gistfile1.txt

\documentclass{article}
\usepackage{graphicx}
\begin{document}

\noindent\textbf{5:}
	{\center \[ f(x) = \frac{(x^2 - 8x + 7)}{(x - 6)} \]}
	\includegraphics[width=5in,height=3.1in]{question5.pdf}
	
	\noindent\emph{The Asymptotes:}
	
	Since the denominator of \( f(x) \) is 0 at \( x = 6\), there is a vertical asymptote at that point.
	
	Because the degree of the numerator is a degree higher than the denominator, we have a slant asymptote. To find the equation of the slant asymptote, we must take the limit of the function.
	
	\[ \lim_{x\to\infty} f(x) = (x^2 - 8x + 7) \div (x - 6) = x - 2 \]
	
	% This is where I will hand-write the synthetic division.
	
	\newpage
	\noindent\emph{The Points:}
	
	Next, simply pick a few points and plot them on the graph, then connect the points with a curve.
	\begin{center}
			\begin{tabular}{r|r}
				$x$&$y$\\
				\hline
				0&\(\frac{-7}{6}\)\\
				1&0\\
				5&8\\
				7&0\\
				6.5&-5.5\\
			\end{tabular}
	\end{center}

\noindent\textbf{14:}

	\emph{D} is the correct answer because that function has a range of $(\infty , 0)$.

\noindent \textbf{20:}

	This requires the change of base formula. 
	
	\[ \frac{\log_{n} 0.0123}{\log_{n} 4} = -3.1726 \]
	
	Where $n$ is any base.

\noindent \textbf{29:}

	\[\log_2(x+5) - \log_2(x-1) = 2\log_22\]
	
	The first thing to do in this case is to condense both sides of the equation. Recall that:
	
	\[\log_n(x) - \log_n(y) = \log_n(\frac{x}{y}) \]
	
	and that:
	
	\[a \log_n(x) = \log_n(x^a)\]
	
	The rest is straightforward algebra.
	
	\begin{eqnarray*}
		\log_2(x+5) - \log_2(x-1) &=& 2\log_22 \\
		\log_2(\frac{x+5}{x-1}) &=& \log_22^2 \\
		\frac{x+5}{x-1} &=& 4 \\
		x+5 &=& 4(x-1) \\
		x+5 &=& 4x - 4 \\
		9 &=& 3x \\
		3 &=& x\\
	\end{eqnarray*}
		
\newpage

\noindent \textbf{35:}


	\[ f(x) = 29 + 49 log(x+1) \]
	
	\emph{a:} This is simply a matter of plugging in 9 for $x$. 
	
	\[ f(9) = 29 + 49 log(9+1) = 78 \]
	
	\emph{b:} Here we know that $f(x) = 90$. So:
	
	\begin{eqnarray*}
		90 &=& 29 + 49 log( x + 1 )\\
		\frac{61}{49} &=& log( x + 1 )\\
		10^{\frac{61}{49}} &=& x + 1\\
		10^{\frac{61}{49}} - 1 &=& x\\
		16.58 &\approx& x
	\end{eqnarray*}

\noindent \textbf{59:}

	The intercept of an axis occurs when all other axis are equal to 0. therefore, to find the $x$-intercept, the $y$- and $z$- axis must be equal to 0. It therefore becomes extremely simple to solve for this sort of thing.
	
	\begin{center}
		\begin{tabular}{r|r|r}
			Intercept&Formula&Solution\\
			\hline
			 $x$& \(5x = 20\) &\(x = 4\) \\
			 $y$& \(2y = 20\) &\(y = 10\)\\
			 $z$& \(-4z = 20\)&\(z = -5\)\\
		 \end{tabular}
	\end{center}

\noindent \textbf{65:}

Since we know that $x = r cos(\theta)$ and that $y = r sin(\theta)$, this is a simple case of plugging in $-4$ for $r$ and $\frac{-7\pi}{4}$ for $\theta$.

\[(-4cos(\frac{-7\pi}{4}), -4sin(\frac{-7\pi}{4})) = (-2\sqrt{2}, -2\sqrt{2})\]

\noindent \textbf{74:}



Because this series does not have a common difference, it is a geometric function. The standard form for a geometric series that increases without bound is simply:

\[\sum_{n =1}^{\infty} a_0(r^{n-1})\]

 It starts at 8 and continues without bound. Because it doubles each time, we know that $r = 2$. We also know that $a_0 = 8$. Therefore, the summation notation is:

\[ \sum_{n=1}^{\infty} 8(2^{n-1}) \]
\end{document}