] $\dfrac{2+3i}{1-i}$ \\
			\textit{Solution:} Calculate
			$$\begin{array}{ll}
			\dfrac{2+3i}{1-i} &= \left( \dfrac{2+3i}{1-i} \right) \left( \dfrac{1+i}{1+i} \right) \\
			&= \dfrac{(2+3i)(1+i)}{(1-i)(1+i)} \\
			&= \dfrac{2+2i+3i+3i^2}{1+i-i-i^2} \\
			&=\dfrac{2+5i-3}{1-(-1)} \\
			&= \dfrac{-1+5i}{2} \\
			&= -\dfrac{1}{2} + \dfrac{5}{2}i.
			\end{array}$$
		\end{parts}
		\question[15] Simplify.
		\begin{parts}
		\part[5] $i^5$ \\
		\textit{Solution:} Compute
		$$i^5 = i i^4 = i (i^2)^2 = i(-1)^2 = i.$$
		\part[5] $i^{53}$ \\
		\textit{Solution:} Compute
		$$i^{53} = i i^{52} = i (i^2)^{26} = i (-1)^{26} = i.$$
		\part[5] $i^{103}$ \\
		\textit{Solution:} Compute
		$$i^{103} = i i^{102} = i \left(i^{2}\right)^{51} = i(-1)^{51} = -i.$$
		\end{parts}
		\question[22] Solve the following equations in any you wish. 		\begin{parts}
			\part[5] $(x+3)(x-1)=0$ \\
			\textit{Solution:} Using the zero-product property of algebra, we get $x=-3$ and $x=1$. 
			\part[5] $x^2+16=0$ \\
			\textit{Solution:} Subtract $16$ to get 
			$$x^2=-16.$$
			Now take square roots to get
			$$x = \pm \sqrt{-16} = \pm 4i.$$
			\part[6] $x^2+5x+1=0$ \\
			\textit{Solution:} Use the quadratic formula with $a=1$, $b=5$, and $c=1$ to see
			$$x = \dfrac{-5 \pm \sqrt{25-4(1)(1)}}{2}=-\dfrac{5}{2} \pm \dfrac{\sqrt{21}}{2}.$$
			\part[6] $3x^2-x+5=0$ \\
			\textit{Solution:} Use the quadratic formula with $a=3, b=-1,$ and $c=5$ to get
			$$x = \dfrac{-(-1) \pm \sqrt{(-1)^2-4(3)(5)}}{2(3)}= \dfrac{1 \pm \sqrt{-59}}{6} = \dfrac{1}{6} \pm \dfrac{\sqrt{59}}{6} i.$$
		\end{parts}		
\newpage
		\question[21] Analyze the following function:
		 $$f(x)=-3\left(x-4\right)^2-5$$
		\begin{parts}
			\part[4] What is the vertex of $f$? \\
			\textit{Solution:} The vertex is $(4,-5)$. 
			\part[3] What is the axis of symmetry?\\
			\textit{Solution:} The axis of symmetry is $x=4$.
			\part[4] What is the minimum (if it exists)? \\
			\textit{Solution:} There is no minimum.
			\part[4] What is the maximum (if it exists)? \\
			\textit{Solution:} The maximum occurs at $x=4$ and it has value $-5$.
			\part[6] Graph $y=f(x)$ based on the information found in parts (a)--(d).
			\textit{Solution:} \\
			\begin{tikzpicture}
      \draw[->] (-1,0) -- (6,0) node[right] {$x$};
      \draw[->] (0,-8) -- (0,2) node[above] {$y$};
      \draw[scale=1,domain=3:5,smooth,variable=\x,black,very thick] plot ({\x},{-3*(\x-4)*(\x-4)-5});
      \draw (4,-4.2) node {$(4,-5)$};
      \draw[fill=black] (4,-5) circle (0.5ex);
    \end{tikzpicture}
		\end{parts}
\newpage		
		\question[21] Analyze the following function: 
		$$f(x)=2x^2-3x+1$$
		\begin{parts}
			\part[4] What is the vertex of $f$? \\
			\textit{Solution:} Completing the square on $f$ yields
			$$f(x) = 2 \left( x - \dfrac{3}{4} \right)^2 - \dfrac{1}{8}.$$
			From this it is clear that the vertex is $\left( \dfrac{3}{4}, -\dfrac{1}{8}\right)$. 
			\part[3] What is the axis of symmetry?\\
			\textit{Solution:} The axis of symmetry is $x=\dfrac{3}{4}$. 
			\part[4] What is the minimum (if it exists)? \\
			\textit{Solution:} The minimum is $-\dfrac{1}{8}$ and it occurs as $x=\dfrac{3}{4}$. 
			\part[4] What is the maximum (if it exists)? \\
			\textit{Solution:} There is no maximum.
			\part[6] Graph $y=f(x)$ based on the information found in parts (a)--(d). \\
			\begin{tikzpicture}[scale=2]
      \draw[->] (-1,0) -- (3,0) node[right] {$x$};
      \draw[->] (0,-1) -- (0,2) node[above] {$y$};
      \draw[scale=1,domain=-0.3:1.8,smooth,variable=\x,black,very thick] plot ({\x},{2*(\x)*(\x)-3*(\x)+1});
      \draw (0.75,-0.5) node {$\left(\dfrac{3}{4},-\dfrac{1}{8} \right)$};
      \draw[fill=black] (0.75,-0.125) circle (0.25ex);
    \end{tikzpicture}
		\end{parts}

\end{questions}
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