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Fri Oct 7 16:42:13 2011 UTC (19 months, 2 weeks ago) by glarose
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LoyolaChicago 9.5: 4e tagging, updates.

    1 # DESCRIPTION
2 # Problem from Functions Modeling Change, Connally et al., 3rd ed.
3 # WeBWorK problem written by Adam Spiegler, <aspiegler@luc.edu>
4 # ENDDESCRIPTION
5
6 ## DBsubject('Precalculus')
7 ## DBchapter('Polynomial And Rational Functions')
8 ## DBsection('The Short-Run Behavior Of Rational Functions')
9 ## KEYWORDS('rational','fraction','polynomial,'asymptote','intercept')
10 ## TitleText1('Functions Modeling Change')
11 ## EditionText1('3')
12 ## AuthorText1('Connally')
13 ## Section1('9.5)
14 ## Problem1('36')
15 ## TitleText2('Functions Modeling Change');
16 ## EditionText2('4')
17 ## AuthorText2('Connally')
18 ## Section2('11.5')
19 ## Problem2('38')
20 ## Author('Adam Spiegler and Paul Pearson')
21 ## Institution('Loyola University Chicago and Fort Lewis College')
22
23 DOCUMENT();
24
26            "PGbasicmacros.pl",
27 #           "PGchoicemacros.pl",
29            "PGgraphmacros.pl",
30            "PGauxiliaryFunctions.pl",
32 "MathObjects.pl",
34 "PGcourse.pl",
35            );
36
37 TEXT(beginproblem());
38
39 Context("Numeric");
40
41 $showPartialCorrectAnswers = 1; 42 43$r = random(1,5,2);
44 $s = random(-6,-2,2); 45$x0 = $r+1; 46 47 48$top1 = $r+.01; 49$bot1 = $r - .01; 50$top2 = $s+.01; 51$bot2 = $s - .01; 52 53 #######$i is vertical reflection ############
54 $i = random(-1,1,2); 55 56 if ($i == -1){
57      valign1 = 'right';dx1 = -.1; $yvert2 = 1; 58$valign2 = 'left'; $dx2 = .1;$yvert1 = -1;
59      $disp_ans = "\frac{-x}{(x-$r)(x-$s)}"; 60$ans = "-x/((x-$r)(x-$s))";
61      $y0 = -$x0/($x0-$s);
62      $y0_bot =$x0-$s;$y0_top = -$x0; 63 64 ($n,$d) = reduce($y0_top,$y0_bot); 65$disp_y0 = ($d == 1 ) ?$n : "\frac{$n}{$d}";
66      $txt_y0 = ($d == 1 ) ? $n : "$n/$d"; 67 68$pos = "below"; $neg = "above";$sign = "negative";
69      $otb = 'top'} 70 else { 71$valign2 = 'left'; $dx2 = .1;$yvert1 = 1;
72      valign1 = 'right';dx1 = -.1; $yvert2 = -1; 73$disp_ans = "\frac{x}{(x-$r)(x-$s)}";
74      $ans = "x/((x-$r)(x-$s))"; 75$y0 = $x0/($x0-$s); 76$y0_bot = $x0-$s; $y0_top =$x0;
77
78      ($n,$d) = reduce($x0,$y0_bot);
79      $disp_y0 = ($d == 1 ) ? $n : "\frac{$n}{$d}"; 80$txt_y0 = ( $d == 1 ) ?$n : "$n/$d";
81
82      $pos = "above";$neg = "below"; $sign = "positive"; 83$otb = 'bottom'};
84
85 $f[0] = "$i*x/((x-$r)(x-$s)) for x in <-10,$bot2> using color:blue and weight:2"; 86$f[1] = "$i*x/((x-$r)(x-$s)) for x in <$top2,$bot1> using color:blue and weight:2"; 87$f[2] = "$i*x/((x-$r)(x-$s)) for x in <$top1,10> using color:blue and weight:2";
88
89 $graph = init_graph(-10,-2,10,2,'axes'=>[0,0],'ticks'=>[1,1]); 90$graph->lb('reset');
91 #for ($i = 1;$i <= 4; $i++) { 92 #$graph->lb(new Label(2*$i,-.1,2*$i,'black','center','top'));
93 #   $graph->lb(new Label(-2*$i,-.1,-2*$i,'black','center','top')); 94 #$graph->lb(new Label(-.1,2*$i,2*$i,'black','right','middle'));
95 #   $graph->lb(new Label(-.1,-2*$i,-2*$i,'black','right','middle'))}; 96$graph->lb(new Label($s+$dx1,$yvert1,"x=$s",'red',valign1,'bottom')); 97graph->lb(new Label($r+$dx2,$yvert2,"x=$r",'red',valign2,'bottom')); 98graph->lb(new Label(9.8,0.1,"x",'black','right','bottom'));
99 $graph->lb(new Label(-.1,1.9,"y",'black','right','top')); 100$graph->lb(new Label($x0+.1,$y0,"($x0,$txt_y0)",'black','left',$otb)); 101$graph->moveTo($r,10); 102$graph->lineTo($r,-10,'red'); 103$graph->moveTo($s,10); 104$graph->lineTo($s,-10,'red'); 105$point[0] = closed_circle( 0,0, black );
106 $point[1] = closed_circle($x0,$y0, black ); 107$graph -> stamps(@point);
108 plot_functions( $graph,$f[0],$f[1],$f[2] );
109 $fig = image(insertGraph($graph), width => 400, height => 400, tex_size => 700);
110
111 Context()->texStrings;
112 BEGIN_TEXT
113 Find a possible formula for the function graphed below.  Assume the function has only one $$x$$-intercept at the origin, and the point marked on the graph below is located at $$\left( x0, disp_y0 \right)$$.  The asymptotes are $$x = s$$ and $$x = r$$.  Give your formula as a reduced rational function.
114 $PAR 115 $$f(x) =$$ \{ ans_rule(30) \} 116 \{ AnswerFormatHelp("formulas") \} 117$PAR
118 $BCENTER 119$fig
120 $BR 121 (Click on graph to enlarge) 122$ECENTER
123 END_TEXT
124 Context()->normalStrings;
125
126 ANS( Compute("$ans")->cmp() ); 127 128 #ANS(fun_cmp($ans, vars=>'x' ) );
129
130
131 Context()->texStrings;
132 SOLUTION(EV3(<<'END_SOLUTION'));
133 $PAR 134$BBOLD  SOLUTION $EBOLD 135$PAR
136 Since the graph has a vertical asymptotes at $$x=r$$ and $$x = s$$, let the denominator be $$(x-r)(x-s)$$.
137 $BR 138 Since the graph a zero at $$x= 0$$ let the numerator be $$x$$. 139$BR
140 Since the long-run behavior tends toward $$y = 0$$ as $$x \to \pm \infty$$, the degree of the numerator must be less than the degree of the denominator, which is true based on our work.
141 $BR 142 Since for large positive values of $$x$$ the graph approaches the $$x$$-axis from$pos while for large negative values of $$x$$ the graph approaches the $$x$$-axis from $neg, the leading coefficient must be$sign.
143 \$PAR
144 So a possible formula is $$\displaystyle y = f(x) = disp_ans$$. You can check that the when $$x = x0$$ we have $$y = disp_y0$$, as we should.
145
146 END_SOLUTION
147 Context()->normalStrings;
148
149
150 COMMENT('MathObject version');
151 ENDDOCUMENT();