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\begin{document}
\part*{MTH122:13 Notes}
These notes have been digitally typeset and reassembled from scratch notes. 
\tableofcontents

\section{2025-09-03}
Notes on domain and range\footnote{Dr. Filiz Dogru was our sub on Friday (Sep 5) because Mrs. Brott was out. She has been a math professor at GV for over 23 years.}
Find domain and range. Domain is x-axis, and range is y-axis.
y-intercept = $f(0)$ or $(0, c)$ 
x-intercept = $\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}=0$

Direction (end behavior) = if a > 0, as $x \longrightarrow \infty$, graph goes up.

Axis of symmetry and $x_{vertex} = \frac{-b}{2a}$ in standard form but simply "h" in vertex form.
$y_{vertex} = f(x)$ where $x = \frac{-b}{2a}$

\begin{equation}
f(x) = 2x^{2}-x-1
\end{equation}
\begin{equation}
f(x) = \frac{1}{2*2} = \frac{1}{4} = 0.25
\end{equation}
\begin{equation}
Vertex = (0.25, -1.125)
\end{equation}
\begin{equation}
\frac{+1\pm\sqrt{-1^{2}-4(2)(-1)}}{2(2)}
\end{equation}
\begin{equation}
\frac{1\pm\sqrt{1+8}}{4}
\end{equation}
\begin{equation}
\frac{1\pm\sqrt{9}}{4} = \frac{1\pm3}{4} = \frac{-2}{4} and \frac{4}{4}\ so\ x=\{-0.5, 1\}
\end{equation}

\section{2025-09-10}
\begin{equation}(x+2)(x+3)=0\end{equation} $\\6x^{2}+x-1=0\\
x^{2}+5x+6x=0$
\hrule
$5x^{2}-7x+2=0$
\begin{enumerate}
\item When $a \neq 1$, multiply first and last term, ($5x^{2}\cdot2=10x$)
\item Find two numbers that multiply to (first * last) and add to the middle term
\item Replace middle term with the found numbers
\item Split the polynomial in half, like $5x^{2}-5x-2x+2$ becomes $(5x^{2}-5x) | (-2x+2)$
\item Factor each group
\end{enumerate}
$\frac{(5x^{2}}{5x}-\frac{5x)}{5x}$
\end{document}