2.E: Linear and Quadratic Functions (Exercises)
- Page ID
- 4768
2.1: Linear Functions
In Exercises \ref{pointslopegivenlinefirst} - \ref{pointslopegivenlinelast}, find both the point-slope form and the slope-intercept form of the line with the given slope which passes through the given point.
\begin{multicols}{2}
\begin{enumerate}
\item $m = 3, \;\; P(3, -1)$ \label{pointslopegivenlinefirst}
\item $m = -2, \;\; P(-5, 8)$
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\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
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\item $m = -1, \;\; P(-7, -1)$
\item $m = \frac{2}{3}, \;\; P(-2, 1)$
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\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
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\item $m = -\frac{1}{5}, \;\; P(10, 4)$
\item $m = \frac{1}{7}, \;\; P(-1, 4)$
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\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
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\item $m = 0, \;\; P(3, 117)$
\item $m = -\sqrt{2}, \;\; P(0, -3)$
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\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
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\item $m = -5, \;\; P(\sqrt{3}, 2\sqrt{3})$
\item $m = 678, \;\; P(-1, -12)$ \label{pointslopegivenlinelast}
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\end{enumerate}
\end{multicols}
In Exercises \ref{twopointsgivenlinefirst} - \ref{twopointsgivenlinelast}, find the slope-intercept form of the line which passes through the given points.
\begin{multicols}{2}
\begin{enumerate}
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\item $P(0, 0), \; Q(-3, 5)$ \label{twopointsgivenlinefirst}
\item $P(-1, -2), \; Q(3, -2)$
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\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
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\item $P(5, 0), \; Q(0, -8)$
\item $P(3, -5), \; Q(7, 4)$
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\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
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\item $P(-1,5), \; Q(7, 5)$
\item $P(4, -8), \; Q(5, -8)$
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\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
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\item $P\left(\frac{1}{2}, \frac{3}{4} \right), \; Q\left(\frac{5}{2}, -\frac{7}{4} \right)$
\item $P\left(\frac{2}{3}, \frac{7}{2} \right), \; Q\left(-\frac{1}{3}, \frac{3}{2} \right)$
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\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
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\item $P\left(\sqrt{2}, -\sqrt{2} \right), \; Q\left(-\sqrt{2}, \sqrt{2} \right)$
\item $P\left(-\sqrt{3}, -1 \right), \; Q\left(\sqrt{3}, 1 \right)$ \label{twopointsgivenlinelast}
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\end{enumerate}
\end{multicols}
In Exercises \ref{graphlineexerfirst} - \ref{graphlineexerlast}, graph the function. Find the slope, $y$-intercept and $x$-intercept, if any exist.
\begin{multicols}{2}
\begin{enumerate}
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\item $f(x) = 2x - 1$ \label{graphlineexerfirst}
\item $f(x) = 3 - x$
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\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
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\item $f(x) = 3$
\item $f(x) = 0$
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\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
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\item $f(x) = \frac{2}{3} x + \frac{1}{3}$ \vphantom{$\dfrac{1-x}{2}$}
\item $f(x) = \dfrac{1-x}{2}$ \label{graphlineexerlast}
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\end{enumerate}
\end{multicols}
\begin{enumerate}
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\item Find all of the points on the line $y=2x+1$ which are $4$ units from the point $(-1,3)$.
\item Jeff can walk comfortably at $3$ miles per hour. Find a linear function $d$ that represents the total distance Jeff can walk in $t$ hours, assuming he doesn't take any breaks.
\item Carl can stuff $6$ envelopes per \textit{minute}. Find a linear function $E$ that represents the total number of envelopes Carl can stuff after $t$ \textit{hours}, assuming he doesn't take any breaks.
\item A landscaping company charges $\$45$ per cubic yard of mulch plus a delivery charge of $\$20$. Find a linear function which computes the total cost $C$ (in dollars) to deliver $x$ cubic yards of mulch.
\item A plumber charges $\$50$ for a service call plus $\$80$ per hour. If she spends no longer than 8 hours a day at any one site, find a linear function that represents her total daily charges $C$ (in dollars) as a function of time $t$ (in hours) spent at any one given location.
\item A salesperson is paid \$200 per week plus 5\% commission on her weekly sales of $x$ dollars. Find a linear function that represents her total weekly pay, $W$ (in dollars) in terms of $x$. What must her weekly sales be in order for her to earn \$475.00 for the week?
\item An on-demand publisher charges $\$22.50$ to print a 600 page book and $\$15.50$ to print a 400 page book. Find a linear function which models the cost of a book $C$ as a function of the number of pages $p$. Interpret the slope of the linear function and find and interpret $C(0)$.
\item The Topology Taxi Company charges $\$2.50$ for the first fifth of a mile and $\$0.45$ for each additional fifth of a mile. Find a linear function which models the taxi fare $F$ as a function of the number of miles driven, $m$. Interpret the slope of the linear function and find and interpret $F(0)$.
\item Water freezes at $0^{\circ}$ Celsius and $32^{\circ}$ Fahrenheit and it boils at $100^{\circ}$C and $212^{\circ}$F.
\label{celsiustofahr}
\begin{enumerate}
\item Find a linear function $F$ that expresses temperature in the Fahrenheit scale in terms of degrees Celsius. Use this function to convert $20^{\circ}$C into Fahrenheit.
\item Find a linear function $C$ that expresses temperature in the Celsius scale in terms of degrees Fahrenheit. Use this function to convert $110^{\circ}$F into Celsius.
\item Is there a temperature $n$ such that $F(n) = C(n)$?
\end{enumerate}
\item Legend has it that a bull Sasquatch in rut will howl approximately 9 times per hour when it is $40^{\circ}F$ outside and only 5 times per hour if it's $70^{\circ}F$. Assuming that the number of howls per hour, $N$, can be represented by a linear function of temperature Fahrenheit, find the number of howls per hour he'll make when it's only $20^{\circ}F$ outside. What is the applied domain of this function? Why?
\item \label{exerredoportaboy} Economic forces beyond anyone's control have changed the cost function for PortaBoys to $C(x) = 105x + 175$. Rework Example \ref{PortaBoyCost} with this new cost function.
\item In response to the economic forces in Exercise \ref{exerredoportaboy} above, the local retailer sets the selling price of a PortaBoy at \$250. Remarkably, 30 units were sold each week. When the systems went on sale for \$220, 40 units per week were sold. Rework Examples \ref{PortaBoyDemand} and \ref{PortaBoyRevenue} with this new data. What difficulties do you encounter?
\item A local pizza store offers medium two-topping pizzas delivered for $\$6.00$ per pizza plus a $\$1.50$ delivery charge per order. On weekends, the store runs a `game day' special: if six or more medium two-topping pizzas are ordered, they are $\$5.50$ each with no delivery charge. Write a piecewise-defined linear function which calculates the cost $C$ (in dollars) of $p$ medium two-topping pizzas delivered during a weekend.
\item A restaurant offers a buffet which costs $\$15$ per person. For parties of $10$ or more people, a group discount applies, and the cost is $\$12.50$ per person. Write a piecewise-defined linear function which calculates the total bill $T$ of a party of $n$ people who all choose the buffet.
\item A mobile plan charges a base monthly rate of $\$10$ for the first $500$ minutes of air time plus a charge of $15$\textcent \, for each additional minute. Write a piecewise-defined linear function which calculates the monthly cost $C$ (in dollars) for using $m$ minutes of air time.
\textbf{HINT:} You may want to revisit Exercise \ref{piecewisemobile} in Section \ref{FunctionNotation}
\item The local pet shop charges $12$\textcent \, per cricket up to 100 crickets, and $10$\textcent \, per cricket thereafter. Write a piecewise-defined linear function which calculates the price $P$, in dollars, of purchasing $c$ crickets.
\item The cross-section of a swimming pool is below. Write a piecewise-defined linear function which describes the depth of the pool, $D$ (in feet) as a function of:
\begin{enumerate}
\item the distance (in feet) from the edge of the shallow end of the pool, $d$.
\item the distance (in feet) from the edge of the deep end of the pool, $s$.
\item Graph each of the functions in (a) and (b). Discuss with your classmates how to transform one into the other and how they relate to the diagram of the pool.
\end{enumerate}
\begin{center}
\begin{mfpic}[25]{-1}{13}{-1}{4}
\point[3pt]{(0,4), (12,4)}
\arrow \polyline{(0,4), (4,4)}
\arrow \polyline{(12,4), (8,4)}
\arrow \reverse \arrow \polyline{(-0.25,0), (-0.25,3)}
\arrow \reverse \arrow \polyline{(0,-0.35), (5,-0.35)}
\arrow \reverse \arrow \polyline{(0,3.35), (12,3.35)}
\arrow \reverse \arrow \polyline{(9,1.65), (12,1.65)}
\arrow \reverse \arrow \polyline{(12.25,2), (12.25,3)}
\gclear \tlabelrect(2, 4){\,$d$ ft.}
\gclear \tlabelrect(10, 4){\, $s$ ft.}
\gclear \tlabelrect(6, 3.35){37 ft.}
\gclear \tlabelrect(2.5, -0.35){15 ft.}
\gclear \tlabelrect(10.5, 1.65){10 ft.}
\gclear \tlabelrect(-1, 1.5){8 ft.}
\gclear \tlabelrect(13, 2.5){2 ft.}
\penwd{1.5pt}
\polyline{(0,0), (5,0), (9, 2), (12, 2), (12,3), (0,3), (0,0)}
\end{mfpic}
\end{center}
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\end{enumerate}
In Exercises \ref{averagerateexerfirst} - \ref{averagerateexerlast}, compute the average rate of change of the function over the specified interval.
\begin{multicols}{2}
\begin{enumerate}
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\item $f(x) = x^{3}, \; [-1, 2]$ \vphantom{$\dfrac{1}{x}$} \label{averagerateexerfirst}
\item $f(x) = \dfrac{1}{x}, \; [1, 5]$
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\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
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\item $f(x) = \sqrt{x}, \; [0, 16]$
\item $f(x) = x^{2}, \; [-3, 3]$
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\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
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\item $f(x) = \dfrac{x + 4}{x - 3}, \; [5, 7]$
\item $f(x) = 3x^{2} + 2x - 7, \; [-4, 2]$ \vphantom{$\dfrac{4}{x}$} \label{averagerateexerlast}
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\end{enumerate}
\end{multicols}
\pagebreak
In Exercises \ref{differexerfirst} - \ref{differexerlast}, compute the average rate of change of the given function over the interval $[x, x + h]$. Here we assume $[x, x + h]$ is in the domain of the function.
\begin{multicols}{2}
\begin{enumerate}
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\item $f(x) = x^{3}$ \vphantom{$\dfrac{1}{x}$} \label{differexerfirst}
\item $f(x) = \dfrac{1}{x}$
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\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
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\item $f(x) = \dfrac{x + 4}{x - 3}$
\item $f(x) = 3x^{2} + 2x - 7$ \vphantom{$\dfrac{4}{x}$} \label{differexerlast}
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\end{enumerate}
\end{multicols}
\begin{enumerate}
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\item The height of an object dropped from the roof of an eight story building is modeled by: $h(t) = -16t^2 + 64$, $0 \leq t \leq 2$. Here, $h$ is the height of the object off the ground in feet, $t$ seconds after the object is dropped. Find and interpret the average rate of change of $h$ over the interval $[0,2]$.
\item Using data from \href{www.bts.gov/publications/nati...derline{Bureau of Transportation Statistics}}, the average fuel economy $F$ in miles per gallon for passenger cars in the US can be modeled by $F(t) = -0.0076t^2+0.45t + 16$, $0 \leq t \leq 28$, where $t$ is the number of years since $1980$. Find and interpret the average rate of change of $F$ over the interval $[0,28]$.
\item The temperature $T$ in degrees Fahrenheit $t$ hours after 6 AM is given by:
\[ T(t) = -\frac{1}{2} t^2 + 8t+32, \quad 0 \leq t \leq 12\]
\begin{enumerate}
\item Find and interpret $T(4)$, $T(8)$ and $T(12)$.
\item Find and interpret the average rate of change of $T$ over the interval $[4,8]$.
\item Find and interpret the average rate of change of $T$ from $t=8$ to $t=12$.
\item Find and interpret the average rate of temperature change between 10 AM and 6 PM.
\end{enumerate}
\item Suppose $C(x) = x^2-10x+27$ represents the costs, in \textit{hundreds}, to produce $x$ \textit{thousand} pens. Find and interpret the average rate of change as production is increased from making 3000 to 5000 pens.
\item With the help of your classmates find several other ``real-world'' examples of rates of change that are used to describe non-linear phenomena.
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\phantomsection
\label{parallellines}
(Parallel Lines) \index{line ! parallel}Recall from Intermediate Algebra that parallel lines have the same slope. (Please note that two vertical lines are also parallel to one another even though they have an undefined slope.) In Exercises \ref{parallelfirst} - \ref{parallellast}, you are given a line and a point which is not on that line. Find the line parallel to the given line which passes through the given point.
\begin{multicols}{2}
\begin{enumerate}
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\item $y = 3x + 2, \; P(0, 0)$ \label{parallelfirst}
\item $y = -6x + 5, \; P(3, 2)$
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\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
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\item $y = \frac{2}{3} x - 7, \; P(6, 0)$
\item $y = \dfrac{4-x}{3}, \; P(1, -1)$
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\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
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\item $y = 6, \; P(3, -2)$
\item $x=1, \; P(-5,0)$ \label{parallellast}
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\end{enumerate}
\end{multicols}
\phantomsection
\label{perpendicularlines}
(Perpendicular Lines) \index{line ! perpendicular} Recall from Intermediate Algebra that two non-vertical lines are perpendicular if and only if they have negative reciprocal slopes. That is to say, if one line has slope $m_{\mbox{\tiny$1$}}$ and the other has slope $m_{\mbox{\tiny$2$}}$ then $m_{\mbox{\tiny$1$}} \cdot m_{\mbox{\tiny$2$}} = -1$. (You will be guided through a proof of this result in Exercise \ref{perpendicularlineproof}.) Please note that a horizontal line is perpendicular to a vertical line and vice versa, so we assume $m_{\mbox{\tiny$1$}} \neq 0$ and $m_{\mbox{\tiny$2$}} \neq 0$. In Exercises \ref{perpendlinefirst} - \ref{perpendlinelast}, you are given a line and a point which is not on that line. Find the line perpendicular to the given line which passes through the given point.
\begin{multicols}{2}
\begin{enumerate}
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\item $y = \frac{1}{3}x + 2, \; P(0, 0)$ \label{perpendlinefirst}
\item $y = -6x + 5, \; P(3, 2)$
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\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
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\item $y = \frac{2}{3} x - 7, \; P(6, 0)$
\item $y = \dfrac{4-x}{3}, \; P(1, -1)$
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\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
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\item $y = 6, \; P(3, -2)$
\item $x=1, \; P(-5,0)$ \label{perpendlinelast}
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\end{enumerate}
\end{multicols}
\begin{enumerate}
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\item We shall now prove that $y = m_{\mbox{\tiny$1$}}x + b_{\mbox{\tiny$1$}}$ is perpendicular to $y = m_{\mbox{\tiny$2$}}x + b_{\mbox{\tiny$2$}}$ if and only if $m_{\mbox{\tiny$1$}} \cdot m_{\mbox{\tiny$2$}} = -1$. To make our lives easier we shall assume that $m_{\mbox{\tiny$1$}} > 0$ and $m_{\mbox{\tiny$2$}} < 0$. We can also ``move'' the lines so that their point of intersection is the origin without messing things up, so we'll assume $b_{\mbox{\tiny$1$}} = b_{\mbox{\tiny$2$}} = 0.$ (Take a moment with your classmates to discuss why this is okay.) Graphing the lines and plotting the points $O(0, 0)\;$, $P(1, m_{\mbox{\tiny$1$}})\;$ and $Q(1, m_{\mbox{\tiny$2$}})$ gives us the following set up. \label{perpendicularlineproof}
\begin{center}
\begin{mfpic}[18]{-5}{5}{-5}{5}
\point[3pt]{(0, 0), (1.5, 0.75), (1.5, -3)}
\arrow \reverse \arrow \polyline{( -4, -2), (4, 2)}
\arrow \reverse \arrow \polyline{( -2, 4), (2, -4)}
\polyline{(1.5, 0.75), (1.5, -3)}
\tlabel(1.2, 1){\scriptsize $P$}
\tlabel(-.5,-.6){\scriptsize $O$}
\tlabel(1.2,-3.55){\scriptsize $Q$}
\axes
\tlabel[cc](5,-0.5){\scriptsize $x$}
\tlabel[cc](0.5,5){\scriptsize $y$}
\end{mfpic}
\end{center}
The line $y = m_{\mbox{\tiny$1$}}x$ will be perpendicular to the line $y = m_{\mbox{\tiny$2$}}x$ if and only if $\bigtriangleup OPQ$ is a right triangle. Let $d_{\mbox{\tiny$1$}}$ be the distance from $O$ to $P$, let $d_{\mbox{\tiny$2$}}$ be the distance from $O$ to $Q$ and let $d_{\mbox{\tiny$3$}}$ be the distance from $P$ to $Q$. Use the Pythagorean Theorem to show that $\bigtriangleup OPQ$ is a right triangle if and only if $m_{\mbox{\tiny$1$}} \cdot m_{\mbox{\tiny$2$}} = -1$ by showing $d_{\mbox{\tiny$1$}}^{2} + d_{\mbox{\tiny$2$}}^{2} = d_{\mbox{\tiny$3$}}^2$ if and only if $m_{\mbox{\tiny$1$}} \cdot m_{\mbox{\tiny$2$}} = -1$.
\item \label{inversemidpointex2} Show that if $a \neq b$, the line containing the points $(a,b)$ and $(b,a)$ is perpendicular to the line $y = x$. (Coupled with the result from Example \ref{inversemidpointex1} on page \pageref{inversemidpoint}, we have now shown that the line $y = x$ is a \textit{perpendicular} bisector of the line segment connecting $(a,b)$ and $(b,a)$. This means the points $(a,b)$ and $(b,a)$ are symmetric about the line $y = x$. We will revisit this symmetry in section \ref{InverseFunctions}.)
\item \label{identityexercise} The function defined by $I(x) = x$ is called the \index{function ! identity} Identity Function.
\begin{enumerate}
\item Discuss with your classmates why this name makes sense.
\item Show that the point-slope form of a line (Equation \ref{pointslope}) can be obtained from $I$ using a sequence of the transformations defined in Section \ref{Transformations}.
\end{enumerate}
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\end{enumerate}
\subsection{Answers}
\begin{multicols}{2}
\begin{enumerate}
\item $y+1 = 3(x-3)$ \\ $y = 3x-10$
\item $y-8 = -2(x+5)$ \\ $y = -2x-2$
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\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
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\item $y + 1 = -(x+7)$ \\ $y = -x-8$
\item $y - 1 = \frac{2}{3} (x+2)$ \\ $y = \frac{2}{3} x + \frac{7}{3}$
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\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
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\item $y - 4 = -\frac{1}{5} (x-10)$ \\ $y = -\frac{1}{5} x + 6$
\item $y - 4 = \frac{1}{7}(x + 1)$ \\ $y = \frac{1}{7}x + \frac{29}{7}$
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\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
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\item $y - 117 = 0$ \\ $y = 117$
\item $y + 3 = -\sqrt{2}(x - 0)$ \\ $y = -\sqrt{2}x - 3$
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\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
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\item $y - 2\sqrt{3} = -5(x - \sqrt{3})$ \\ $y = -5x + 7\sqrt{3}$
\item $y + 12 = 678(x + 1)$ \\ $y = 678x + 666$
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\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
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\item $y = -\frac{5}{3}x$
\item $y = -2$
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\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
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\item $y = \frac{8}{5}x - 8$
\item $y = \frac{9}{4}x - \frac{47}{4}$
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\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
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\item $y = 5$
\item $y = -8$
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\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
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\item $y = -\frac{5}{4} x + \frac{11}{8}$
\item $y = 2x + \frac{13}{6}$
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\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
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\item $y = -x$
\item $y = \frac{\sqrt{3}}{3} x$
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\end{enumerate}
\end{multicols}
\begin{enumerate}
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\item \begin{multicols}{2} \raggedcolumns
$f(x) =2x-1$
slope: $m = 2$
$y$-intercept: $(0,-1)$
$x$-intercept: $\left(\frac{1}{2}, 0 \right)$
\vfill
\columnbreak
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\tiny
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\normalsize
\arrow \reverse \arrow \function{-1,2, 0.1}{2*x-1}
\end{mfpic}
\end{multicols}
\item \begin{multicols}{2} \raggedcolumns
$f(x) =3-x$
slope: $m = -1$
$y$-intercept: $(0,3)$
$x$-intercept: $(3, 0)$
\vfill
\columnbreak
\begin{mfpic}[15]{-2}{5}{-2}{5}
\point[3pt]{(0,3), (3,0)}
\axes
\tlabel[cc](5,-0.5){\scriptsize $x$}
\tlabel[cc](0.5,5){\scriptsize $y$}
\xmarks{-1,1,2,3,4}
\ymarks{-1,1,2,3,4}
\tlpointsep{4pt}
\tiny
\axislabels {x}{{$-1 \hspace{6pt}$} -1, {$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4}
\axislabels {y}{{$-1$} -1, {$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4}
\normalsize
\arrow \reverse \arrow \function{-1,4, 0.1}{3-x}
\end{mfpic}
\end{multicols}
\item \begin{multicols}{2} \raggedcolumns
$f(x) = 3$
slope: $m =0$
$y$-intercept: $(0,3)$
$x$-intercept: none
\vfill
\columnbreak
\begin{mfpic}[15]{-3}{3}{-1}{5}
\point[3pt]{(0,3)}
\axes
\tlabel[cc](3,-0.5){\scriptsize $x$}
\tlabel[cc](0.5,5){\scriptsize $y$}
\xmarks{-2,-1,1,2}
\ymarks{1,2,3,4}
\tlpointsep{4pt}
\tiny
\axislabels {x}{{$-2 \hspace{6pt}$} -2,{$-1 \hspace{6pt}$} -1, {$1$} 1, {$2$} 2}
\axislabels {y}{{$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4}
\normalsize
\arrow \reverse \arrow \function{-3,3, 0.1}{3}
\end{mfpic}
\end{multicols}
\item \begin{multicols}{2} \raggedcolumns
$f(x) = 0$
slope: $m =0$
$y$-intercept: $(0,0)$
$x$-intercept: $\{ (x,0) \, | \, \text{$x$ is a real number} \}$
\vfill
\columnbreak
\begin{mfpic}[15]{-3}{3}{-2}{2}
\arrow \polyline{(0,-2), (0,2)}
\tlabel[cc](3,-0.5){\scriptsize $x$}
\tlabel[cc](0.5,2){\scriptsize $y$}
\xmarks{-2,-1,1,2}
\ymarks{-1,1}
\tlpointsep{4pt}
\tiny
\axislabels {x}{{$-2 \hspace{6pt}$} -2,{$-1 \hspace{6pt}$} -1, {$1$} 1, {$2$} 2}
\axislabels {y}{{$-1$} -1,{$1$} 1}
\normalsize
\penwd{1.15pt}
\arrow \reverse \arrow \function{-3,3, 0.1}{0}
\end{mfpic}
\end{multicols}
\item \begin{multicols}{2} \raggedcolumns
$f(x) = \frac{2}{3} x + \frac{1}{3}$
slope: $m = \frac{2}{3}$
$y$-intercept: $\left(0, \frac{1}{3}\right)$
$x$-intercept: $\left(-\frac{1}{2}, 0\right)$
\vfill
\columnbreak
\begin{mfpic}[15]{-3}{3}{-2}{3}
\point[3pt]{(0,0.33333), (-0.5,0)}
\axes
\tlabel[cc](3,-0.5){\scriptsize $x$}
\tlabel[cc](0.5,3){\scriptsize $y$}
\xmarks{-2,-1,1,2}
\ymarks{-1,1,2}
\tlpointsep{4pt}
\tiny
\axislabels {x}{{$-2 \hspace{6pt}$} -2, {$1$} 1, {$2$} 2}
\axislabels {y}{{$-1$} -1, {$1$} 1, {$2$} 2}
\normalsize
\arrow \reverse \arrow \function{-3,3, 0.1}{0.66667*x+0.33333}
\end{mfpic}
\end{multicols}
\item \begin{multicols}{2} \raggedcolumns
$f(x) = \dfrac{1-x}{2}$
slope: $m = -\frac{1}{2}$
$y$-intercept: $\left(0, \frac{1}{2}\right)$
$x$-intercept: $\left(1, 0\right)$
\vfill
\columnbreak
\begin{mfpic}[15]{-3}{3}{-2}{3}
\point[3pt]{(0,0.5), (1,0)}
\axes
\tlabel[cc](3,-0.5){\scriptsize $x$}
\tlabel[cc](0.5,3){\scriptsize $y$}
\xmarks{-2,-1,1,2}
\ymarks{-1,1,2}
\tlpointsep{4pt}
\tiny
\axislabels {x}{{$-2 \hspace{6pt}$} -2,{$-1 \hspace{6pt}$} -1, {$1$} 1, {$2$} 2}
\axislabels {y}{{$-1$} -1, {$1$} 1, {$2$} 2}
\normalsize
\arrow \reverse \arrow \function{-3,3, 0.1}{0.5-0.5*x}
\end{mfpic}
\end{multicols}
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $(-1,-1)$ and $\left(\frac{11}{5}, \frac{27}{5}\right)$
\item $d(t) = 3t$, $t \geq 0$.
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $E(t) = 360t$, $t \geq 0$.
\item $C(x) = 45x+20$, $x \geq 0$.
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $C(t) = 80t + 50$, $0 \leq t \leq 8$.
\item $W(x) = 200 + .05x,\, x \geq 0\;\;$ She must make \$5500 in weekly sales.
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $C(p) = 0.035p + 1.5 \;$ The slope $0.035$ means it costs $3.5$\textcent \, per page. $C(0) = 1.5$ means there is a fixed, or start-up, cost of $\$1.50$ to make each book.
\item $F(m) = 2.25m + 2.05 \;$ The slope $2.25$ means it costs an additional $\$2.25$ for each mile beyond the first 0.2 miles. $F(0) = 2.05$, so according to the model, it would cost $\$2.05$ for a trip of $0$ miles. Would this ever really happen? Depends on the driver and the passenger, we suppose.
\pagebreak
\item \begin{multicols}{2}
\begin{enumerate}
\item $F(C) = \frac{9}{5}C + 32$
\item $C(F) = \frac{5}{9}(F - 32) = \frac{5}{9}F - \frac{160}{9}$
\setcounter{HWindent}{\value{enumii}}
\end{enumerate}
\end{multicols}
\begin{enumerate}
\setcounter{enumii}{\value{HWindent}}
\item $F(-40) = -40 = C(-40)$.
\end{enumerate}
\item $N(T) = -\frac{2}{15}T + \frac{43}{3}$
Having a negative number of howls makes no sense and since $N(107.5) = 0$ we can put an upper bound of $107.5^{\circ}F$ on the domain. The lower bound is trickier because there's nothing other than common sense to go on. As it gets colder, he howls more often. At some point it will either be so cold that he freezes to death or he's howling non-stop. So we're going to say that he can withstand temperatures no lower than $-60^{\circ}F$ so that the applied domain is $[-60, 107.5]$.
\addtocounter{enumi}{2}
\item ${\displaystyle C(p) = \left\{ \begin{array}{rcl} 6p + 1.5 & \mbox{ if } & 1 \leq p \leq 5 \\
5.5p & \mbox{ if } & p\geq 6
\end{array} \right. }$
\item ${\displaystyle T(n) = \left\{ \begin{array}{rcl} 15n & \mbox{ if } & 1 \leq n \leq 9 \\
12.5n & \mbox{ if } & n \geq 10 \\
\end{array} \right. }$
\item ${\displaystyle C(m) = \left\{ \begin{array}{rcl} 10 & \mbox{ if } & 0 \leq m \leq 500 \\
10+0.15(m-500) & \mbox{ if } & m > 500
\end{array} \right. }$
\item ${\displaystyle P(c) = \left\{ \begin{array}{rcl} 0.12c & \mbox{ if } & 1 \leq c \leq 100 \\
12 + 0.1(c-100) & \mbox{ if } & c > 100
\end{array} \right. }$
\item
\begin{enumerate}
\item \[{\displaystyle D(d) = \left\{ \begin{array}{rcl} 8 & \mbox{ if } & 0 \leq d \leq 15 \\
-\frac{1}{2} \, d + \frac{31}{2} & \mbox{ if } & 15 \leq d \leq 27 \\
2 & \mbox{ if } & 27 \leq d \leq 37 \\
\end{array} \right. }\]
\item \[{\displaystyle D(s) = \left\{ \begin{array}{rcl} 2 & \mbox{ if } & 0 \leq s \leq 10 \\
\frac{1}{2} \, s -3 & \mbox{ if } & 10 \leq s \leq 22 \\
8 & \mbox{ if } & 22 \leq s \leq 37 \\
\end{array} \right. }\]
\item $~$
\begin{center}
\begin{tabular}{cc}
\begin{mfpic}[10][15]{-1}{13}{-1}{4}
\axes
\polyline{(0,3), (5,3), (9,1), (12,1)}
\point[3pt]{(0,3), (5,3), (9,1), (12,1)}
\xmarks{5,9,12}
\ymarks{0,1,3}
\tlpointsep{5pt}
\axislabels{x}{ {$15$} 5, {$27$} 9, {$37$} 12}
\axislabels{y}{ {$2$} 1, {$8$} 3}
\tcaption{$y = D(d)$}
\end{mfpic}
&
\hspace{.5in}
\begin{mfpic}[10][15]{-1}{13}{-1}{4}
\axes
\polyline{(12,3), (7,3), (3,1), (0,1)}
\point[3pt]{(12,3), (7,3), (3,1), (0,1)}
\xmarks{3,7,12}
\ymarks{0,1,3}
\tlpointsep{5pt}
\axislabels{x}{ {$10$} 3, {$22$} 7, {$37$} 12}
\axislabels{y}{ {$2$} 1, {$8$} 3}
\tcaption{$y = D(s)$}
\end{mfpic} \\
\end{tabular}
\end{center}
\end{enumerate}
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $\dfrac{2^{3} - (-1)^{3}}{2 - (-1)} = 3$
\item $\dfrac{\frac{1}{5} - \frac{1}{1}}{5 - 1} = -\dfrac{1}{5}$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $\dfrac{\sqrt{16} - \sqrt{0}}{16 - 0} = \dfrac{1}{4}$
\item $\dfrac{3^{2} - (-3)^{2}}{3 - (-3)} = 0$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $\dfrac{\frac{7 + 4}{7 - 3} - \frac{5 + 4}{5 - 3}}{7 - 5} = -\dfrac{7}{8}$
\item \scriptsize $\dfrac{(3(2)^{2}+2(2)-7)-(3(-4)^{2}+2(-4)-7)}{2-(-4)}=-4$ \normalsize
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $3x^{2} + 3xh + h^{2}$ \vphantom{$\dfrac{4}{x}$}
\item $\dfrac{-1}{x(x + h)}$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $\dfrac{-7}{(x - 3)(x + h - 3)}$
\item $6x + 3h + 2$ \vphantom{$\dfrac{4}{x}$}
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item The average rate of change is $\frac{h(2) - h(0)}{2-0}=-32$. During the first two seconds after it is dropped, the object has fallen at an average rate of $32$ feet per second. (This is called the \textit{average velocity} of the object.)
\item The average rate of change is $\frac{F(28) - F(0)}{28-0}=0.2372$. During the years from 1980 to 2008, the average fuel economy of passenger cars in the US increased, on average, at a rate of $0.2372$ miles per gallon per year.
\item
\begin{enumerate}
\item $T(4) = 56$, so at 10 AM (4 hours after 6 AM), it is $56^{\circ}$F. $T(8) = 64$, so at 2 PM (8 hours after 6 AM), it is $64^{\circ}$F. $T(12) = 56$, so at 6 PM (12 hours after 6 AM), it is $56^{\circ}$F.
\item The average rate of change is $\frac{T(8)-T(4)}{8-4}=2$. Between 10 AM and 2 PM, the temperature increases, on average, at a rate of $2^{\circ}$F per hour.
\item The average rate of change is $\frac{T(12)-T(8)}{12-8}=-2$. Between 2 PM and 6 PM, the temperature decreases, on average, at a rate of $2^{\circ}$F per hour.
\item The average rate of change is $\frac{T(12)-T(4)}{12-4}=0$. Between 10 AM and 6 PM, the temperature, on average, remains constant.
\end{enumerate}
\item The average rate of change is $\frac{C(5)-C(3)}{5-3}=-2$. As production is increased from 3000 to 5000 pens, the cost decreases at an average rate of $\$200$ per 1000 pens produced (20\textcent \, per pen.)
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\begin{multicols}{3}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\addtocounter{enumi}{1}
\item $y = 3x$
\item $y = -6x + 20$
\item $y = \frac{2}{3} x - 4$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $y = -\frac{1}{3} x - \frac{2}{3}$
\item $y=-2$
\item $x=-5$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $y = -3x$
\item $y = \frac{1}{6}x + \frac{3}{2}$
\item $y = -\frac{3}{2} x +9$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $y = 3x-4$
\item $x=3$
\item $y=0$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
2.2: Absolute Value Functions
newpage
In Exercises \ref{solveabsvalequfirst} - \ref{solveabsvalequlast}, solve the equation.
\begin{multicols}{3}
\begin{enumerate}
\item $|x| = 6$ \label{solveabsvalequfirst}
\item $|3x-1| = 10$
\item $|4-x| = 7$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $4 - |x| = 3$
\item $2|5x+1| - 3 = 0$
\item $|7x-1| + 2 = 0$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $\dfrac{5 - |x|}{2} = 1$
\item $\frac{2}{3} |5-2x| - \frac{1}{2} = 5$ \vphantom{$\dfrac{5 - |x|}{2} = 1$}
\item $|x| = x + 3$ \vphantom{$\dfrac{5 - |x|}{2} = 1$}
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $|2x-1| = x+1$
\item $4 - |x| = 2x+1$
\item $|x-4| = x-5$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $|x| = x^2$
\item $|x| = 12 - x^2$
\item $|x^2 - 1| = 3$ \label{solveabsvalequlast}
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
Prove that if $|f(x)| = |g(x)|$ then either $f(x) = g(x)$ or $f(x) = -g(x)$. Use that result to solve the equations in Exercises \ref{moreabsvalequfirst} - \ref{moreabsvalequlast}.
\begin{multicols}{3}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $|3x - 2| = |2x + 7|$ \label{moreabsvalequfirst}
\item $|3x+1| = |4x|$
\item $|1-2x| = |x+1|$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $|4-x| - |x+2| = 0$
\item $|2-5x| = 5 |x+1|$
\item $3|x-1| = 2|x+1|$ \label{moreabsvalequlast}
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
In Exercises \ref{graphabsvalexerfirst} - \ref{graphabsvalexerlast}, graph the function. Find the zeros of each function and the $x$- and $y$-intercepts of each graph, if any exist. From the graph, determine the domain and range of each function, list the intervals on which the function is increasing, decreasing or constant, and find the relative and absolute extrema, if they exist.
\begin{multicols}{3}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $f(x) = |x + 4|$ \label{graphabsvalexerfirst}
\item $f(x) = |x| + 4$
\item $f(x) = |4x|$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $f(x) = -3|x|$ \vphantom{$\dfrac{1}{3}$}
\item $f(x) = 3|x + 4| - 4$ \vphantom{$\dfrac{1}{3}$}
\item $f(x) = \dfrac{1}{3}|2x - 1|$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $f(x) = \dfrac{|x + 4|}{x + 4}$
\item $f(x) = \dfrac{|2 - x|}{2 - x}$
\item $f(x) = x + |x| - 3$ \vphantom{$\dfrac{|x + 4|}{x + 4}$}
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $f(x) = |x+2| - x$
\item $f(x) = |x+2| - |x|$
\item $f(x) = |x + 4| + |x - 2|$ \label{graphabsvalexerlast}
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item With the help of your classmates, find an absolute value function whose graph is given below.
\begin{center}
\begin{mfpic}[10]{-10}{10}{-1}{5}
\arrow \polyline{(-4,0), (-9,5)}
\arrow \polyline{(4,0), (9, 5)}
\polyline{(-4,0), (0,4), (4,0)}
\point[3pt]{(-4, 0), (0, 4), (4, 0)}
\axes
\tlabel[cc](10,-0.5){\scriptsize $x$}
\tlabel[cc](0.5,5){\scriptsize $y$}
\xmarks{-8,-7,-6,-5,-4,-3,-2,-1,1,2,3,4,5,6,7,8}
\ymarks{1,2,3,4}
\tlpointsep{5pt}
\scriptsize
\axislabels {x}{{$-8 \hspace{7pt}$} -8, {$-7 \hspace{7pt}$} -7, {$-6 \hspace{7pt}$} -6, {$-5 \hspace{7pt}$} -5, {$-4 \hspace{7pt}$} -4, {$-3 \hspace{7pt}$} -3, {$-2 \hspace{7pt}$} -2, {$-1 \hspace{7pt}$} -1, {$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4, {$5$} 5, {$6$} 6, {$7$} 7, {$8$} 8}
\axislabels {y}{{$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4}
\normalsize
\end{mfpic}
\end{center}
\item With help from your classmates, prove the second, third and fifth parts of Theorem \ref{absolutevalueprops}.
\item \label{triangleinequalityreals} Prove \index{Triangle Inequality} \textbf{The Triangle Inequality:} For all real numbers $a$ and $b,\;\;$ $|a+b| \leq |a| + |b|$.
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\subsection{Answers}
\begin{multicols}{3}
\begin{enumerate}
\item $x = -6$ or $x=6$
\item $x = -3$ or $x= \frac{11}{3}$
\item $x = -3$ or $x= 11$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $x = -1$ or $x= 1$
\item $x=-\frac{1}{2}$ or $x= \frac{1}{10}$
\item no solution
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $x=-3$ or $x= 3$
\item $x = -\frac{13}{8}$ or $x= \frac{53}{8}$
\item $x=-\frac{3}{2}$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $x=0$ or $x= 2$
\item $x=1$
\item no solution
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $x=-1$, $x= 0$ or $x= 1$
\item $x=-3$ or $x=3$
\item $x=-2$ or $x=2$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $x = -1$ or $x = 9$
\item $x = -\frac{1}{7}$ or $x = 1$
\item $x = 0$ or $x = 2$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $x=1$
\item $x = -\frac{3}{10}$
\item $x = \frac{1}{5}$ or $x = 5$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item \begin{multicols}{2} \raggedcolumns
$f(x) = |x + 4|$ \\ $f(-4) = 0$ \\ $x$-intercept $(-4, 0)$ \\ $y$-intercept $(0, 4)$ \\ Domain $(-\infty, \infty)$ \\ Range $[0, \infty)$ \\ Decreasing on $(-\infty, -4]$ \\ Increasing on $[-4, \infty)$ \\ Relative and absolute min.~at~$(-4,0)$ \\ No relative or absolute maximum
\begin{mfpic}[18][20]{-9}{2}{-1}{5}
\arrow \polyline{(-4,0), (-8.5,4.5)}
\arrow \polyline{(-4,0), (1,5)}
\point[3pt]{(-4, 0), (0,4)}
\axes
\tlabel[cc](2,-0.5){\scriptsize $x$}
\tlabel[cc](0.5,4.75){\scriptsize $y$}
\xmarks{-8,-7,-6,-5,-4,-3,-2,-1,1}
\ymarks{1,2,3,4}
\tlpointsep{4pt}
\scriptsize
\axislabels {x}{{$-8 \hspace{7pt}$} -8, {$-7 \hspace{7pt}$} -7, {$-6 \hspace{7pt}$} -6, {$-5 \hspace{7pt}$} -5, {$-4 \hspace{7pt}$} -4, {$-3 \hspace{7pt}$} -3, {$-2 \hspace{7pt}$} -2, {$-1 \hspace{7pt}$} -1, {$1$} 1}
\axislabels {y}{{$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4}
\normalsize
\end{mfpic}
\end{multicols}
\item \begin{multicols}{2} \raggedcolumns
$f(x) = |x| + 4$ \\ No zeros \\ No $x$-intercepts \\ $y$-intercept $(0, 4)$ \\ Domain $(-\infty, \infty)$ \\ Range $[4, \infty)$ \\ Decreasing on $(-\infty, 0]$ \\ Increasing on $[0, \infty)$ \\ Relative and absolute minimum at $(0, 4)$ \\ No relative or absolute maximum
\begin{mfpic}[15]{-5}{5}{-1}{9}
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\axes
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\tlabel[cc](0.5,8.75){\scriptsize $y$}
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\axislabels {y}{{$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4, {$5$} 5, {$6$} 6, {$7$} 7, {$8$} 8}
\normalsize
\end{mfpic}
\end{multicols}
\item \begin{multicols}{2} \raggedcolumns
$f(x) = |4x|$ \\ $f(0) = 0$ \\ $x$-intercept $(0, 0)$ \\ $y$-intercept $(0, 0)$ \\ Domain $(-\infty, \infty)$ \\ Range $[0, \infty)$ \\ Decreasing on $(-\infty, 0]$ \\ Increasing on $[0, \infty)$ \\ Relative and absolute minimum at $(0, 0)$ \\ No relative or absolute maximum
\begin{mfpic}[15]{-3}{3}{-1}{9}
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\scriptsize
\axislabels {x}{{$-2 \hspace{7pt}$} -2, {$-1 \hspace{7pt}$} -1, {$1$} 1, {$2$} 2}
\axislabels {y}{{$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4, {$5$} 5, {$6$} 6, {$7$} 7, {$8$} 8}
\normalsize
\end{mfpic}
\end{multicols}
\item \begin{multicols}{2} \raggedcolumns
$f(x) = -3|x|$ \\ $f(0) = 0$ \\ $x$-intercept $(0, 0)$ \\ $y$-intercept $(0, 0)$ \\ Domain $(-\infty, \infty)$ \\ Range $(-\infty, 0]$ \\ Increasing on $(-\infty, 0]$ \\ Decreasing on $[0, \infty)$ \\ Relative and absolute maximum at $(0, 0)$ \\ No relative or absolute minimum
\begin{mfpic}[15]{-3}{3}{-7}{1}
\arrow \polyline{(0, 0), (-2.1,-6.3)}
\arrow \polyline{(0,0), (2.1,-6.3)}
\point[3pt]{(0,0)}
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\tlabel[cc](3,-0.5){\scriptsize $x$}
\tlabel[cc](0.5,.75){\scriptsize $y$}
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\scriptsize
\axislabels {x}{{$-2 \hspace{7pt}$} -2, {$-1 \hspace{7pt}$} -1, {$1$} 1, {$2$} 2}
\axislabels {y}{{$-6$} -6, {$-5$} -5, {$-4$} -4, {$-3$} -3, {$-2$} -2, {$-1$} -1}
\normalsize
\end{mfpic}
\end{multicols}
\item \begin{multicols}{2} \raggedcolumns
$f(x) = 3|x + 4| - 4$ \\ $f\left(-\frac{16}{3}\right) = 0$, $f\left(-\frac{8}{3}\right) = 0$\\ $x$-intercepts $\left(-\frac{16}{3}, 0\right)$, $\left(-\frac{8}{3}, 0\right)$ \\ $y$-intercept $(0, 8)$ \\ Domain $(-\infty, \infty)$ \\ Range $[-4, \infty)$ \\ Decreasing on $(-\infty, -4]$ \\ Increasing on $[-4, \infty)$ \\ Relative and absolute min.~at~$(-4,-4)$ \\ No relative or absolute maximum
\begin{mfpic}[10]{-9}{2}{-5}{10}
\arrow \polyline{(-4,-4), (-8.5,9.5)}
\arrow \polyline{(-4,-4), (0.5,9.5)}
\point[3pt]{(-4, -4), (0,8), (-5.333, 0), (-2.6667, 0)}
\axes
\tlabel[cc](2,-0.5){\scriptsize $x$}
\tlabel[cc](0.5,9.75){\scriptsize $y$}
\xmarks{-8,-7,-6,-5,-4,-3,-2,-1,1}
\ymarks{-4,-3,-2,-1,1,2,3,4,5,6,7,8}
\tlpointsep{4pt}
\tiny
\axislabels {x}{{$-8 \hspace{6pt}$} -8, {$-7 \hspace{6pt}$} -7, {$-6 \hspace{6pt}$} -6, {$-5 \hspace{6pt}$} -5, {$-4 \hspace{6pt}$} -4, {$-3 \hspace{6pt}$} -3, {$-2 \hspace{6pt}$} -2, {$-1 \hspace{6pt}$} -1, {$1$} 1}
\axislabels {y}{{$-4$} -4, {$-3$} -3, {$-2$} -2, {$-1$} -1, {$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4, {$5$} 5, {$6$} 6, {$7$} 7, {$8$} 8}
\normalsize
\end{mfpic}
\end{multicols}
\item \begin{multicols}{2} \raggedcolumns
$f(x) = \frac{1}{3}|2x - 1|$ \\ $f\left(\frac{1}{2}\right) = 0$ \\ $x$-intercepts $\left(\frac{1}{2}, 0\right)$ \\ $y$-intercept $\left(0, \frac{1}{3}\right)$ \\ Domain $(-\infty, \infty)$ \\ Range $[0, \infty)$ \\ Decreasing on $\left(-\infty, \frac{1}{2}\right]$ \\ Increasing on $\left[\frac{1}{2}, \infty\right)$ \\ Relative and absolute min. at $\left(\frac{1}{2},0\right)$ \\ No relative or absolute maximum
\begin{mfpic}[15]{-4}{5}{-1}{3}
\arrow \polyline{(0.5,0), (-4,3)}
\arrow \polyline{(0.5,0), (5,3)}
\point[3pt]{(0,0.3333), (0.5, 0)}
\axes
\tlabel[cc](5,-0.5){\scriptsize $x$}
\tlabel[cc](0.5,2.75){\scriptsize $y$}
\xmarks{-3,-2,-1,1,2,3,4}
\ymarks{1,2}
\tlpointsep{4pt}
\tiny
\axislabels {x}{{$-3 \hspace{6pt}$} -3, {$-2 \hspace{6pt}$} -2, {$-1 \hspace{6pt}$} -1, {$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4}
\axislabels {y}{{$1$} 1, {$2$} 2}
\normalsize
\end{mfpic}
\end{multicols}
\item \begin{multicols}{2} \raggedcolumns
$f(x) = \dfrac{|x + 4|}{x + 4}$ \\ No zeros \\ No $x$-intercept \\ $y$-intercept $(0, 1)$ \\ Domain $(-\infty, -4) \cup (-4, \infty)$ \\ Range $\{-1, 1\}$ \\ Constant on $(-\infty, -4)$ \\ Constant on $(-4, \infty)$ \\ Absolute minimum at every point $(x, -1)$ where $x < -4$ \\ Absolute maximum at every point $(x, 1)$ where $x > -4$ \\ Relative maximum AND minimum at every point on the graph
\begin{mfpic}[15]{-8}{2}{-2}{2}
\arrow \polyline{(-4,1), (2,1)}
\arrow \polyline{(-4,-1), (-8,-1)}
\point[3pt]{(0,1)}
\axes
\tlabel[cc](2,-0.5){\scriptsize $x$}
\tlabel[cc](0.5,1.75){\scriptsize $y$}
\xmarks{-8,-7,-6,-5,-4,-3,-2,-1,1}
\ymarks{-1,1}
\tlpointsep{4pt}
\scriptsize
\axislabels {x}{{$-8 \hspace{7pt}$} -8, {$-7 \hspace{7pt}$} -7, {$-6 \hspace{7pt}$} -6, {$-5 \hspace{7pt}$} -5, {$-4 \hspace{7pt}$} -4, {$-3 \hspace{7pt}$} -3, {$-2 \hspace{7pt}$} -2, {$-1 \hspace{7pt}$} -1, {$1$} 1}
\axislabels {y}{{$-1$} -1, {$1$} 1}
\normalsize
\gclear \circle{(-4,1),0.1}
\circle{(-4,1),0.1}
\gclear \circle{(-4,-1),0.1}
\circle{(-4,-1),0.1}
\end{mfpic}
\end{multicols}
\item \begin{multicols}{2} \raggedcolumns
$f(x) = \dfrac{|2 - x|}{2 - x}$ \\ No zeros \\ No $x$-intercept \\ $y$-intercept $(0, 1)$ \\ Domain $(-\infty, 2) \cup (2, \infty)$ \\ Range $\{-1, 1\}$ \\ Constant on $(-\infty, 2)$ \\ Constant on $(2, \infty)$ \\ Absolute minimum at every point $(x, -1)$ where $x > 2$ \\ Absolute maximum at every point $(x, 1)$ where $x < 2$ \\ Relative maximum AND minimum at every point on the graph
\begin{mfpic}[13]{-4}{6}{-2}{2}
\arrow \polyline{(2,1), (-4,1)}
\arrow \polyline{(2,-1), (6,-1)}
\point[3pt]{(0,1)}
\axes
\tlabel[cc](6,-0.5){\scriptsize $x$}
\tlabel[cc](0.5,1.75){\scriptsize $y$}
\xmarks{-3,-2,-1,1,2,3,4,5}
\ymarks{-1,1}
\tlpointsep{4pt}
\scriptsize
\axislabels {x}{{$-3 \hspace{7pt}$} -3, {$-2 \hspace{7pt}$} -2, {$-1 \hspace{7pt}$} -1, {$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4, {$5$} 5}
\axislabels {y}{{$-1$} -1, {$1$} 1}
\normalsize
\gclear \circle{(2,1),0.1}
\circle{(2,1),0.1}
\gclear \circle{(2,-1),0.1}
\circle{(2,-1),0.1}
\end{mfpic}
\end{multicols}
\item \begin{multicols}{2} \raggedcolumns
Re-write $f(x) = x+|x| - 3$ as \\ ${\displaystyle f(x) = \left\{ \begin{array}{rcl}
-3 & \mbox{ if } & x < 0\\
2x -3 & \mbox{ if } & x \geq 0 \\ \end{array} \right. }$ \\ $f\left(\frac{3}{2}\right) = 0$ \\ $x$-intercept $\left(\frac{3}{2}, 0\right)$ \\ $y$-intercept $(0,-3)$ \\ Domain $(-\infty, \infty)$ \\ Range $[-3, \infty)$ \\ Increasing on $[0,\infty)$ \\ Constant on $(-\infty, 0]$ \\ Absolute minimum at every point $(x,-3)$ where $x \leq 0$ \\ No absolute maximum \\ Relative minimum at every point $(x, -3)$ where $x \leq 0$ \\ Relative maximum at every point $(x, -3)$ where $x < 0$
\begin{mfpic}[15]{-3}{3}{-4}{3}
\arrow \polyline{(0,-3), (-3,-3)}
\arrow \polyline{(0,-3), (3,3)}
\point[3pt]{(1.5,0), (0,-3)}
\axes
\tlabel[cc](3,-0.5){\scriptsize $x$}
\tlabel[cc](0.5,3){\scriptsize $y$}
\xmarks{-2,-1,1,2}
\ymarks{-4,-3,-2,-1,1,2}
\tlpointsep{4pt}
\scriptsize
\axislabels {x}{{$-2 \hspace{7pt}$} -2,{$-1 \hspace{7pt}$} -1, {$1$} 1, {$2$} 2}
\axislabels {y}{{$-4$} -4,{$-3$} -3,{$-2$} -2,{$-1$} -1, {$1$} 1, {$2$} 2}
\normalsize
\end{mfpic}
\end{multicols}
\small
\item \begin{multicols}{2} \raggedcolumns
Re-write $f(x) = |x+2| - x$ as \\ ${\displaystyle f(x) = \left\{ \begin{array}{rcl}
-2x-2 & \mbox{ if } & x < -2\\
2 & \mbox{ if } & x \geq -2 \\ \end{array} \right. }$ \\ No zeros \\ No $x$-intercepts \\ $y$-intercept $(0,2)$ \\ Domain $(-\infty, \infty)$ \\ Range $[2, \infty)$ \\ Decreasing on $(-\infty, -2]$ \\ Constant on $[-2,\infty)$ \\ Absolute minimum at every point $(x,2)$ where $x \geq -2$ \\ No absolute maximum \\ Relative minimum at every point $(x, 2)$ where $x \geq -2$ \\ Relative maximum at every point $(x, 2)$ where $x > -2$
\begin{mfpic}[15]{-4}{3}{-1}{4}
\arrow \polyline{(-2,2), (-3,4)}
\arrow \polyline{(-2,2), (3,2)}
\point[3pt]{(-2,2), (0,2)}
\axes
\tlabel[cc](3,-0.5){\scriptsize $x$}
\tlabel[cc](0.5,4){\scriptsize $y$}
\xmarks{-3,-2,-1,1,2}
\ymarks{1,2,3}
\tlpointsep{4pt}
\scriptsize
\axislabels {x}{{$-3 \hspace{7pt}$} -3,{$-2 \hspace{7pt}$} -2,{$-1 \hspace{7pt}$} -1, {$1$} 1, {$2$} 2}
\axislabels {y}{{$1$} 1, {$2$} 2, {$3$} 3}
\normalsize
\end{mfpic}
\end{multicols}
\item \begin{multicols}{2} \raggedcolumns
Re-write $f(x) = |x+2|-|x|$ as \\ ${\displaystyle f(x) = \left\{ \begin{array}{rcl}
-2 & \mbox{ if } & x < -2\\
2x+2 & \mbox{ if } & -2 \leq x < 0 \\
2 & \mbox{ if } & x \geq 0 \\ \end{array} \right. }$ \\ $f\left(-1\right) = 0$ \\ $x$-intercept $\left(-1, 0\right)$ \\ $y$-intercept $(0,2)$ \\ Domain $(-\infty, \infty)$ \\ Range $[-2,2]$ \\ Increasing on $[-2,0]$ \\ Constant on $(-\infty, -2]$ \\ Constant on $[0,\infty)$ \\ Absolute minimum at every point $(x,-2)$ where $x \leq -2$ \\ Absolute maximum at every point $(x,2)$ where $x \geq 0$ \\ Relative minimum at every point $(x, -2)$ where $x \leq -2$ and at every point $(x,2)$ where $x>0$ \\ Relative maximum at every point $(x, -2)$ where $x < -2$ and at every point $(x,2)$ where $x \geq 0$
\begin{mfpic}[15]{-4}{3}{-3}{3}
\arrow \polyline{(-2,-2), (-4,-2)}
\arrow \polyline{(0,2), (3,2)}
\polyline{(-2,-2), (0,2)}
\point[3pt]{(-2,-2), (-1,0), (0,2)}
\axes
\tlabel[cc](3,-0.5){\scriptsize $x$}
\tlabel[cc](0.5,3){\scriptsize $y$}
\xmarks{-3,-2,-1,1,2}
\ymarks{-2,-1,1,2}
\tlpointsep{4pt}
\scriptsize
\axislabels {x}{{$-3 \hspace{7pt}$} -3,{$-2 \hspace{7pt}$} -2,{$-1 \hspace{7pt}$} -1, {$1$} 1, {$2$} 2}
\axislabels {y}{{$-2$} -2,{$-1$} -1, {$1$} 1, {$2$} 2}
\normalsize
\end{mfpic}
\end{multicols}
%
\item \begin{multicols}{2} \raggedcolumns
Re-write $f(x) = |x + 4| + |x - 2|$ as \\ ${\displaystyle f(x) = \left\{ \begin{array}{rcl}
-2x - 2 & \mbox{ if } & x < -4\\
6 & \mbox{ if } & -4 \leq x < 2\\
2x + 2 & \mbox{ if } & x \geq 2 \end{array} \right. }$ \\ No zeros \\ No $x$-intercept \\ $y$-intercept $(0, 6)$ \\ Domain $(-\infty, \infty)$ \\ Range $[6, \infty)$ \\ Decreasing on $(-\infty, -4]$ \\ Constant on $[-4, 2]$ \\ Increasing on $[2, \infty)$ \\ Absolute minimum at every point $(x, 6)$ where $-4 \leq x \leq 2$ \\ No absolute maximum \\ Relative minimum at every point $(x, 6)$ where $-4 \leq x \leq 2$ \\ Relative maximum at every point $(x, 6)$ where $-4 < x < 2$
\begin{mfpic}[15]{-6}{4}{-1}{9}
\arrow \polyline{(-4,6), (-5.5,9)}
\arrow \polyline{(2,6), (3.5,9)}
\polyline{(2,6), (-4,6)}
\point[3pt]{(2,6), (-4,6), (0,6)}
\axes
\tlabel[cc](4,-0.5){\scriptsize $x$}
\tlabel[cc](0.5,8.75){\scriptsize $y$}
\xmarks{-5,-4,-3,-2,-1,1,2,3}
\ymarks{1,2,3,4,5,6,7,8}
\tlpointsep{4pt}
\scriptsize
\axislabels {x}{{$-5 \hspace{7pt}$} -5, {$-4 \hspace{7pt}$} -4, {$-3 \hspace{7pt}$} -3, {$-2 \hspace{7pt}$} -2, {$-1 \hspace{7pt}$} -1, {$1$} 1, {$2$} 2, {$3$} 3}
\axislabels {y}{{$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4, {$5$} 5, {$6$} 6, {$7$} 7, {$8$} 8}
\normalsize
\end{mfpic}
\end{multicols}
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\normalsize
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\addtocounter{enumi}{1}
\item $f(x) = ||x| - 4|$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
2.3: Quadratic Functions
In Exercises \ref{graphquadfuncfirst} - \ref{graphquadfunclast}, graph the quadratic function. Find the $x$- and $y$-intercepts of each graph, if any exist. If it is given in general form, convert it into standard form; if it is given in standard form, convert it into general form. Find the domain and range of the function and list the intervals on which the function is increasing or decreasing. Identify the vertex and the axis of symmetry and determine whether the vertex yields a relative and absolute maximum or minimum.
\begin{multicols}{3}
\begin{enumerate}
\item $f(x) = x^{2} + 2$ \label{graphquadfuncfirst}
\item $f(x) = -(x + 2)^{2}$
\item $f(x) = x^{2} - 2x - 8$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $f(x) = -2(x + 1)^{2} + 4$
\item $f(x) = 2x^2 - 4x - 1$
\item $f(x) = -3x^{2} + 4x - 7$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $f(x) = x^2 + x + 1$
\item $f(x) = -3x^2+5x+4$
\item \hspace{-0.15in} \footnote{We have already seen the graph of this function. It was used as an example in Section \ref{GraphsofFunctions} to show how the graphing calculator can be misleading.}
$f(x) = x^{2} - \dfrac{1}{100} x - 1$ \label{graphquadfunclast}
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
In Exercises \ref{maxprofitfirst} - \ref{maxprofitlast}, the cost and price-demand functions are given for different scenarios. For each scenario,
\begin{itemize}
\item Find the profit function $P(x)$.
\item Find the number of items which need to be sold in order to maximize profit.
\item Find the maximum profit.
\item Find the price to charge per item in order to maximize profit.
\item Find and interpret break-even points.
\end{itemize}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item The cost, in dollars, to produce $x$ ``I'd rather be a Sasquatch'' T-Shirts is $C(x) = 2x+26$, $x \geq 0$ and the price-demand function, in dollars per shirt, is $p(x) = 30 - 2x$, $0 \leq x \leq 15$. \label{maxprofitfirst}
\item The cost, in dollars, to produce $x$ bottles of $100 \%$ All-Natural Certified Free-Trade Organic Sasquatch Tonic is $C(x) = 10x+100$, $x \geq 0$ and the price-demand function, in dollars per bottle, is $p(x) = 35 - x$, $0 \leq x \leq 35$.
\item The cost, in cents, to produce $x$ cups of Mountain Thunder Lemonade at Junior's Lemonade Stand is $C(x) = 18x + 240$, $x \geq 0$ and the price-demand function, in cents per cup, is $p(x) = 90-3x$, $0 \leq x \leq 30$.
\item The daily cost, in dollars, to produce $x$ Sasquatch Berry Pies is $C(x) = 3x + 36$, $x \geq 0$ and the price-demand function, in dollars per pie, is $p(x) = 12-0.5x$, $0 \leq x \leq 24$.
\item The monthly cost, in \emph{hundreds} of dollars, to produce $x$ custom built electric scooters is $C(x) = 20x + 1000$, $x \geq 0$ and the price-demand function, in \emph{hundreds} of dollars per scooter, is $p(x) = 140-2x$, $0 \leq x \leq 70$. \label{maxprofitlast}
%\setcounter{HW}{\value{enumi}}
%\end{enumerate}
%\begin{enumerate}
%\setcounter{enumi}{\value{HW}}
\item The International Silver Strings Submarine Band holds a bake sale each year to fund their trip to the National Sasquatch Convention. It has been determined that the cost in dollars of baking $x$ cookies is $C(x) = 0.1x + 25$ and that the demand function for their cookies is $p = 10 - .01x.$ How many cookies should they bake in order to maximize their profit?
\item Using data from \href{www.bts.gov/publications/nati...derline{Bureau of Transportation Statistics}}, the average fuel economy $F$ in miles per gallon for passenger cars in the US can be modeled by $F(t) = -0.0076t^2+0.45t + 16$, $0 \leq t \leq 28$, where $t$ is the number of years since $1980$. Find and interpret the coordinates of the vertex of the graph of $y = F(t)$.
\item The temperature $T$, in degrees Fahrenheit, $t$ hours after 6 AM is given by: \[ T(t) = -\frac{1}{2} t^2 + 8t+32, \quad 0 \leq t \leq 12\]
What is the warmest temperature of the day? When does this happen?
\item Suppose $C(x) = x^2-10x+27$ represents the costs, in \textit{hundreds}, to produce $x$ \textit{thousand} pens. How many pens should be produced to minimize the cost? What is this minimum cost?
\item \label{fixedperimetermaxareagarden} Skippy wishes to plant a vegetable garden along one side of his house. In his garage, he found 32 linear feet of fencing. Since one side of the garden will border the house, Skippy doesn't need fencing along that side. What are the dimensions of the garden which will maximize the area of the garden? What is the maximum area of the garden?
\item In the situation of Example \ref{donniealpaca}, Donnie has a nightmare that one of his alpaca herd fell into the river and drowned. To avoid this, he wants to move his rectangular pasture \textit{away} from the river. This means that all four sides of the pasture require fencing. If the total amount of fencing available is still 200 linear feet, what dimensions maximize the area of the pasture now? What is the maximum area? Assuming an average alpaca requires 25 square feet of pasture, how many alpaca can he raise now?
\item What is the largest rectangular area one can enclose with 14 inches of string?
\item The height of an object dropped from the roof of an eight story building is modeled by $h(t) = -16t^2 + 64$, $0 \leq t \leq 2$. Here, $h$ is the height of the object off the ground, in feet, $t$ seconds after the object is dropped. How long before the object hits the ground?
\item The height $h$ in feet of a model rocket above the ground $t$ seconds after lift-off is given by $h(t) = -5t^2+100t$, for $0 \leq t \leq 20$. When does the rocket reach its maximum height above the ground? What is its maximum height?
\item Carl's friend Jason participates in the Highland Games. In one event, the hammer throw, the height $h$ in feet of the hammer above the ground $t$ seconds after Jason lets it go is modeled by $h(t) = -16t^2 + 22.08t + 6$. What is the hammer's maximum height? What is the hammer's total time in the air? Round your answers to two decimal places.
\item Assuming no air resistance or forces other than the Earth's gravity, the height above the ground at time $t$ of a falling object is given by $s(t) = -4.9t^{2} + v_{\mbox{\tiny $0$}}t + s_{\mbox{\tiny $0$}}$ where $s$ is in meters, $t$ is in seconds, $v_{\mbox{\tiny $0$}}$ is the object's initial velocity in meters per second and $s_{\mbox{\tiny $0$}}$ is its initial position in meters.
\label{whatgoesup}
\begin{enumerate}
\item What is the applied domain of this function?
\item Discuss with your classmates what each of $v_{\mbox{\tiny $0$}} > 0, \; v_{\mbox{\tiny $0$}} = 0$ and $v_{\mbox{\tiny $0$}} < 0$ would mean.
\item Come up with a scenario in which $s_{\mbox{\tiny $0$}} < 0$.
\item Let's say a slingshot is used to shoot a marble straight up from the ground $(s_{\mbox{\tiny $0$}} = 0)$ with an initial velocity of 15 meters per second. What is the marble's maximum height above the ground? At what time will it hit the ground?
\item Now shoot the marble from the top of a tower which is 25 meters tall. When does it hit the ground?
\item What would the height function be if instead of shooting the marble up off of the tower, you were to shoot it straight DOWN from the top of the tower?
\end{enumerate}
\item \label{parabolicbridgecable} The two towers of a suspension bridge are 400 feet apart. The parabolic cable\footnote{The weight of the bridge deck forces the bridge cable into a parabola and a free hanging cable such as a power line does not form a parabola. We shall see in Exercise \ref{catenary} in Section \ref{ExpLogApplications} what shape a free hanging cable makes.} attached to the tops of the towers is 10 feet above the point on the bridge deck that is midway between the towers. If the towers are 100 feet tall, find the height of the cable directly above a point of the bridge deck that is 50 feet to the right of the left-hand tower.
\item Graph $f(x) = |1 - x^{2}|$
\item Find all of the points on the line $y=1-x$ which are $2$ units from $(1,-1)$.
\item Let $L$ be the line $y = 2x+1$. Find a function $D(x)$ which measures the distance \textit{squared} from a point on $L$ to $(0,0)$. Use this to find the point on $L$ closest to $(0,0)$.
\item With the help of your classmates, show that if a quadratic function $f(x) = ax^{2} + bx + c$ has two real zeros then the $x$-coordinate of the vertex is the midpoint of the zeros.
\setcounter{HW}{\value{enumi}}
\end{enumerate}
In Exercises \ref{solvequadvarifirst} - \ref{solvequadvarilast}, solve the quadratic equation for the indicated variable.
\begin{multicols}{3}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $x^{2} - 10y^{2} = 0$ for $x$ \label{solvequadvarifirst}
\item $y^{2} - 4y = x^{2} - 4$ for $x$
\item $x^{2} - mx = 1$ for $x$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $y^{2} - 3y = 4x$ for $y$
\item $y^{2} - 4y = x^{2} - 4$ for $y$
\item $-gt^{2} + v_{\mbox{\tiny $0$}}t + s_{\mbox{\tiny $0$}} = 0$ for $t$ (Assume $g \neq 0$.) \label{solvequadvarilast}
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\subsection{Answers}
\begin{enumerate}
\item \begin{multicols}{2} \raggedcolumns
$f(x) = x^{2} + 2$ (this is both forms!) \\
No $x$-intercepts \\
$y$-intercept $(0, 2)$\\
Domain: $(-\infty, \infty)$ \\
Range: $[2, \infty)$ \\
Decreasing on $(-\infty, 0]$ \\
Increasing on $[0, \infty)$ \\
Vertex $(0, 2)$ is a minimum \\
Axis of symmetry $x = 0$ \\
\begin{mfpic}[15][10]{-3}{3}{-1}{11}
\point[3pt]{(0,2)}
\arrow \reverse \arrow \function{-3,3,0.1}{x**2 + 2}
\axes
\tlabel[cc](3,-0.5){\scriptsize $x$}
\tlabel[cc](0.5,11){\scriptsize $y$}
\xmarks{-2,-1,1,2}
\ymarks{1 step 1 until 10}
\tlpointsep{4pt}
\tiny
\axislabels {x}{{$-2 \hspace{6pt}$} -2, {$-1 \hspace{6pt}$} -1, {$1$} 1, {$2$} 2}
\axislabels {y}{{$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4, {$5$} 5, {$6$} 6, {$7$} 7, {$8$} 8, {$9$} 9, {$10$} 10}
\normalsize
\end{mfpic}
\end{multicols}
\item \begin{multicols}{2} \raggedcolumns
$f(x) = -(x + 2)^{2} = -x^2-4x-4$\\
$x$-intercept $(-2, 0)$ \\
$y$-intercept $(0, -4)$\\
Domain: $(-\infty, \infty)$ \\
Range: $(-\infty, 0]$ \\
Increasing on $(-\infty, -2]$ \\
Decreasing on $[-2, \infty)$ \\
Vertex $(-2, 0)$ is a maximum \\
Axis of symmetry $x = -2$ \\
\begin{mfpic}[15][10]{-5}{1}{-9}{1}
\point[3pt]{(-2,0), (0,-4)}
\arrow \reverse \arrow \function{-5,1,0.1}{-((x + 2)**2)}
\axes
\tlabel[cc](1,-0.5){\scriptsize $x$}
\tlabel[cc](0.5,1){\scriptsize $y$}
\xmarks{-4,-3,-2,-1}
\ymarks{-8 step 1 until -1}
\tlpointsep{4pt}
\tiny
\axislabels {x}{{$-4 \hspace{6pt}$} -4, {$-3 \hspace{6pt}$} -3, {$-2 \hspace{6pt}$} -2, {$-1 \hspace{6pt}$} -1}
\axislabels {y}{{$-8$} -8, {$-7$} -7, {$-6$} -6, {$-5$} -5, {$-4$} -4, {$-3$} -3, {$-2$} -2, {$-1$} -1}
\normalsize
\end{mfpic}
\end{multicols}
\item \begin{multicols}{2} \raggedcolumns
$f(x) = x^{2} - 2x - 8 = (x - 1)^{2} - 9$\\
$x$-intercepts $(-2, 0)$ and $(4, 0)$\\
$y$-intercept $(0, -8)$\\
Domain: $(-\infty, \infty)$ \\
Range: $[-9, \infty)$ \\
Decreasing on $(-\infty, 1]$ \\
Increasing on $[1, \infty)$ \\
Vertex $(1, -9)$ is a minimum \\
Axis of symmetry $x = 1$ \\
\begin{mfpic}[15][10]{-3}{5}{-10}{3}
\point[3pt]{(-2,0),(0,-8),(1,-9),(4,0)}
\arrow \reverse \arrow \function{-2.4,4.4,0.1}{x**2 - 2*x - 8}
\axes
\tlabel[cc](5,-0.5){\scriptsize $x$}
\tlabel[cc](0.5,3){\scriptsize $y$}
\xmarks{-2 step 1 until 4}
\ymarks{-9 step 1 until 2}
\tlpointsep{4pt}
\tiny
\axislabels {x}{{$-2 \hspace{6pt}$} -2, {$-1 \hspace{6pt}$} -1, {$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4}
\axislabels {y}{{$-9$} -9, {$-8$} -8, {$-7$} -7, {$-6$} -6, {$-5$} -5, {$-4$} -4, {$-3$} -3, {$-2$} -2, {$-1$} -1, {$1$} 1, {$2$} 2}
\normalsize
\end{mfpic}
\end{multicols}
\item \begin{multicols}{2} \raggedcolumns
$f(x) = -2(x + 1)^{2} + 4 = -2x^2-4x+2$\\
$x$-intercepts {\small $(-1 - \sqrt{2}, 0)$ and $(-1 + \sqrt{2}, 0)$}\\
$y$-intercept $(0, 2)$\\
Domain: $(-\infty, \infty)$ \\
Range: $(-\infty, 4]$ \\
Increasing on $(-\infty, -1]$ \\
Decreasing on $[-1, \infty)$ \\
Vertex $(-1, 4)$ is a maximum \\
Axis of symmetry $x = -1$ \\
\begin{mfpic}[20][10]{-3.5}{2}{-5}{5}
\point[3pt]{(-2.4142,0),(0,2),(-1,4),(.4142,0)}
\arrow \reverse \arrow \function{-3.1,1.1,0.1}{4- 2*((x + 1)**2)}
\axes
\tlabel[cc](2,-0.5){\scriptsize $x$}
\tlabel[cc](0.5,5){\scriptsize $y$}
\xmarks{-3 step 1 until 1}
\ymarks{-4 step 1 until 4}
\tlpointsep{4pt}
\tiny
\axislabels {x}{{$-3 \hspace{6pt}$} -3, {$-2 \hspace{6pt}$} -2, {$-1 \hspace{6pt}$} -1, {$1$} 1}
\axislabels {y}{{$-4$} -4, {$-3$} -3, {$-2$} -2, {$-1$} -1, {$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4}
\normalsize
\end{mfpic}
\end{multicols}
\item \begin{multicols}{2} \raggedcolumns
$f(x) = 2x^2-4x-1 = 2(x-1)^2-3$\\
$x$-intercepts {\small $\left(\frac{2-\sqrt{6}}{2}, 0\right)$ and $\left(\frac{2+\sqrt{6}}{2}, 0\right)$}\\
$y$-intercept $(0, -1)$\\
Domain: $(-\infty, \infty)$ \\
Range: $[-3, \infty)$ \\
Increasing on $[1,\infty)$ \\
Decreasing on $(-\infty,1]$ \\
Vertex $(1, -3)$ is a minimum \\
Axis of symmetry $x = 1$ \\
\begin{mfpic}[15]{-2}{4}{-4}{5}
\point[3pt]{(-0.2247,0),(0,-1),(1,-3),(2.2247,0)}
\arrow \reverse \arrow \function{-0.8,2.8,0.1}{2*(x**2)-4*x-1}
\axes
\tlabel[cc](4,-0.5){\scriptsize $x$}
\tlabel[cc](0.5,5){\scriptsize $y$}
\xmarks{-1 step 1 until 3}
\ymarks{-3 step 1 until 4}
\tlpointsep{4pt}
\tiny
\axislabels {x}{{$-1 \hspace{6pt}$} -1, {$1$} 1, {$2$} 2, {$3$} 3}
\axislabels {y}{{$-3$} -3, {$-2$} -2, {$-1$} -1, {$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4}
\normalsize
\end{mfpic}
\end{multicols}
\item \begin{multicols}{2} \raggedcolumns
$f(x) = -3x^{2} + 4x - 7 = -3\left(x - \frac{2}{3} \right)^{2} - \frac{17}{3}$\\
No $x$-intercepts \\
$y$-intercept $(0, -7)$\\
Domain: $(-\infty, \infty)$ \\
Range: $\left(-\infty, -\frac{17}{3}\right]$ \\
Increasing on $\left(-\infty, \frac{2}{3}\right]$ \\
Decreasing on $\left[\frac{2}{3}, \infty\right)$ \\
Vertex $\left(\frac{2}{3}, -\frac{17}{3}\right)$ is a maximum \\
Axis of symmetry $x = \frac{2}{3}$ \\
\begin{mfpic}[20][10]{-1}{3}{-15}{1}
\point[3pt]{(0,-7),(.66667,-5.66667)}
\arrow \reverse \arrow \function{-1,2.33,0.1}{-3*(x**2) + 4*x - 7}
\axes
\tlabel[cc](3,-0.5){\scriptsize $x$}
\tlabel[cc](0.25,1){\scriptsize $y$}
\xmarks{1,2}
\ymarks{-14 step 1 until -1}
\tlpointsep{4pt}
\tiny
\axislabels {x}{{$1$} 1, {$2$} 2}
\axislabels {y}{{$-14$} -14, {$-13$} -13, {$-12$} -12, {$-11$} -11, {$-10$} -10, {$-9$} -9, {$-8$} -8, {$-7$} -7, {$-6$} -6, {$-5$} -5, {$-4$} -4, {$-3$} -3, {$-2$} -2, {$-1$} -1}
\normalsize
\end{mfpic}
\end{multicols}
\item \begin{multicols}{2} \raggedcolumns
$f(x) = x^2+x+1 = \left(x + \frac{1}{2}\right)^{2} + \frac{3}{4}$\\
No $x$-intercepts \\
$y$-intercept $(0, 1)$\\
Domain: $(-\infty, \infty)$ \\
Range: $\left[ \frac{3}{4}, \infty\right)$ \\
Increasing on $\left[-\frac{1}{2}, \infty\right)$ \\
Decreasing on $\left(-\infty, -\frac{1}{2}\right]$ \\
Vertex $\left(-\frac{1}{2}, \frac{3}{4}\right)$ is a minimum \\
Axis of symmetry $x = -\frac{1}{2}$ \\
\begin{mfpic}[18]{-3}{2}{-1}{5}
\point[3pt]{(0,1),(-0.5,0.75)}
\arrow \reverse \arrow \function{-2.5,1.5,0.1}{(x**2)+x+1}
\axes
\tlabel[cc](2,-0.5){\scriptsize $x$}
\tlabel[cc](0.5,5){\scriptsize $y$}
\xmarks{-2,-1,1}
\ymarks{1,2,3,4}
\tlpointsep{4pt}
\tiny
\axislabels {x}{{$-2 \hspace{6pt}$} -2,{$-1 \hspace{6pt}$} -1,{$1$} 1}
\axislabels {y}{{$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4}
\normalsize
\end{mfpic}
\end{multicols}
\pagebreak
\item \begin{multicols}{2} \raggedcolumns
$f(x) = -3x^2+5x+4 = -3\left(x-\frac{5}{6}\right)^2 + \frac{73}{12}$\\
$x$-intercepts {\small $\left(\frac{5 - \sqrt{73}}{6}, 0\right)$ and $\left(\frac{5+\sqrt{73}}{6}, 0\right)$}\\
$y$-intercept $(0, 4)$\\
Domain: $(-\infty, \infty)$ \\
Range: $\left(-\infty, \frac{73}{12} \right]$ \\
Increasing on $\left(-\infty, \frac{5}{6}\right]$ \\
Decreasing on $\left[ \frac{5}{6}, \infty\right)$ \\
Vertex $\left(\frac{5}{6}, \frac{73}{12} \right)$ is a maximum \\
Axis of symmetry $x = \frac{5}{6}$ \\
\begin{mfpic}[15]{-2}{4}{-4}{7}
\point[3pt]{(-0.5907,0),(0,4),(0.8333,6.0833),(2.2573,0)}
\arrow \reverse \arrow \function{-1,2.62,0.1}{0-3*(x**2)+5*x+4}
\axes
\tlabel[cc](4,-0.5){\scriptsize $x$}
\tlabel[cc](0.5,7){\scriptsize $y$}
\xmarks{-1 step 1 until 3}
\ymarks{-3 step 1 until 6}
\tlpointsep{4pt}
\tiny
\axislabels {x}{{$-1 \hspace{6pt}$} -1, {$1$} 1, {$2$} 2, {$3$} 3}
\axislabels {y}{{$-3$} -3, {$-2$} -2, {$-1$} -1, {$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4, {$5$} 5, {$6$} 6}
\normalsize
\end{mfpic}
\end{multicols}
\item \begin{multicols}{2} \raggedcolumns
$f(x) = x^{2} - \frac{1}{100} x - 1 = \left(x - \frac{1}{200}\right)^{2} - \frac{40001}{40000}$\\
$x$-intercepts $\left(\frac{1 + \sqrt{40001}}{200}\right)$ and $\left(\frac{1 - \sqrt{40001}}{200}\right)$\\
$y$-intercept $(0, -1)$\\
Domain: $(-\infty, \infty)$ \\
Range: $\left[-\frac{40001}{40000}, \infty \right)$ \\
Decreasing on $\left(-\infty, \frac{1}{200}\right]$ \\
Increasing on $\left[\frac{1}{200}, \infty \right)$ \\
Vertex $\left(\frac{1}{200}, -\frac{40001}{40000}\right)$ is a minimum\footnote{You'll need to use your calculator to zoom in far enough to see that the vertex is not the $y$-intercept.} \\
Axis of symmetry $x = \frac{1}{200}$ \\
\begin{mfpic}[15][10]{-3}{3}{-2}{9}
\point[3pt]{(0,-1), (0.005, -1.000025)}
\arrow \reverse \arrow \function{-3,3,0.1}{x**2 - (x/100) - 1}
\axes
\tlabel[cc](3,-0.5){\scriptsize $x$}
\tlabel[cc](0.5,9){\scriptsize $y$}
\xmarks{-2,-1,1,2}
\ymarks{1 step 1 until 8}
\tlpointsep{4pt}
\tiny
\axislabels {x}{{$-2 \hspace{6pt}$} -2, {$-1 \hspace{6pt}$} -1, {$1$} 1, {$2$} 2}
\axislabels {y}{{$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4, {$5$} 5, {$6$} 6, {$7$} 7, {$8$} 8}
\normalsize
\end{mfpic}
\end{multicols}
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item \begin{itemize}
\item $P(x) = -2x^2+28x-26$, for $0 \leq x \leq 15$.
\item $7$ T-shirts should be made and sold to maximize profit.
\item The maximum profit is $\$72$.
\item The price per T-shirt should be set at $\$16$ to maximize profit.
\item The break even points are $x=1$ and $x=13$, so to make a profit, between 1 and 13 T-shirts need to be made and sold.
\end{itemize}
\item \begin{itemize}
\item $P(x) = -x^2+25x-100$, for $0 \leq x \leq 35$
\item Since the vertex occurs at $x=12.5$, and it is impossible to make or sell $12.5$ bottles of tonic, maximum profit occurs when either $12$ or $13$ bottles of tonic are made and sold.
\item The maximum profit is $\$56$.
\item The price per bottle can be either $\$23$ (to sell 12 bottles) or $\$22$ (to sell 13 bottles.) Both will result in the maximum profit.
\item The break even points are $x=5$ and $x=20$, so to make a profit, between 5 and 20 bottles of tonic need to be made and sold.
\end{itemize}
\item \begin{itemize}
\item $P(x) = -3x^2+72x-240$, for $0 \leq x \leq 30$
\item $12$ cups of lemonade need to be made and sold to maximize profit.
\item The maximum profit is $192$\textcent \, or $\$1.92$.
\item The price per cup should be set at $54$\textcent \, per cup to maximize profit.
\item The break even points are $x=4$ and $x=20$, so to make a profit, between 4 and 20 cups of lemonade need to be made and sold.
\end{itemize}
\item \begin{itemize}
\item $P(x) = -0.5 x^2+9x-36$, for $0 \leq x \leq 24$
\item $9$ pies should be made and sold to maximize the daily profit.
\item The maximum daily profit is $\$4.50$.
\item The price per pie should be set at $\$7.50$ to maximize profit.
\item The break even points are $x=6$ and $x=12$, so to make a profit, between 6 and 12 pies need to be made and sold daily.
\end{itemize}
\item \begin{itemize}
\item $P(x) = -2x^2+120x-1000$, for $0 \leq x \leq 70$
\item $30$ scooters need to be made and sold to maximize profit.
\item The maximum monthly profit is $800$ hundred dollars, or $\$80,\!000$.
\item The price per scooter should be set at $80$ hundred dollars, or $\$8000$ per scooter.
\item The break even points are $x=10$ and $x=50$, so to make a profit, between 10 and 50 scooters need to be made and sold monthly.
\end{itemize}
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item 495 cookies
\item The vertex is (approximately) $(29.60, 22.66)$, which corresponds to a maximum fuel economy of 22.66 miles per gallon, reached sometime between 2009 and 2010 (29 -- 30 years after 1980.) Unfortunately, the model is only valid up until 2008 (28 years after 1908.) So, at this point, we are using the model to \textit{predict} the maximum fuel economy.
\item $64^{\circ}$ at 2 PM (8 hours after 6 AM.)
\item 5000 pens should be produced for a cost of $\$200$.
\item 8 feet by 16 feet; maximum area is 128 square feet.
\item 50 feet by 50 feet; maximum area is 2500 feet; he can raise 100 average alpacas.
\item The largest rectangle has area $12.25$ square inches.
\item $2$ seconds.
\item The rocket reaches its maximum height of $500$ feet $10$ seconds after lift-off.
\item The hammer reaches a maximum height of approximately $13.62$ feet. The hammer is in the air approximately $1.61$ seconds.
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item \begin{enumerate}
\item The applied domain is $[0, \infty)$.
\addtocounter{enumii}{2}
\item The height function is this case is $s(t) = -4.9t^{2} + 15t$. The vertex of this parabola is approximately $(1.53, 11.48)$ so the maximum height reached by the marble is $11.48$ meters. It hits the ground again when $t \approx 3.06$ seconds.
\item The revised height function is $s(t) = -4.9t^{2} + 15t + 25$ which has zeros at $t \approx -1.20$ and $t \approx 4.26$. We ignore the negative value and claim that the marble will hit the ground after $4.26$ seconds.
\item Shooting down means the initial velocity is negative so the height functions becomes $s(t) = -4.9t^{2} - 15t + 25$.
\end{enumerate}
\item Make the vertex of the parabola $(0, 10)$ so that the point on the top of the left-hand tower where the cable connects is $(-200, 100)$ and the point on the top of the right-hand tower is $(200, 100)$. Then the parabola is given by $p(x) = \frac{9}{4000}x^{2} + 10$. Standing $50$ feet to the right of the left-hand tower means you're standing at $x= -150$ and $p(-150) = 60.625$. So the cable is 60.625 feet above the bridge deck there.
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $y = |1 -x^{2}|$
\begin{mfpic}[12]{-3}{3}{-1}{8}
\axes
\tlabel[cc](3,-0.5){\scriptsize $x$}
\tlabel[cc](0.5,8){\scriptsize $y$}
\arrow \reverse \function{-3,-1,0.1}{x**2 - 1}
\function{-1,1,0.1}{1-x**2}
\arrow \function{1,3,0.1}{x**2-1}
\point[3pt]{(-1,0),(1,0),(0,1)}
\xmarks{-2 step 1 until 2}
\ymarks{1,2,3,4,5,6,7}
\tiny
\tlpointsep{4pt}
\axislabels {x}{{$-2 \hspace{6pt}$} -2, {$-1 \hspace{6pt}$} -1, {$1$} 1, {$2$} 2}
\axislabels {y}{{$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4, {$5$} 5, {$6$} 6, {$7$} 7}
\normalsize
\end{mfpic}
\item $\left(\dfrac{3 - \sqrt{7}}{2}, \dfrac{-1 + \sqrt{7}}{2} \right)$, $\left(\dfrac{3 + \sqrt{7}}{2}, \dfrac{-1 - \sqrt{7}}{2} \right)$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $D(x) = x^2 + (2x+1)^2 = 5x^2+4x+1$, $D$ is minimized when $x=-\frac{2}{5}$, so the point on $y=2x+1$ closest to $(0,0)$ is $\left(-\frac{2}{5}, \frac{1}{5}\right)$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\begin{multicols}{3}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\addtocounter{enumi}{1}
\item $x = \pm y\sqrt{10}$ \vphantom{$\dfrac{m \pm \sqrt{m^{2} + 4}}{2}$}
\item $x = \pm (y - 2) $ \vphantom{$\dfrac{m \pm \sqrt{m^{2} + 4}}{2}$}
\item $x = \dfrac{m \pm \sqrt{m^{2} + 4}}{2}$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $y = \dfrac{3 \pm \sqrt{16x + 9}}{2}$ \vphantom{$\dfrac{m \pm \sqrt{m^{2} + 4}}{2g}$}
\item $y = 2 \pm x$ \vphantom{$\dfrac{m \pm \sqrt{m^{2} + 4}}{2g}$}
\item $t = \dfrac{v_{\mbox{\tiny $0$}} \pm \sqrt{v_{\mbox{\tiny $0$}}^{2} + 4gs_{\mbox{\tiny $0$}}}}{2g} $
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
2.4: Inequalities with Absolute Value and Quadratic Functions
In Exercises \ref{solveinequabsquadfirst} - \ref{solveinequabsquadlast}, solve the inequality. Write your answer using interval notation.
\begin{multicols}{2}
\begin{enumerate}
\item $|3x - 5| \leq 4$ \label{solveinequabsquadfirst}
\item $|7x + 2| > 10$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $|2x+1| - 5 < 0$
\item $|2-x| - 4 \geq -3$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $|3x+5| + 2 < 1$
\item $2|7-x| +4 > 1$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $2 \leq |4-x| < 7$
\item $1 < |2x - 9| \leq 3$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $|x + 3| \geq |6x + 9|$
\item $|x-3| - |2x+1| < 0$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $|1-2x| \geq x + 5$
\item $x + 5 < |x+5|$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $x \geq |x+1|$
\item $|2x + 1| \leq 6-x$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $x + |2x-3| < 2$
\item $|3-x| \geq x-5$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $x^{2} + 2x - 3 \geq 0$
\item $16x^2+8x+1 > 0$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $x^2+9 < 6x$
\item $9x^2 + 16 \geq 24x$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $x^2+4 \leq 4x$
\item $x^{2} + 1 < 0$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $3x^{2} \leq 11x + 4$
\item $x > x^{2}$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $2x^2-4x-1 > 0$
\item $5x+4 \leq 3x^2$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $2 \leq |x^{2} - 9| < 9$
\item $x^{2} \leq |4x - 3|$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $x^{2} + x + 1 \geq 0$
\item $x^2 \geq |x|$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $x |x+5| \geq -6$
\item $x |x-3| < 2$ \label{solveinequabsquadlast}
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item The profit, in dollars, made by selling $x$ bottles of $100 \%$ All-Natural Certified Free-Trade Organic Sasquatch Tonic is given by $P(x) = -x^2+25x-100$, for $0 \leq x \leq 35$. How many bottles of tonic must be sold to make at least $\$50$ in profit?
\item Suppose $C(x) = x^2-10x+27$, $x \geq 0$ represents the costs, in \textit{hundreds} of dollars, to produce $x$ \textit{thousand} pens. Find the number of pens which can be produced for no more than $\$1100$.
\item The temperature $T$, in degrees Fahrenheit, $t$ hours after 6 AM is given by $T(t) = -\frac{1}{2} t^2 + 8t+32$, for $0 \leq t \leq 12$. When is it warmer than $42^{\circ}$ Fahrenheit?
\item The height $h$ in feet of a model rocket above the ground $t$ seconds after lift-off is given by $h(t) = -5t^2+100t$, for $0 \leq t \leq 20$. When is the rocket at least $250$ feet off the ground? Round your answer to two decimal places.
\item If a slingshot is used to shoot a marble straight up into the air from 2 meters above the ground with an initial velocity of 30 meters per second, for what values of time $t$ will the marble be over 35 meters above the ground? (Refer to Exercise \ref{whatgoesup} in Section \ref{QuadraticFunctions} for assistance if needed.) Round your answers to two decimal places.
\item What temperature values in degrees Celsius are equivalent to the temperature range $50^{\circ}F$ to $95^{\circ}F$? (Refer to Exercise \ref{celsiustofahr} in Section \ref{LinearFunctions} for assistance if needed.)
\setcounter{HW}{\value{enumi}}
\end{enumerate}
In Exercises \ref{writeabsinequfirst} - \ref{writeabsinequlast}, write and solve an inequality involving absolute values for the given statement.
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item Find all real numbers $x$ so that $x$ is within $4$ units of $2$. \label{writeabsinequfirst}
\item Find all real numbers $x$ so that $3x$ is within $2$ units of $-1$.
\item Find all real numbers $x$ so that $x^2$ is within $1$ unit of $3$.
\item Find all real numbers $x$ so that $x^2$ is at least $7$ units away from $4$. \label{writeabsinequlast}
\item The surface area $S$ of a cube with edge length $x$ is given by $S(x) = 6x^{2}$ for $x > 0$. Suppose the cubes your company manufactures are supposed to have a surface area of exactly 42 square centimeters, but the machines you own are old and cannot always make a cube with the precise surface area desired. Write an inequality using absolute value that says the surface area of a given cube is no more than 3 square centimeters away (high or low) from the target of 42 square centimeters. Solve the inequality and write your answer using interval notation.
\item Suppose $f$ is a function, $L$ is a real number and $\varepsilon$ is a positive number. Discuss with your classmates what the inequality $|f(x) - L| < \varepsilon$ means algebraically and graphically.\footnote{Understanding this type of inequality is really important in Calculus.}
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\end{enumerate}
In Exercises \ref{sketchregionineqfirst} - \ref{sketchregionineqlast}, sketch the graph of the relation.
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $R = \left\{ (x,y) : y \leq x-1 \right\}$ \label{sketchregionineqfirst}
\item $R = \left\{ (x,y) : y > x^2 +1 \right\}$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $R = \left\{ (x,y) : -1 < y \leq 2x + 1 \right\}$
\item $R = \left\{ (x,y) : x^2 \leq y < x+2 \right\}$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $R = \left\{ (x,y) : |x|-4 < y < 2-x \right\}$
\item $R = \left\{ (x,y) : x^{2} < y \leq |4x - 3| \right\}$ \label{sketchregionineqlast}
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item Prove the second, third and fourth parts of Theorem \ref{absolutevalueineq}.
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\subsection{Answers}
\begin{multicols}{2}
\begin{enumerate}
\item $\left[\frac{1}{3}, 3\right]$
\item $\left(-\infty, -\frac{12}{7} \right) \cup \left(\frac{8}{7}, \infty\right)$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $(-3,2)$
\item $(-\infty,1] \cup [3,\infty)$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item No solution
\item $(-\infty, \infty)$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $(-3,2] \cup [6,11)$
\item $[3, 4) \cup (5, 6]$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $\left[-\frac{12}{7}, -\frac{6}{5}\right]$
\item $(-\infty, -4) \cup \left( \frac{2}{3}, \infty\right)$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $\left(-\infty, -\frac{4}{3} \right] \cup [6, \infty)$
\item $(-\infty, -5)$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item No Solution.
\item $\left[ -7, \frac{5}{3}\right]$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $\left( 1, \frac{5}{3} \right)$
\item $(-\infty, \infty)$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $(-\infty, -3] \cup [1, \infty)$
\item $\left(-\infty, -\frac{1}{4}\right) \cup \left(-\frac{1}{4}, \infty \right)$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item No solution
\item $(-\infty, \infty)$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $\left\{2 \right\}$
\item No solution
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $\left[-\frac{1}{3}, 4 \right]$
\item $(0, 1)$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $\left(-\infty, 1-\frac{\sqrt{6}}{2} \right) \cup \left(1+\frac{\sqrt{6}}{2}, \infty \right)$
\item $\left(-\infty, \frac{5 - \sqrt{73}}{6} \right] \cup \left[\frac{5 + \sqrt{73}}{6}, \infty \right)$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item {\scriptsize $\left(-3\sqrt{2}, -\sqrt{11} \right] \cup \left[-\sqrt{7}, 0 \right) \cup \left(0, \sqrt{7} \right] \cup \left[\sqrt{11}, 3\sqrt{2} \right)$}
\item $\left[-2-\sqrt{7}, -2+\sqrt{7} \right] \cup [1, 3]$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $(-\infty, \infty)$
\item $(-\infty, -1] \cup \left\{ 0 \right\} \cup [1,\infty)$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $[-6,-3] \cup [-2, \infty)$
\item $(-\infty, 1) \cup \left(2, \frac{3+\sqrt{17}}{2}\right)$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $P(x) \geq 50$ on $[10,15]$. This means anywhere between 10 and 15 bottles of tonic need to be sold to earn at least $\$50$ in profit.
\item $C(x) \leq 11$ on $[2,8]$. This means anywhere between 2000 and 8000 pens can be produced and the cost will not exceed $\$1100$.
\item $T(t) > 42$ on $(8-2\sqrt{11}, 8+2\sqrt{11}) \approx (1.37, 14.63)$, which corresponds to between 7:22 AM (1.37 hours after 6 AM) to 8:38 PM (14.63 hours after 6 AM.) However, since the model is valid only for $t$, $0 \leq t \leq 12$, we restrict our answer and find it is warmer than $42^{\circ}$ Fahrenheit from 7:22 AM to 6 PM.
\item $h(t) \geq 250$ on $[10-5\sqrt{2}, 10+5\sqrt{2}] \approx [2.93, 17.07]$. This means the rocket is at least 250 feet off the ground between 2.93 and 17.07 seconds after lift off.
\item $s(t) = -4.9t^2 + 30t + 2$. $s(t) > 35$ on (approximately) $(1.44, 4.68)$. This means between 1.44 and 4.68 seconds after it is launched into the air, the marble is more than 35 feet off the ground.
\item From our previous work $C(F) = \frac{5}{9}(F - 32)$ so $50 \leq F \leq 95$ becomes $10 \leq C \leq 35$.
\item $|x-2| \leq 4$, $[-2,6]$
\item $|3x+1| \leq 2$, $\left[-1, \frac{1}{3}\right]$
\item $|x^2-3| \leq 1$, $[-2, -\sqrt{2} \,] \cup [\sqrt{2}, 2]$
\item $|x^2 -4| \geq 7$, $(-\infty, -\sqrt{11} \,] \cup [\sqrt{11}, \infty)$
\item Solving $|S(x) - 42| \leq 3$, and disregarding the negative solutions yields $\left[\sqrt{\frac{13}{2}}, \sqrt{\frac{15}{2}}\right] \approx [2.550, 2.739]$. The edge length must be within $2.550$ and $2.739$ centimeters.
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\addtocounter{enumi}{1}
\item $~$
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\begin{multicols}{2}
\begin{enumerate}
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\item $~$
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\begin{multicols}{2}
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\item $~$
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2.5: Regression
\begin{enumerate}
\item According to this \href{www.ohiobiz.com/census/Lake.p...us/Lake.pdf}}}, the census data for Lake County, Ohio is:
\noindent \begin{tabular}{|l|r|r|r|r|} \hline
Year & 1970 & 1980 & 1990 & 2000 \\
\hline
Population & 197200 & 212801 & 215499 & 227511 \\ \hline
\end{tabular}
\begin{enumerate}
\item Find the least squares regression line for these data and comment on the goodness of fit.\footnote{We'll develop more sophisticated models for the growth of populations in Chapter \ref{ExpLogs}. For the moment, we use a theorem from Calculus to approximate those functions with lines.} Interpret the slope of the line of best fit.
\item Use the regression line to predict the population of Lake County in 2010. (The recorded figure from the 2010 census is $230,\!041$)
\item Use the regression line to predict when the population of Lake County will reach $250,\!000$.
\end{enumerate}
\item According to this \href{www.ohiobiz.com/census/Lorain.../Lorain.pdf}}}, the census data for Lorain County, Ohio is:
\noindent \begin{tabular}{|l|r|r|r|r|} \hline
Year & 1970 & 1980 & 1990 & 2000 \\
\hline
Population & 256843 & 274909 & 271126 & 284664 \\ \hline
\end{tabular}
\begin{enumerate}
\item Find the least squares regression line for these data and comment on the goodness of fit. Interpret the slope of the line of best fit.
\item Use the regression line to predict the population of Lorain County in 2010. (The recorded figure from the 2010 census is $301,\!356$)
\item Use the regression line to predict when the population of Lake County will reach $325,\!000$.
\end{enumerate}
\item Using the energy production data given below
\noindent \begin{tabular}{|l|r|r|r|r|r|r|} \hline
Year & 1950 & 1960 & 1970 & 1980 & 1990 & 2000 \\
\hline
Production & & & & & & \\
(in Quads) & 35.6 & 42.8 & 63.5 & 67.2 & 70.7 & 71.2 \\ \hline
\end{tabular}
\begin{enumerate}
\item Plot the data using a graphing calculator and explain why it does not appear to be linear.
\item Discuss with your classmates why ignoring the first two data points may be justified from a historical perspective.
\item Find the least squares regression line for the last four data points and comment on the goodness of fit. Interpret the slope of the line of best fit.
\item Use the regression line to predict the annual US energy production in the year $2010$.
\item Use the regression line to predict when the annual US energy production will reach $100$ Quads.
\end{enumerate}
\item The chart below contains a portion of the fuel consumption information for a 2002 Toyota Echo that I (Jeff) used to own. The first row is the cumulative number of gallons of gasoline that I had used and the second row is the odometer reading when I refilled the gas tank. So, for example, the fourth entry is the point (28.25, 1051) which says that I had used a total of 28.25 gallons of gasoline when the odometer read 1051 miles.
\medskip
\small
\noindent \begin{tabular}{|l|r|r|r|r|r|r|r|r|r|r|r|} \hline
Gasoline Used & & & & & & & & & & & \\
(Gallons) & 0 & 9.26 & 19.03 & 28.25 & 36.45 & 44.64 & 53.57 & 62.62 & 71.93 & 81.69 & 90.43\\
\hline
Odometer & & & & & & & & & & & \\
(Miles) & 41 & 356 & 731 & 1051 & 1347 & 1631 & 1966 & 2310 & 2670 & 3030 & 3371\\ \hline
\end{tabular}
\normalsize
\medskip
\noindent Find the least squares line for this data. Is it a good fit? What does the slope of the line represent? Do you and your classmates believe this model would have held for ten years had I not crashed the car on the Turnpike a few years ago? (I'm keeping a fuel log for my 2006 Scion xA for future College Algebra books so I hope not to crash it, too.)
\item On New Year's Day, I (Jeff, again) started weighing myself every morning in order to have an interesting data set for this section of the book. (Discuss with your classmates if that makes me a nerd or a geek. Also, the professionals in the field of weight management strongly discourage weighing yourself every day. When you focus on the number and not your overall health, you tend to lose sight of your objectives. I was making a noble sacrifice for science, but you should \underline{not} try this at home.) The whole chart would be too big to put into the book neatly, so I've decided to give only a small portion of the data to you. This then becomes a Civics lesson in honesty, as you shall soon see. There are two charts given below. One has my weight for the first eight Thursdays of the year (January 1, 2009 was a Thursday and we'll count it as Day 1.) and the other has my weight for the first 10 Saturdays of the year.
\medskip
\small
\noindent \begin{tabular}{|l|r|r|r|r|r|r|r|r|} \hline
Day \# & & & & & & & & \\
(Thursday) & 1 & 8 & 15 & 22 & 29 & 36 & 43 & 50 \\
\hline
My weight & & & & & & & & \\
in pounds & 238.2 & 237.0 & 235.6 & 234.4 & 233.0 & 233.8 & 232.8 & 232.0\\ \hline
\end{tabular}
\medskip
\noindent \begin{tabular}{|l|r|r|r|r|r|r|r|r|r|r|} \hline
Day \# & & & & & & & & & & \\
(Saturday) & 3 & 10 & 17 & 24 & 31 & 38 & 45 & 52 & 59 & 66 \\
\hline
My weight & & & & & & & & & & \\
in pounds & 238.4 & 235.8 & 235.0 & 234.2 & 236.2 & 236.2 & 235.2 & 233.2 & 236.8 & 238.2\\ \hline
\end{tabular}
\normalsize
\medskip
\begin{enumerate}
\item Find the least squares line for the Thursday data and comment on its goodness of fit.
\item Find the least squares line for the Saturday data and comment on its goodness of fit.
\item Use Quadratic Regression to find a parabola which models the Saturday data and comment on its goodness of fit.
\item Compare and contrast the predictions the three models make for my weight on January 1, 2010 (Day \#366). Can any of these models be used to make a prediction of my weight 20 years from now? Explain your answer.
\item Why is this a Civics lesson in honesty? Well, compare the two linear models you obtained above. One was a good fit and the other was not, yet both came from careful selections of real data. In presenting the tables to you, I have not lied about my weight, nor have you used any bad math to falsify the predictions. The word we're looking for here is `disingenuous'. Look it up and then discuss the implications this type of data manipulation could have in a larger, more complex, politically motivated setting. (Even Obi-Wan presented the truth to Luke only ``from a certain point of view.'')
\end{enumerate}
\item (Data that is neither linear nor quadratic.) We'll close this exercise set with two data sets that, for reasons presented later in the book, cannot be modeled correctly by lines or parabolas. It is a good exercise, though, to see what happens when you attempt to use a linear or quadratic model when it's not appropriate.
\begin{enumerate}
\item \label{APLcats} This first data set came from a Summer 2003 publication of the Portage County Animal Protective League called ``Tattle Tails''. They make the following statement and then have a chart of data that supports it. ``It doesn't take long for two cats to turn into 80 million. If two cats and their surviving offspring reproduced for ten years, you'd end up with 80,399,780 cats.'' We assume $N(0) = 2$.
\medskip
\scriptsize
\noindent \begin{tabular}{|l|r|r|r|r|r|r|r|r|r|r|} \hline
Year $x$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\
\hline
Number of & & & & & & & & & & \\
Cats $N(x)$ & 12 & 66 & 382 & 2201 & 12680 & 73041 & 420715 & 2423316 & 13968290 & 80399780 \\ \hline
\end{tabular}
\normalsize
\medskip
\noindent Use Quadratic Regression to find a parabola which models this data and comment on its goodness of fit. (Spoiler Alert: Does anyone know what type of function we need here?)
\medskip
\item \label{regsunlight} This next data set comes from the \href{aa.usno.navy.mil/data/docs/RS...\underline{U.S. Naval Observatory}}. That site has loads of awesome stuff on it, but for this exercise I used the sunrise/sunset times in Fairbanks, Alaska for 2009 to give you a chart of the number of hours of daylight they get on the $21^{\mbox{st}}$ of each month. We'll let $x = 1$ represent January 21, 2009, $x = 2$ represent February 21, 2009, and so on.
\medskip
\small
\noindent \begin{tabular}{|l|r|r|r|r|r|r|r|r|r|r|r|r|} \hline
Month & & & & & & & & & & & & \\
Number & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12\\
\hline
Hours of & & & & & & & & & & & & \\
Daylight & 5.8 & 9.3 & 12.4 & 15.9 & 19.4 & 21.8 & 19.4 & 15.6 & 12.4 & 9.1 & 5.6 & 3.3 \\ \hline
\end{tabular}
\normalsize
\medskip
\noindent Use Quadratic Regression to find a parabola which models this data and comment on its goodness of fit. (Spoiler Alert: Does anyone know what type of function we need here?)
\end{enumerate}
\end{enumerate}
\subsection{Answers}
\begin{enumerate}
\item \begin{enumerate}
\item $y = 936.31x - 1645322.6$ with $r=0.9696$ which indicates a good fit. The slope $936.31$ indicates Lake County's population is increasing at a rate of (approximately) 936 people per year.
\item According to the model, the population in 2010 will be $236, \!660$.
\item According to the model, the population of Lake County will reach $250,\!000$ sometime between 2024 and 2025.
\end{enumerate}
\item \begin{enumerate}
\item $y = 796.8x - 1309762.5$ with $r=0.8916$ which indicates a reasonable fit. The slope $796.8$ indicates Lorain County's population is increasing at a rate of (approximately) 797 people per year.
\item According to the model, the population in 2010 will be $291, \! 805$.
\item According to the model, the population of Lake County will reach $325,\!000$ sometime between 2051 and 2052.
\end{enumerate}
\item \begin{enumerate}
\setcounter{enumii}{2}
\item $y = 0.266x - 459.86$ with $r = 0.9607$ which indicates a good fit. The slope $0.266$ indicates the country's energy production is increasing at a rate of $0.266$ Quad per year.
\item According to the model, the production in 2010 will be $74.8$ Quad.
\item According to the model, the production will reach $100$ Quad in the year 2105.
\end{enumerate}
\item The line is $y = 36.8x + 16.39$. We have $r = .99987$ and $r^{2} = .9997$ so this is an excellent fit to the data. The slope $36.8$ represents miles per gallon.
\item \begin{enumerate}
\item The line for the Thursday data is $y = -.12x + 237.69$. We have $r = -.9568$ and $r^{2} = .9155$ so this is a really good fit.
\item The line for the Saturday data is $y = -0.000693x + 235.94$. We have $r = -0.008986$ and $r^{2} = 0.0000807$ which is horrible. This data is not even close to linear.
\item The parabola for the Saturday data is $y = 0.003x^{2} - 0.21x + 238.30$. We have $R^{2} = .47497$ which isn't good. Thus the data isn't modeled well by a quadratic function, either.
\item The Thursday linear model had my weight on January 1, 2010 at 193.77 pounds. The Saturday models give 235.69 and 563.31 pounds, respectively. The Thursday line has my weight going below 0 pounds in about five and a half years, so that's no good. The quadratic has a positive leading coefficient which would mean unbounded weight gain for the rest of my life. The Saturday line, which mathematically does not fit the data at all, yields a plausible weight prediction in the end. I think this is why grown-ups talk about ``Lies, Damned Lies and Statistics.''
\end{enumerate}
\item \begin{enumerate}
\item The quadratic model for the cats in Portage county is $y = 1917803.54x^{2} - 16036408.29x + 24094857.7$. Although $R^{2} = .70888$ this is not a good model because it's so far off for small values of $x$. Case in point, the model gives us 24,094,858 cats when $x = 0$ but we know $N(0) = 2$.
\item The quadratic model for the hours of daylight in Fairbanks, Alaska is $y = .51x^{2} + 6.23x - .36$. Even with $R^{2} = .92295$ we should be wary of making predictions beyond the data. Case in point, the model gives $-4.84$ hours of daylight when $x = 13$. So January 21, 2010 will be ``extra dark''? Obviously a parabola pointing down isn't telling us the whole story.
\end{enumerate}
\end{enumerate}