\documentclass[11pt]{article} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \usepackage{amssymb} \usepackage[nomarginpar]{geometry} \usepackage{color} \usepackage{amsfonts} \usepackage{amsmath} \usepackage{fancyhdr} \usepackage{times} \usepackage{multicol} \usepackage{setspace} \usepackage{points} \usepackage{boxedminipage} \usepackage[none]{hyphenat} \usepackage{hyperref} \setcounter{MaxMatrixCols}{10} %TCIDATA{OutputFilter=LATEX.DLL} %TCIDATA{Version=5.00.0.2570} %TCIDATA{} %TCIDATA{Created=Wednesday, July 12, 2006 00:27:03} %TCIDATA{LastRevised=Sunday, October 28, 2018 10:11:19} %TCIDATA{} %TCIDATA{} %TCIDATA{} %TCIDATA{CSTFile=40 LaTeX article.cst} %TCIDATA{PageSetup=72,72,72,72,1} %TCIDATA{Counters=arabic,1} %TCIDATA{} %TCIDATA{AllPages= %H=36 %F=36,\PARA{038

\hfill \hfill } %} \newtheorem{theorem}{Theorem} \newtheorem{acknowledgement}[theorem]{Acknowledgement} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{axiom}[theorem]{Axiom} \newtheorem{case}[theorem]{Case} \newtheorem{claim}[theorem]{Claim} \newtheorem{conclusion}[theorem]{Conclusion} \newtheorem{condition}[theorem]{Condition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{criterion}[theorem]{Criterion} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{exercise}{Exercise} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{notation}[theorem]{Notation} \newtheorem{problem}[theorem]{Problem} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \newtheorem{solution}[theorem]{Solution} \newtheorem{summary}[theorem]{Summary} \newenvironment{proof}[1][Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \geometry{left=0.5in,right=0.5in,top=0.5in,bottom=0.4in} \pagestyle{fancy} \lhead{\color{blue}\large Lecture Notes } \chead{{\color{black}}{\Large{Graphs of Equations}}} \rhead{\small page \ \thepage} \cfoot{} \lfoot{\footnotesize \copyright $\;$ Hidegkuti, Powell, 2007} \rfoot{\footnotesize Last revised: October 28, 2018} \textwidth 7.6in \textheight 9.6in \setlength{\headheight}{28pt} \setlength{\parindent}{0in} \newcommand{\defstart}{ \begin{boxedminipage}{7in} \setlength{\fboxrule}{15pt} \setlength{\fboxsep}{15pt} \setlength{\rightskip}{20pt} \setlength{\leftskip}{20pt} \color{white} . \color{black} } \newcommand{\defcont}{ \setlength{\hangindent}{57pt} } \newcommand{\definside}{ \setlength{\hangindent}{57pt} \setlength{\parindent}{57pt} } \newcommand{\defend}{ \end{boxedminipage} \setlength{\parindent}{0pt} \setlength{\hangindent}{0pt} \setlength{\parskip}{0.06in} \setstretch{1.08} } \newcommand{\exstart}{ \setlength{\parindent}{0pt} \setlength{\hangindent}{60pt} \setlength{\parskip}{0.04in} \setstretch{1.08} } \newcommand{\excont}{ \setlength{\parindent}{60pt} \setlength{\hangindent}{75pt} } \newcommand{\solstart}{ \setlength{\parindent}{0pt} \setlength{\hangindent}{63pt} } \newcommand{\solcont}{ \setlength{\parindent}{63pt} \setlength{\hangindent}{63pt} } \newcommand{\solrestart}{ \setlength{\parindent}{48pt} \setlength{\hangindent}{63pt} } \newcommand{\sol}{ \setlength{\parindent}{0pt} \setlength{\hangindent}{50pt} } \newcommand{\solinside}{ \setlength{\parindent}{50pt} \setlength{\hangindent}{50pt} \setstretch{1.08} } \newcommand{\vissza}{ \setlength{\parindent}{0pt} \setlength{\hangindent}{0pt} \setlength{\parskip}{0.1in} \setstretch{1.08} \normalsize \setcounter{enumi}{0}\RESUME } \newcommand{\matek}{ \setstretch{1.04} \normalsize \setlength{\parskip}{0.04in} \setlength{\parindent}{0pt} \setlength{\hangindent}{0pt} \setcounter{enumi}{0}\RESUME } \setstretch{1.06} \begin{document} %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion As we have seen more and more algebraic statements, the solution sets became increasingly more complex. \ A linear equation usually has a single number solution. \ In case of linear inequalities, we often have infinitely many solutions. \ To express those solution sets, we developed interval notation. Suppose we have an equation in two variables, $x$ and $y$. \ The equations $% y=2x-3$ or \ $x^{2}-y^{2}=5$ or $xy=-2$ are examples for such equations. \ A solution for such equations is a set of ordered pairs of numbers, $\left( x,y\right) $. \ For example, $\left( 5,7\right) $ is short for $x=5$ and $% y=7 $ and this ordered pair is a solution of the equation $y=2x-3$. \ The ordered pair $\left( 3,-2\right) $ is a solution of $x^{2}-y^{2}=5,$ and the ordered pair $\left( 2,-1\right) $ is a solution of $xy=-2$. Equations in two variables often have infintely many solutions, where that can no longer meaningfully represented on a number line. \ We step out into two dimensions, and use a coordinate system to depict solution sets. \ On a coordinate system, each ordered pair $\left( x,y\right) $ can be represented as a point. \ %TCIMACRO{\TeXButton{\defstart}{\defstart}}% %BeginExpansion \defstart% %EndExpansion %TCIMACRO{\TeXButton{\defcont}{\defcont}}% %BeginExpansion \defcont% %EndExpansion \textbf{Definition}: \ $\ $The \textbf{graph} of an equation in $x$, in $y$, or both in $x$ and $y$ is the set of all points $P\left( x,y\right) $ \ whose coordinates are solution of the equation. %TCIMACRO{\TeXButton{\defend}{\defend}}% %BeginExpansion \defend% %EndExpansion \vspace{0.1in} In short, the graph of an equation is a solution set of an equation in $x$ and $y$. \ The shape of graphs depends on the type of equation. \ Before we started to graph equations, it is useful to know that we can do quite a lot just using the definition of graphs.\vspace{0.1in} %TCIMACRO{\TeXButton{\begin{minipage}{4.7in}}{\begin{minipage}{4.7in}}}% %BeginExpansion \begin{minipage}{4.7in}% %EndExpansion %TCIMACRO{\TeXButton{\exstart}{\exstart}}% %BeginExpansion \exstart% %EndExpansion \textbf{Example 1.} \ Consider the graph shown. \ Three points on the graph are marked. \ These are $A\left( -2,0\right) $, \ $B\left( -1,5\right) $, \ and $C\left( 1,9\right) $. \ Use these points to determine, which of the given equations is the one whose graph is the shape we see. \vspace{0.03in} %TCIMACRO{\TeXButton{\excont}{\excont}}% %BeginExpansion \excont% %EndExpansion The possible equations offered are: \vspace{0.03in} %TCIMACRO{\TeXButton{\excont}{\excont}}% %BeginExpansion \excont% %EndExpansion $\qquad y=3x+6$ %TCIMACRO{\TeXButton{\excont}{\excont}}% %BeginExpansion \excont% %EndExpansion $\qquad \left( x-4\right) ^{2}+\left( y-5\right) ^{2}=25$ %TCIMACRO{\TeXButton{\excont}{\excont}}% %BeginExpansion \excont% %EndExpansion $\qquad y=-x^{2}+2x+8$\ \vspace{0.37in}% %TCIMACRO{\TeXButton{\end{minipage}}{\end{minipage}}}% %BeginExpansion \end{minipage}% %EndExpansion %TCIMACRO{% %\TeXButton{\begin{minipage}{3.3in}}{\begin{minipage}{3.3in} %\setlength{\leftskip}{24pt}} }% %BeginExpansion \begin{minipage}{3.3in} \setlength{\leftskip}{24pt} %EndExpansion \FRAME{itbpF}{2.3091in}{2.3091in}{0in}{}{}{Plot}{\special{language "Scientific Word";type "MAPLEPLOT";width 2.3091in;height 2.3091in;depth 0in;display "USEDEF";plot_snapshots TRUE;mustRecompute FALSE;lastEngine "MuPAD";xmin "-10.01";xmax "10.04";xviewmin "-10.01";xviewmax "10.04";yviewmin "-10.4";yviewmax "10.4";viewset"XY";rangeset"X";plottype 4;plotticks 1;num-x-ticks 11;num-y-ticks 11;constrained TRUE;numpoints 100;plotstyle "patch";axesstyle "normal";xis \TEXUX{x};yis \TEXUX{y};var1name \TEXUX{$x$};var2name \TEXUX{$y$};function \TEXUX{$1$};linecolor "gray";linestyle 1;pointstyle "point";linethickness 1;lineAttributes "Solid";var1range 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"Line";rangeset"X";function \TEXUX{$\sqrt{0.04-\left( x+2\right) ^{2}}$};linecolor "blue";linestyle 1;pointstyle "point";linethickness 3;lineAttributes "Solid";var1range "-10.01,10.04";num-x-gridlines 100;curveColor "[flat::RGB:0x000000ff]";curveStyle "Line";valid_file "T";tempfilename 'PGRUEG00.wmf';tempfile-properties "XPR";}}% %TCIMACRO{\TeXButton{\end{minipage}}{\end{minipage}}}% %BeginExpansion \end{minipage}% %EndExpansion %TCIMACRO{\TeXButton{\sol}{\sol}}% %BeginExpansion \sol% %EndExpansion \textbf{Solution:} \ Let us consider first the equation $y=3x+6$. \ If the graph belongs to this equation, then the coordinates of \textit{all} points on the graph are solutions of the equation, including those of $A$, $B$, and $C$. \ Let's check. \ %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion Point $A\left( -2,0\right) $ is on the graph if and only if its coordinates are a solution of $y=3x+6$.\vspace{0.06in} %TCIMACRO{% %\TeXButton{\solinside parskip}{\solinside %\setlength{\parskip}{0.04in}}}% %BeginExpansion \solinside \setlength{\parskip}{0.04in}% %EndExpansion \qquad \qquad \qquad Check $y=3x+6$ \ \ with $x=-2$ and $y=0$. %TCIMACRO{% %\TeXButton{\solinside parskip}{\solinside %\setlength{\parskip}{0.04in}}}% %BeginExpansion \solinside \setlength{\parskip}{0.04in}% %EndExpansion \qquad \qquad \qquad The left-hand side is: \ \ LHS$\,=0$ \ \ %TCIMACRO{% %\TeXButton{\solinside parskip}{\solinside %\setlength{\parskip}{0.04in}}}% %BeginExpansion \solinside \setlength{\parskip}{0.04in}% %EndExpansion \qquad \qquad \qquad and \ the right-hand side is: \ RHS$\,=3\left( -2\right) +6=0.$ \ \ RHS$\,=\,$LHS $\checkmark $ %TCIMACRO{% %\TeXButton{\solinside parskip}{\solinside %\setlength{\parskip}{0.04in}}}% %BeginExpansion \solinside \setlength{\parskip}{0.04in}% %EndExpansion Point $A$ is on \ the graph of $y=3x+6$. \ This does not mean that $y=3x+6$ is the right equation. \ It only means that we didn't rule it out based on point $A$ alone.\vspace{0.06in} %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion Let's see about point \ $B\left( -1,5\right) $. \ Is this pont on the graph of $y=3x+6$?\vspace{0.06in} %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion \qquad \qquad \qquad Check $y=3x+6$ \ \ with $x=-1$ and $y=5$. %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion \qquad \qquad \qquad LHS$\,=5$ \ \ and \ RHS$\,=3\left( -1\right) +6=-3+6=3$ \ \ RHS$\,\not=\,$LHS %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion At this point, we can conclude that\ the graph shown is not of the equation of $y=3x+6,$ because point $B$ is on the graph but its coordinates are not a solution of this equation. \ \ So, we can move on to the next equation. \pagebreak %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion Consider now the equation $\left( x-4\right) ^{2}+\left( y-5\right) ^{2}=25$% . \ If the graph belongs to this equation, then the coordinates of \textit{% all} points on the graph are solutions of the equation, including those of $% A $, $B$, and $C$. \ Let's check.\vspace{0.06in} %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion Point $A\left( -2,0\right) $ is on the graph if and only if its coordinates are a solution of $\left( x-4\right) ^{2}+\left( y-5\right) ^{2}=25$.\vspace{% 0.06in} %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion \qquad \qquad \qquad Check $\ \left( x-4\right) ^{2}+\left( y-5\right) ^{2}=25$ \ \ with $x=-2$ and $y=0$. %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion \qquad \qquad \qquad LHS$\,=\left( -2-4\right) ^{2}+\left( 0-5\right) ^{2}=\left( -6\right) ^{2}+\left( -5\right) ^{2}=36+25=61$ \ \ %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion \qquad \qquad \qquad RHS$\,=25$. \ \ RHS$\,\not=\,$LHS %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion We can conclude that\ the graph shown is not of the equation of $\left( x-4\right) ^{2}+\left( y-5\right) ^{2}=25$, because point $A$ is on the graph but its coordinates are not a solution of this equation. \ \ So, we can move on to the next equation.\vspace{0.1in} %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion Consider now the equation $y=-x^{2}+2x+8$. \ If the graph belongs to this equation, then the coordinates of \textit{all} points on the graph are solutions of the equation, including those of $A$, $B$, and $C$. \ Let's check.\vspace{0.06in} %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion Point $A\left( -2,0\right) $ is on the graph if and only if its coordinates are a solution of $y=-x^{2}+2x+8$.\vspace{0.06in} %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion \qquad \qquad \qquad Check $\ y=-x^{2}+2x+8$ \ \ with $x=-2$ and $y=0$. %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion \qquad \qquad \qquad LHS$\,=0$ \ and RHS$\,=-\left( -2\right) ^{2}+2\left( -2\right) +8=-4-4+8=0$. \ \ RHS$\,=\,$LHS $\checkmark $ %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion This does not mean that $y=-x^{2}+2x+8$ is the right equation. \ It only means that we didn't rule it out based on point $A$ alone. \ Let's see point $B$.\vspace{0.06in} %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion Point $B\left( -1,5\right) $ is on the graph if and only if its coordinates are a solution of $y=-x^{2}+2x+8$.\vspace{0.06in} %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion \qquad \qquad \qquad Check $\ y=-x^{2}+2x+8$ \ \ with $x=-1$ and $y=5$. %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion \qquad \qquad \qquad LHS$\,=5$ \ and \ RHS$\,=-\left( -1\right) ^{2}+2\left( -1\right) +8=-1-2+8=5$. \ \ RHS$\,=\,$LHS $\checkmark $ %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion This does not mean that $y=-x^{2}+2x+8$ is the right equation. \ It only means that we didn't rule it out based on points $A$ and $B$. \ Let's see point $C$.\vspace{0.1in} %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion Point $C\left( 1,9\right) $ is on the graph if and only if its coordinates are a solution of $y=-x^{2}+2x+8$.\vspace{0.06in} %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion \qquad \qquad \qquad Check $\ y=-x^{2}+2x+8$ \ \ with $x=1$ and $y=9$. %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion \qquad \qquad \qquad LHS$\,=9$ \ and RHS$\,=-1^{2}+2\cdot 1+8=-1+2+8=9$. \ \ RHS$\,=\,$LHS $\checkmark $\vspace{0.06in} %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion We found that all three points are on the graph of this equation. \ This still does not mean that $y=-x^{2}+2x+8$ is the right equation. \ Given that we were given three equations with the assumption that the correct equation is among them, it can only be this one. \ So, our answer is that the graph shown is of the equation $y=-x^{2}+2x+8$. \ We can find additional nice points on the graph (for example, $\left( 4,0\right) $ or $\left( 2,8\right) $) and test them against the equation. \ Soon we will learn how to graph such shapes. \ \bigskip \pagebreak %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion %TCIMACRO{\TeXButton{\begin{minipage}{4.7in}}{\begin{minipage}{4.7in}}}% %BeginExpansion \begin{minipage}{4.7in}% %EndExpansion %TCIMACRO{\TeXButton{\exstart}{\exstart}}% %BeginExpansion \exstart% %EndExpansion \textbf{Example 2.} \ Consider the graph shown. \ Three points on the graph are marked. \ These are $A\left( -8,1\right) $, \ $B\left( -6,5\right) $, \ and \ $C\left( 1,4\right) $. \ Use these points to determine, which of the given equations is the one whose graph is the shape we see. \vspace{0.03in} %TCIMACRO{\TeXButton{\excont}{\excont}}% %BeginExpansion \excont% %EndExpansion The possible equations offered are: \vspace{0.03in} %TCIMACRO{\TeXButton{\excont}{\excont}}% %BeginExpansion \excont% %EndExpansion $\qquad 3y=x+11$ %TCIMACRO{\TeXButton{\excont}{\excont}}% %BeginExpansion \excont% %EndExpansion $\qquad 3y+x^{2}=-8x+3$ %TCIMACRO{\TeXButton{\excont}{\excont}}% %BeginExpansion \excont% %EndExpansion $\qquad \left( x+3\right) ^{2}+\left( y-1\right) ^{2}=25$\ \vspace{0.37in}% %TCIMACRO{\TeXButton{\end{minipage}}{\end{minipage}}}% %BeginExpansion \end{minipage}% %EndExpansion %TCIMACRO{% %\TeXButton{\begin{minipage}{3.3in}}{\begin{minipage}{3.3in} 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"Line";rangeset"X";function \TEXUX{$4+\sqrt{0.04-\left( x-1\right) ^{2}}$};linecolor "blue";linestyle 1;pointstyle "point";linethickness 3;lineAttributes "Solid";var1range "0.8,1.2";num-x-gridlines 100;curveColor "[flat::RGB:0x000000ff]";curveStyle "Line";rangeset"X";valid_file "T";tempfilename 'PGRVFT02.wmf';tempfile-properties "XPR";}}% %TCIMACRO{\TeXButton{\end{minipage}}{\end{minipage}}}% %BeginExpansion \end{minipage}% %EndExpansion %TCIMACRO{\TeXButton{\sol}{\sol}}% %BeginExpansion \sol% %EndExpansion \textbf{Solution: } Let us consider first the equation $3y=x+11$. \ If the graph belongs to this equation, then the coordinates of \textit{all} points on the graph are solutions of the equation, including those of $A$, $B$, and $C$. \ Let's check.\vspace{0.06in} %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion Point $A\left( -8,1\right) $ is on the graph if and only if its coordinates are a solution of $3y=x+11$.\vspace{0.06in} %TCIMACRO{% %\TeXButton{\solinside parskip}{\solinside %\setlength{\parskip}{0.04in}}}% %BeginExpansion \solinside \setlength{\parskip}{0.04in}% %EndExpansion \qquad \qquad \qquad Check $3y=x+11$ \ \ with $x=-8$ and $y=1$. %TCIMACRO{% %\TeXButton{\solinside parskip}{\solinside %\setlength{\parskip}{0.04in}}}% %BeginExpansion \solinside \setlength{\parskip}{0.04in}% %EndExpansion \qquad \qquad \qquad The left-hand side is: \ \ LHS$\,=3\cdot 1=3$ \ \ %TCIMACRO{% %\TeXButton{\solinside parskip}{\solinside %\setlength{\parskip}{0.04in}}}% %BeginExpansion \solinside \setlength{\parskip}{0.04in}% %EndExpansion \qquad \qquad \qquad and \ the right-hand side is: \ RHS$\,=-8+11=3.$ \ \ RHS% $\,=\,$LHS $\checkmark $ %TCIMACRO{% %\TeXButton{\solinside parskip}{\solinside %\setlength{\parskip}{0.04in}}}% %BeginExpansion \solinside \setlength{\parskip}{0.04in}% %EndExpansion Point $A$ is on \ the graph of $3y=x+11$. \ This does not mean that $3y=x+11$ is the right equation. \ It only means that we didn't rule it out based on point $A$ alone. \ Let's see point $B$.\vspace{0.06in} %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion Point $B\left( -6,5\right) $ is on the graph if and only if its coordinates are a solution of $3y=x+11$.\vspace{0.06in} %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion \qquad \qquad \qquad Check $\ 3y=x+11$\ \ with $x=-6$ and $y=5$. %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion \qquad \qquad \qquad LHS$\,=3\left( -6\right) =-18$ \ and \ RHS$\,=-6+11=5$. \ \ RHS$\,\not=\,$LHS %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion We can conclude that\ the graph shown is not of the equation of $3y=x+11$, because point $B$ is on the graph but its coordinates are not a solution of this equation. \ \ So, we can move on to the next equation.\vspace{0.1in} %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion Consider now the equation $3y+x^{2}=-8x+3$. \ If the graph belongs to this equation, then the coordinates of \textit{all} points on the graph are solutions of the equation, including those of $A$, $B$, and $C$. \ Let's check.\vspace{0.06in} %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion Point $A\left( -8,1\right) $ is on the graph if and only if its coordinates are a solution of $3y+x^{2}=-8x+3$.\vspace{0.06in} %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion \qquad \qquad \qquad Check $\ 3y+x^{2}=-8x+3$ \ \ with $x=-8$ and $y=1$. %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion \qquad \qquad \qquad LHS$\,=3\cdot 1+\left( -8\right) ^{2}=3+64=67$ \ and RHS% $\,=-8\left( -8\right) +3=64+3=67$. \ \ RHS$\,=\,$LHS $\checkmark $ %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion This does not mean that $3y+x^{2}=-8x+3$ is the right equation. \ It only means that we didn't rule it out based on point $A$ alone. \ Let's see point $B$.\vspace{0.06in} %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion Point $B\left( -6,5\right) $ is on the graph if and only if its coordinates are a solution of $3y+x^{2}=-8x+3$.\vspace{0.06in} %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion \qquad \qquad \qquad Check $\ 3y+x^{2}=-8x+3$ \ \ with $x=-6$ and $y=5$. %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion \qquad \qquad \qquad LHS$\,=3\cdot 5+\left( -6\right) ^{2}=15+36=51$ \ and \ RHS$\,=-8\left( -6\right) +3=48+3=\allowbreak 51$. \ \ RHS$\,=\,$LHS $% \checkmark $ %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion This does not mean that $3y+x^{2}=-8x+3$ is the right equation. \ It only means that we didn't rule it out based on points $A$ and $B$. \ Let's see point $C$.\vspace{0.1in} \pagebreak %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion Point $C\left( 1,4\right) $ is on the graph if and only if its coordinates are a solution of $3y+x^{2}=-8x+3$.\vspace{0.06in} %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion \qquad \qquad \qquad Check $\ 3y+x^{2}=-8x+3$ \ \ with $x=1$ and $y=4$. %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion \qquad \qquad \qquad LHS$\,=3\cdot 4+4^{2}=12+16=28$ \ and RHS$\,=-8\cdot 1+3=-5$. \ \ RHS$\,\not=\,$LHS\vspace{0.06in} %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion We can conclude that\ the graph shown is not of the equation of $% 3y+x^{2}=-8x+3$, because point $C$ is on the graph but its coordinates are not a solution of this equation. \ \ So, we can move on to the next equation.% \vspace{0.08in} %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion Consider now the equation $\left( x+3\right) ^{2}+\left( y-1\right) ^{2}=25$% . \ If the graph belongs to this equation, then the coordinates of \textit{% all} points on the graph are solutions of the equation, including those of $% A $, $B$, and $C$. \ Let's check.\vspace{0.06in} %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion Point $A\left( -8,1\right) $ is on the graph if and only if its coordinates are a solution of $\left( x+3\right) ^{2}+\left( y-1\right) ^{2}=25$.\vspace{% 0.06in} %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion \qquad \qquad \qquad Check $\ \left( x+3\right) ^{2}+\left( y-1\right) ^{2}=25$ \ with $x=-8$ and $y=1$. %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion \qquad \qquad \qquad LHS$\,=\left( -8+3\right) ^{2}+\left( 1-1\right) ^{2}=\left( -5\right) ^{2}+0^{2}=25$ \ and RHS$\,=25$. \ \ RHS$\,=\,$LHS $% \checkmark $ %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion This does not mean that $\left( x+3\right) ^{2}+\left( y-1\right) ^{2}=25$ is the right equation. \ It only means that we didn't rule it out based on point $A$ alone. \ Let's see point $B$.\vspace{0.06in} %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion Point $B\left( -6,5\right) $ is on the graph if and only if its coordinates are a solution of $\left( x+3\right) ^{2}+\left( y-1\right) ^{2}=25$.\vspace{% 0.06in} %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion \qquad \qquad \qquad Check $\ \left( x+3\right) ^{2}+\left( y-1\right) ^{2}=25$ \ \ with $x=-6$ and $y=5$. %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion \qquad \qquad \qquad LHS$\,=\left( -6+3\right) ^{2}+\left( 5-1\right) ^{2}=\left( -3\right) ^{2}+4^{2}=9+16=25$ \ and \ RHS$\,=25$ \ \ RHS$\,=\,$% LHS $\checkmark $ %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion This does not mean that $\left( x+3\right) ^{2}+\left( y-1\right) ^{2}=25$ is the right equation. \ It only means that we didn't rule it out based on points $A$ and $B$. \ Let's see point $C$.\vspace{0.1in} %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion Point $C\left( 1,4\right) $ is on the graph if and only if its coordinates are a solution of $\left( x+3\right) ^{2}+\left( y-1\right) ^{2}=25$.\vspace{% 0.06in} %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion \qquad \qquad \qquad Check $\ \left( x+3\right) ^{2}+\left( y-1\right) ^{2}=25$ \ \ with $x=1$ and $y=4$. %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion \qquad \qquad \qquad LHS$\,=\left( 1+3\right) ^{2}+\left( 4-1\right) ^{2}=4^{2}+3^{2}=16+9=25$ \ and \ RHS$\,=25$ \ \ RHS$\,=\,$LHS $\checkmark \vspace{0.06in}$ %TCIMACRO{% %\TeXButton{\solinside}{\solinside %\setstretch{1.1}}}% %BeginExpansion \solinside \setstretch{1.1}% %EndExpansion We found that all three points are on the graph of this equation. \ This still does not mean that \newline $\left( x+3\right) ^{2}+\left( y-1\right) ^{2}=25$ is the right equation. \ Given that we were given three equations with the assumption that the correct equation is among them, it can only be this one. \ So, our answer is that the graph shown is of the equation $\left( x+3\right) ^{2}+\left( y-1\right) ^{2}=25$. \ We can find additional nice points on the graph (for example, $\left( 2,1\right) $ or $\left( -3,-4\right) $) and test them against the equation. \pagebreak %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion \FRAME{itbpF}{0.5794in}{0.601in}{0.2006in}{}{}{work.jpg}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 0.5794in;height 0.601in;depth 0.2006in;original-width 4.6977in;original-height 4.875in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'work.jpg';file-properties "XNPEU";}} \ \ \ {\Large Practice Problems} %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion %TCIMACRO{% %\TeXButton{\begin{minipage}{4.7in} hang 17}{\begin{minipage}{4.7in} %\setlength{\hangindent}{17pt}}}% %BeginExpansion \begin{minipage}{4.7in} \setlength{\hangindent}{17pt}% %EndExpansion 1. \ Consider the graph shown. \ Three points on the graph are marked. \ These are $A\left( -3,4\right) $, $B\left( 0,-2\right) $, and $C\left( 7,0\right) $. \ Use these points to determine, which of the given equations is the one whose graph is the shape we see. \vspace{0.03in} %TCIMACRO{% %\TeXButton{ par hang 17}{\setlength{\parindent}{17pt} %\setlength{\hangindent}{17pt}}}% %BeginExpansion \setlength{\parindent}{17pt} \setlength{\hangindent}{17pt}% %EndExpansion The possible equations offered are: \vspace{0.03in} $\qquad y+2=\left\vert 2x\right\vert $\vspace{0.03in} $\qquad 6-y=\left\vert 8-\left\vert 2x\right\vert \right\vert $\vspace{0.03in% } $\qquad x^{2}+y^{2}=1+4\left( x+y+5\right) $\ \vspace{0.5in}% %TCIMACRO{\TeXButton{\end{minipage}}{\end{minipage}}}% %BeginExpansion \end{minipage}% %EndExpansion %TCIMACRO{% %\TeXButton{\begin{minipage}{3.3in}}{\begin{minipage}{3.3in} %\setlength{\leftskip}{24pt}}}% %BeginExpansion \begin{minipage}{3.3in} \setlength{\leftskip}{24pt}% %EndExpansion \FRAME{itbpF}{2.5097in}{2.5097in}{0in}{}{}{Plot}{\special{language "Scientific Word";type "MAPLEPLOT";width 2.5097in;height 2.5097in;depth 0in;display "USEDEF";plot_snapshots TRUE;mustRecompute FALSE;lastEngine "MuPAD";xmin "-10.01";xmax "10.04";xviewmin "-10.01";xviewmax "10.04";yviewmin "-10.4";yviewmax "10.4";viewset"XY";rangeset"X";plottype 4;plotticks 1;num-x-ticks 11;num-y-ticks 11;constrained TRUE;numpoints 100;plotstyle 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\TEXUX{$-\sqrt{0.04-\left( x-7\right) ^{2}}$};linecolor "blue";linestyle 1;pointstyle "point";linethickness 3;lineAttributes "Solid";var1range "6.8,7.2";num-x-gridlines 100;curveColor "[flat::RGB:0x000000ff]";curveStyle "Line";rangeset"X";function \TEXUX{$\sqrt{0.04-\left( x-7\right) ^{2}}$};linecolor "blue";linestyle 1;pointstyle "point";linethickness 3;lineAttributes "Solid";var1range "6.8,7.2";num-x-gridlines 100;curveColor "[flat::RGB:0x000000ff]";curveStyle "Line";rangeset"X";valid_file "T";tempfilename 'PGRYKM06.wmf';tempfile-properties "XPR";}}% %TCIMACRO{\TeXButton{\end{minipage}}{\end{minipage}}}% %BeginExpansion \end{minipage}% %EndExpansion \bigskip %TCIMACRO{% %\TeXButton{\begin{minipage}{4.7in} hang 17}{\begin{minipage}{4.7in} %\setlength{\hangindent}{17pt}}}% %BeginExpansion \begin{minipage}{4.7in} \setlength{\hangindent}{17pt}% %EndExpansion 2. \ Consider the graph shown. \ Three points on the graph are marked. \ These are $A\left( -3,3\right) $,\ $B\left( 1,5\right) $, and $C\left( 4,-2\right) $. \ Use these points to determine, which of the given equations is the one whose graph is the shape we see. \vspace{0.03in} %TCIMACRO{% %\TeXButton{ par hang 17}{\setlength{\parindent}{17pt} %\setlength{\hangindent}{17pt}}}% %BeginExpansion \setlength{\parindent}{17pt} \setlength{\hangindent}{17pt}% %EndExpansion The possible equations offered are: \vspace{0.03in} $\qquad 2y=x+9$\vspace{0.03in} $\qquad \left\vert x\right\vert +\left\vert y\right\vert =6$\vspace{0.03in} $\qquad y+3=9-\left\vert x\right\vert $\vspace{0.5in}% %TCIMACRO{\TeXButton{\end{minipage}}{\end{minipage}}}% %BeginExpansion \end{minipage}% %EndExpansion %TCIMACRO{% %\TeXButton{\begin{minipage}{3.3in}}{\begin{minipage}{3.3in} %\setlength{\leftskip}{24pt}}}% %BeginExpansion \begin{minipage}{3.3in} \setlength{\leftskip}{24pt}% %EndExpansion \FRAME{itbpF}{2.5097in}{2.5097in}{0in}{}{}{Plot}{\special{language "Scientific Word";type "MAPLEPLOT";width 2.5097in;height 2.5097in;depth 0in;display 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"1";cropbottom "0";filename 'answers.jpg';file-properties "XNPEU";}} \ \ \ {\Large Answers\vspace{0.2in}} %TCIMACRO{% %\TeXButton{\begin{minipage}{2.8in} hang 16}{\begin{minipage}{2.8in} %\setlength{\hangindent}{16pt} %\setstretch{1.2} %\setlength{\rightskip}{20pt}}}% %BeginExpansion \begin{minipage}{2.8in} \setlength{\hangindent}{16pt} \setstretch{1.2} \setlength{\rightskip}{20pt}% %EndExpansion 1. \ Consider $y+2=\left\vert 2x\right\vert $\vspace{0.03in} %TCIMACRO{% %\TeXButton{ par hang 16}{\setlength{\parindent}{16pt} %\setlength{\hangindent}{16pt}}}% %BeginExpansion \setlength{\parindent}{16pt} \setlength{\hangindent}{16pt}% %EndExpansion $A\left( -3,4\right) $ is on the graph, $6=6$ $\checkmark $ %TCIMACRO{% %\TeXButton{ par hang 16}{\setlength{\parindent}{16pt} %\setlength{\hangindent}{16pt}}}% %BeginExpansion \setlength{\parindent}{16pt} \setlength{\hangindent}{16pt}% %EndExpansion $B\left( 0,-2\right) $ is on the graph, $0=0$ $\checkmark $ %TCIMACRO{% %\TeXButton{ par hang 16}{\setlength{\parindent}{16pt} %\setlength{\hangindent}{16pt}}}% %BeginExpansion \setlength{\parindent}{16pt} \setlength{\hangindent}{16pt}% %EndExpansion $C\left( 7,0\right) $ is not on the graph, $2\not=14$ \ %TCIMACRO{% %\TeXButton{ par hang 16}{\setlength{\parindent}{16pt} %\setlength{\hangindent}{16pt}}}% %BeginExpansion \setlength{\parindent}{16pt} \setlength{\hangindent}{16pt}% %EndExpansion Therefore, $y+2=\left\vert 2x\right\vert $ is not the equation of the graph.% %TCIMACRO{\TeXButton{\end{minipage}}{\end{minipage}}}% %BeginExpansion \end{minipage}% %EndExpansion %TCIMACRO{% %\TeXButton{\begin{minipage}{3in} hang 16}{\begin{minipage}{3in} %\setlength{\hangindent}{16pt} %\setstretch{1.2} %\setlength{\leftskip}{22pt} %\setlength{\rightskip}{22pt}}}% %BeginExpansion \begin{minipage}{3in} \setlength{\hangindent}{16pt} \setstretch{1.2} \setlength{\leftskip}{22pt} \setlength{\rightskip}{22pt}% %EndExpansion Consider $6-y=\left\vert 8-\left\vert 2x\right\vert \right\vert $\vspace{% 0.03in} $A\left( -3,4\right) $ is on the graph, $2=2$ $\checkmark $ $B\left( 0,-2\right) $ is on the graph, $8=8$ $\checkmark $ $C\left( 7,0\right) $ is not on the graph, $6=6$ $\checkmark $ Therefore, $6-y=\left\vert 8-\left\vert 2x\right\vert \right\vert $ is the equation of the graph.% %TCIMACRO{\TeXButton{\end{minipage}}{\end{minipage}}}% %BeginExpansion \end{minipage}% %EndExpansion \vspace{0.2in} \qquad \qquad \qquad \qquad %TCIMACRO{% %\TeXButton{\begin{minipage}{3.8in} hang 16}{\begin{minipage}{3.8in} %\setlength{\hangindent}{16pt} %\setstretch{1.2} %\setlength{\rightskip}{20pt} %\setlength{\leftskip}{16pt}}}% %BeginExpansion \begin{minipage}{3.8in} \setlength{\hangindent}{16pt} \setstretch{1.2} \setlength{\rightskip}{20pt} \setlength{\leftskip}{16pt}% %EndExpansion Consider $x^{2}+y^{2}=1+4\left( x+y+5\right) $\vspace{0.03in} $A\left( -3,4\right) $ is on the graph, $25=25$ $\checkmark $ $B\left( 0,-2\right) $ is not on the graph, $4\not=13$ $C\left( 7,0\right) $ is on the graph, $49=49$ $\checkmark $ Therefore, $x^{2}+y^{2}=1+4\left( x+y+5\right) $ \newline is not the equation of the graph.% %TCIMACRO{\TeXButton{\end{minipage}}{\end{minipage}}}% %BeginExpansion \end{minipage}% %EndExpansion \bigskip %TCIMACRO{% %\TeXButton{\begin{minipage}{3.2in} hang 16}{\begin{minipage}{3.2in} %\setlength{\hangindent}{16pt} %\setstretch{1.2} %\setlength{\rightskip}{20pt}}}% %BeginExpansion \begin{minipage}{3.2in} \setlength{\hangindent}{16pt} \setstretch{1.2} \setlength{\rightskip}{20pt}% %EndExpansion 2. \ Consider $2y=x+9$\vspace{0.03in} %TCIMACRO{% %\TeXButton{ par hang 16}{\setlength{\parindent}{16pt} %\setlength{\hangindent}{16pt}}}% %BeginExpansion \setlength{\parindent}{16pt} \setlength{\hangindent}{16pt}% %EndExpansion $A\left( -3,3\right) $ is on the graph, $6=6$ $\checkmark $ %TCIMACRO{% %\TeXButton{ par hang 16}{\setlength{\parindent}{16pt} %\setlength{\hangindent}{16pt}}}% %BeginExpansion \setlength{\parindent}{16pt} \setlength{\hangindent}{16pt}% %EndExpansion $B\left( 1,5\right) $ is on the graph, $10=10$ $\checkmark $ %TCIMACRO{% %\TeXButton{ par hang 16}{\setlength{\parindent}{16pt} %\setlength{\hangindent}{16pt}}}% %BeginExpansion \setlength{\parindent}{16pt} \setlength{\hangindent}{16pt}% %EndExpansion $C\left( 4,-2\right) $ is not on the graph, $-4\not=13$ \ %TCIMACRO{% %\TeXButton{ par hang 16}{\setlength{\parindent}{16pt} %\setlength{\hangindent}{16pt}}}% %BeginExpansion \setlength{\parindent}{16pt} \setlength{\hangindent}{16pt}% %EndExpansion Therefore, $2y=x+9$ is not the equation of the graph.% %TCIMACRO{\TeXButton{\end{minipage}}{\end{minipage}}}% %BeginExpansion \end{minipage}% %EndExpansion %TCIMACRO{% %\TeXButton{\begin{minipage}{3in} hang 16}{\begin{minipage}{3in} %\setlength{\hangindent}{16pt} %\setstretch{1.2} %\setlength{\leftskip}{22pt} %\setlength{\rightskip}{22pt}}}% %BeginExpansion \begin{minipage}{3in} \setlength{\hangindent}{16pt} \setstretch{1.2} \setlength{\leftskip}{22pt} \setlength{\rightskip}{22pt}% %EndExpansion Consider $\left\vert x\right\vert +\left\vert y\right\vert =6$\vspace{0.03in} $A\left( -3,3\right) $ is on the graph, $6=6$ $\checkmark $ $B\left( 1,5\right) $ is on the graph, $6=6$ $\checkmark $ $C\left( 4,-2\right) $ is on the graph, $6=6$ $\checkmark $ Therefore, $\left\vert x\right\vert +\left\vert y\right\vert =6$ is probably the equation of the graph.% %TCIMACRO{\TeXButton{\end{minipage}}{\end{minipage}}}% %BeginExpansion \end{minipage}% %EndExpansion \vspace{0.2in} \qquad \qquad \qquad \qquad %TCIMACRO{% %\TeXButton{\begin{minipage}{3.8in} hang 16}{\begin{minipage}{3.8in} %\setlength{\hangindent}{16pt} %\setstretch{1.2} %\setlength{\rightskip}{20pt} %\setlength{\leftskip}{16pt}}}% %BeginExpansion \begin{minipage}{3.8in} \setlength{\hangindent}{16pt} \setstretch{1.2} \setlength{\rightskip}{20pt} \setlength{\leftskip}{16pt}% %EndExpansion Consider $y+3=9-\left\vert x\right\vert $\vspace{0.03in} $A\left( -3,3\right) $ is on the graph, $6=6$ $\checkmark $ $B\left( 1,5\right) $ is on the graph, $8=8$ $\checkmark $ $C\left( 4,-2\right) $ is not on the graph, $1\not=5$ $\checkmark $ Therefore, $y+3=9-\left\vert x\right\vert $ \newline is not the equation of the graph.% %TCIMACRO{\TeXButton{\end{minipage}}{\end{minipage}}}% %BeginExpansion \end{minipage}% %EndExpansion \vspace{0.2in}\vspace{2in} %TCIMACRO{\TeXButton{\small}{\small}}% %BeginExpansion \small% %EndExpansion \href{http://www.teaching.martahidegkuti.com/shared/lnotes/lecturenotes.html% }{For more documents like this, visit our page at\ http://www.teaching.martahidegkuti.com and click on Lecture Notes. \ E-mail questions or comments to mhidegkuti@ccc.edu.} \end{document}