#### Graphing f(x) = ax²

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###### Monitoring Progress
Exercise name Free?
Monitoring Progress 1
Monitoring Progress 2
Monitoring Progress 3
Monitoring Progress 4
Monitoring Progress 5
Monitoring Progress 6
Monitoring Progress 7
Monitoring Progress 8
Monitoring Progress 9
###### Exercises
Exercise name Free?
Exercises 1 Let's review some of the main characteristics of the graph of a quadratic function. A quadratic function is nonlinear. Its values increase faster as ∣x∣ becomes greater, and it always has an axis of symmetry. These characteristics cause it to have a U-shape. This special graph is called a parabola.This shape has special geometric properties that make it useful for many everyday devices, such as the famous parabolic antennas and the mirrors used in car lights.
Exercises 2 Let's start by reviewing the graph of the quadratic parent function y=x2.Notice that this function opens up. Nevertheless, this function can be transformed by multiplying by a factor a, obtaining the offspring function y=ax2. If a<0 the function gets reflected in the x-axis, causing it to open downwards. If a>0 the function opens up, just as the parent function. Notice that the parameter a on the graph of y=ax2 can also cause a vertical stretch or vertical shrink.
Exercises 3 By paying close attention to the given graph, we can identify key characteristics such as the vertex, axis of symmetry, domain, range, and behavior.From the graph we can note the following.The vertex is the point (1,-1). The axis of symmetry is the vertical line x=1. The domain is all real numbers. The range is y≤-1. When x<1, the y-variable increases as the x-variable increases. When x>1, the y-variable decreases as the x-variable increases.
Exercises 4 By paying close attention to the given graph, we can identify key characteristics such as the vertex, axis of symmetry, domain, range, and behavior.From the graph we can note the following.The vertex is the point (-2,4). The axis of symmetry is the vertical line x=-2. The domain is all real numbers. The range is y≥4. When x>-2, the y-variable increases as the x-variable increases. When x<-2, the y-variable decreases as the x-variable increases.
Exercises 5 To graph the function we will make a table of values.x6x2g(x)=6x2 -16(-1)26 -0.56(-0.5)21.5 05(0)20 0.56(0.5)21.5 16(1)26 Let's now draw the parabola that connects the obtained points. We will also draw the parent function f(x)=x2.From the graph above, we can note the following.Both graphs open up. Both graphs have the same axis of symmetry x=0. The graph of the given function is narrower than the graph of the parent function. Both graphs have the same vertex (0,0). From the graph and the observations above, we can conclude that the graph of g is a vertical stretch by a factor of 6 of the graph of f.
Exercises 6 To graph the function we will make a table of values.x2.5x2b(x)=2.5x2 -22.5(-2)210 -12.5(-1)22.5 02.5(0)20 12.5(1)22.5 22.5(2)210 Let's now draw the parabola that connects the obtained points. We will also draw the parent function f(x)=x2.From the graph above, we can note the following.Both graphs open up. Both graphs have the same axis of symmetry x=0. The graph of the given function is narrower than the graph of the parent function. Both graphs have the same vertex (0,0). From the graph and the observations above, we can conclude that the graph of b is a vertical stretch by a factor of 2.5 of the graph of f.
Exercises 7 To graph the function we will make a table of values.x41​x2h(x)=41​x2 -441​(-4)24 -241​(-2)21 041​(0)20 241​(2)21 441​(4)24 Let's now draw the parabola that connects the obtained points. We will also draw the parent function f(x)=x2.From the graph above, we can note the following.Both graphs open up. Both graphs have the same axis of symmetry x=0. The graph of the given function is wider than the graph of the parent function. Both graphs have the same vertex (0,0). From the graph and the observations above, we can conclude that the graph of h is a vertical shrink by a factor of 41​ of the graph of f.
Exercises 8 To graph the function we will make a table of values.x0.75x2j(x)=0.75x2 -20.75(-2)23 -10.75(-1)20.75 00.75(0)20 10.75(1)20.75 20.75(2)23 Let's now draw the parabola that connects the obtained points. We will also draw the parent function f(x)=x2.From the graph above, we can note the following.Both graphs open up. Both graphs have the same axis of symmetry x=0. The graph of the given function is wider than the graph of the parent function. Both graphs have the same vertex (0,0). From the graph and the observations above, we can conclude that the graph of j is a vertical shrink by a factor of 0.75 of the graph of f.
Exercises 9 To graph the function we will make a table of values.x-2x2p(x)=-2x2 -2-2(-2)2-8 -1-2(-1)2-2 0-2(0)20 1-2(1)2-2 2-2(2)2-8 Let's now draw the parabola that connects the obtained points. We will also draw the parent function f(x)=x2.From the graph above, we can note the following.The graph of the given function opens down, and the graph of the parent function opens up. Both graphs have the same axis of symmetry x=0. The graph of the given function is narrower than the graph of the parent function. Both graphs have the same vertex (0,0). From the graph and the observations above, we can conclude that the graph of m is a vertical stretch by a factor of 2 followed by a reflection in the x-axis of the graph of f.
Exercises 10 To graph the function we will make a table of values.x-29​x2q(x)=-29​x2 -34​-29​(-34​)2-8 -1-29​(-1)2-4.5 0-29​(0)20 1-29​(1)2-4.5 34​-29​(34​)2-8 Let's now draw the parabola that connects the obtained points. We will also draw the parent function f(x)=x2.From the graph above, we can note the following.The graph of the given function opens down, and the graph of the parent function opens up. Both graphs have the same axis of symmetry x=0. The graph of the given function is narrower than the graph of the parent function. Both graphs have the same vertex (0,0). From the graph and the observations above, we can conclude that the graph of q is a vertical stretch by a factor of 29​ and the reflection in the x-axis of the graph of f.
Exercises 11 To graph the function we will make a table of values.x-0.2x2q(x)=-0.2x2 -4-0.2(-4)2-3.2 -2-0.2(-2)2-0.8 0-0.2(0)20 2-0.2(2)2-0.8 4-0.2(4)2-3.2 Let's now draw the parabola that connects the obtained points. We will also draw the parent function f(x)=x2.From the graph above, we can note the following.The graph of the given function opens down, and the graph of the parent function opens up. Both graphs have the same axis of symmetry x=0. The graph of the given function is wider than the graph of the parent function. Both graphs have the same vertex (0,0). From the graph and the observations above, we can conclude that the graph of k is a vertical shrink by a factor of 0.2 followed by a reflection in the x-axis of the graph of f.
Exercises 12 To graph the function we will make a table of values.x-32​x2p(x)=-32​x2 -3-32​(-3)2-6 -1-32​(-1)2-0.67 0-32​(0)20 1-32​(1)2-0.67 3-41​(3)2-6 Let's now draw the parabola that connects the obtained points. We will also draw the parent function f(x)=x2.From the graph above, we can note the following.The graph of the given function opens down, and the graph of the parent function opens up. Both graphs have the same axis of symmetry x=0. The graph of the given function is wider than the graph of the parent function. Both graphs have the same vertex (0,0). From the graph and the observations above, we can conclude that the graph of p is a vertical shrink by a factor of 32​ followed by the reflection in the x-axis of the graph of f.
Exercises 13 Let's enter the function in the calculator by pushing Y=​ and typing it in the first row.Next, by pushing GRAPH​, the calculator will draw the graph of the equation.We see that the graph of y=4x2 is a reflection in the x-axis of the graph of y=-4x2.Showing Our Work info Graphing y=-4x2 Let's enter y=-4x2 in the calculator by pushing Y=​ and typing it in the first row.Next, by pushing GRAPH​, the calculator will draw the graph of the equation.
Exercises 14 Let's enter the function in the calculator by pushing Y=​ and typing it in the first row.Next, by pushing GRAPH​, the calculator will draw the graph of the equation.We see that the graph of y=-0.4x2 is a vertical shrink by a factor of 101​ of the graph of y=-4x2.Showing Our Work info Graphing y=-4x2 Let's enter y=-4x2 in the calculator by pushing Y=​ and typing it in the first row.Next, by pushing GRAPH​, the calculator will draw the graph of the equation.
Exercises 15 Let's enter the function in the calculator by pushing Y=​ and typing it in the first row.Next, by pushing GRAPH​, the calculator will draw the graph of the equation.We see that the graph of y=-0.4x2 is a vertical shrink by a factor of 1001​ of the graph of y=-4x2.Showing Our Work info Graphing y=-4x2 Let's enter y=-4x2 in the calculator by pushing Y=​ and typing it in the first row.Next, by pushing GRAPH​, the calculator will draw the graph of the equation.
Exercises 16 Let's enter the function in the calculator by pushing Y=​ and typing it in the first row.Next, by pushing GRAPH​, the calculator will draw the graph of the equation.We may want to zoom out by changing the window settings. To do this, push WINDOW​ and change the settings to a much bigger window. Then push GRAPH​ once more to draw the equation with these new settings.We see that the graph of y=-0.004x2 is a vertical shrink by a factor of 10001​ of the graph of y=-4x2.Showing Our Work info Graphing y=-4x2 Let's enter y=-4x2 in the calculator by pushing Y=​ and typing it in the first row.Next, by pushing GRAPH​, the calculator will draw the graph of the equation.To see the difference between the graphs, let's zoom out by changing the window settings. To do this, push WINDOW​ and change the settings to a much bigger window. Then push GRAPH​ once more to draw the equation with these new settings.
Exercises 17 The graph below shows how the graph of y=ax2 changes when a changes.window.JXQ = window.JXQ || []; window.JXQtable = window.JXQtable || {}; if(!window.JXQtable["Solution20551_1_779560014_l"]) { window.JXQ.push( function () { var code = function (gid) { var images = []; try { var b = mlg.board([-5.5,16.5,5.5,-4.5],{desktopSize:'medium',style:'usa',yScale:3}); b.xaxis(1,0,'x'); b.yaxis(3,0,'y'); var p = b.node(0,1); var m = b.node(1,0); var f1 = b.func('x^2',{opacity:0.6}); b.legend(f1,[-2.5,6],'y=x^2',{}); var graph = b.board.create('functiongraph', [function(x){ return p.Y()*JXG.Math.pow(x*m.X()*m.X(),2);},-10, 10],{strokeWidth:2,doAdvancedPlot:false,numberPointsLow:150,numberPointsHigh:150});var n1 = b.node(-1.5,2.25); var n2 = b.node(-1.5,function(){return graph.Y(-1.5);}); b.arrow(n1,n2,{strokeWidth:1.5});var n3 = b.node(1.5,2.25); var n4 = b.node(1.5,function(){return graph.Y(1.5);}); b.arrow(n3,n4,{strokeWidth:1.5});var l = b.txt(2.45,function(){return graph.Y(2.45);},'y=(' + p.Y().toFixed(2) + ')x^2',{flag:true}); (b.getId(l)).css({ "min-width":"28%", "text-align":"center", "padding":"2px", "background-color":"rgb(175, 175, 255)", "border":"1px solid blue" });mlg.af('villkorgraph645.k1', function(sliderValue){ p.moveTo([0,sliderValue]); if (sliderValue > 1) { b.changeText(l,'y=('+ (sliderValue).toFixed(2) + ')x^2'); graph.setAttribute({strokeColor:'green'});(b.getId(l)).css({ "min-width":"28%", "text-align":"center", "padding":"2px", "background-color":"rgb(51, 204, 51)", "border":"1px solid green"}); } else { if(sliderValue < 1 ){ b.changeText(l,'y=('+ (sliderValue).toFixed(2) + ')x^2'); graph.setAttribute({strokeColor:'red'}); (b.getId(l)).css({ "min-width":"28%", "text-align":"center", "padding":"2px", "background-color":"rgb(255,175,175)", "border":"1px solid red"}); } } });} catch( e ) { mw.hook("ml.error.report").fire(e, { tags: { module: "jsxg", graph: "Solution20551_1_779560014_l" } }); console.error(e); } if(b) {b.board.update()} }; mw.loader.using('ext.MLJSXGraph', function() { window.ml.jsxgraph.buildRealGraph("Solution20551_1_779560014_l", "Solution20551_1_779560014_p", 1, code); }); } ); } window.JXQtable["Solution20551_1_779560014_l"] = true;window.JXQ = window.JXQ || []; window.JXQtable = window.JXQtable || {}; if(!window.JXQtable["Solution20551_1-slider_1548391487_l"]) { window.JXQ.push( function () { var code = function (gid) { var images = []; try { var b = mlg.board([0.15,0.5,2.5,-0.05],{desktopSize:'medium'}); var s = b.slider(1,null,null,{title:{label:'Choose ' + 'a'.italics() + '-value'},snapWidth:-1,precision:2}); var snapMode = false; s.slider.on('drag',function(){ if(Math.abs(s.slider.Value()) < 0.15){ if(!snapMode){ s.slider.setAttribute({snapWidth:0.15}); snapMode = true; } }else{ if(snapMode){ s.slider.setAttribute({snapWidth:-1}); snapMode = false; } } mlg.cf("villkorgraph645.k1", s.slider.Value()); });} catch( e ) { mw.hook("ml.error.report").fire(e, { tags: { module: "jsxg", graph: "Solution20551_1-slider_1548391487_l" } }); console.error(e); } if(b) {b.board.update()} }; mw.loader.using('ext.MLJSXGraph', function() { window.ml.jsxgraph.buildRealGraph("Solution20551_1-slider_1548391487_l", "Solution20551_1-slider_1548391487_p", 1, code); }); } ); } window.JXQtable["Solution20551_1-slider_1548391487_l"] = true;From the graph above, we can deduce the following facts.When ∣a∣<1, the graph y=ax2 is a vertical shrink of the graph of y=x2. When ∣a∣>1, the graph y=ax2 is a a vertical stretch of the graph of y=x2.With these in mind, the graph of y=0.5x2 should be wider than the graph of y=x2 since 0.5<1.We see that the graphs have the same vertex (0,0) and the same axis of symmetry, y-axis. Exercises 18 We will start by determining the domain and range of the function that models the arch support of a bridge.We see that the leftmost point on the graph is (-500,-300), and the rightmost point is (500,-300.) The x-coordinates of these points determine the domain. Domain: -500≤x≤500​ The domain of the function represents the width of the arch, so it is 1000 ft. 500−(-500)=1000​ To find the height, we will use the range. The highest point on the graph is (0,0), and the lowest points on the graph are (-500,-300) and (500,-300). Therefore, the range is -300≤y≤0. Range: -300≤y≤0​ The range of the function represents the height of the arch, so it is 300 ft. 0−(-300)=300​ Exercises 19 Exercises 20 Exercises 21 To find the increasing interval, we will scan the graph from left to right.The violet arrows shows increasing intervals of the functions. From left side of graph to x=0From x=0 to right side of graph​→→​​g is increasingf is increasing​ For all x-values less than x=0, g will be increasing. Additionally, for all x-values greater than x=0, f will be increasing.FunctionIncreasing Interval f(x)x>0 g(x)x<0 Exercises 22 To find the decreasing interval, we will scan the graph from left to right.The orange arrows shows decreasing intervals of the functions. From left side of graph to x=0From x=0 to right side of graph​→→​​f is decreasingg is decreasing​ For all x-values less than x=0, f will be decreasing. Additionally, for all x-values greater than x=0, g will be decreasing.FunctionDecreasing Interval f(x)x<0 g(x)x>0 Exercises 23 The point (-2,3) should be in the second quadrant, since the x-coordinate of the point is negative and the y-coordinate of it is positive. Hence, the function f(x) could include the point.Let's substitute the point into the function f(x)=ax2. f(x)=ax2x=-2, y=33=a(-2)2 Solve for a Calculate power3=a(4)LHS/4=RHS/443​=aRearrange equation a=43​ The value of a is 43​ when the graph passes through (-2,3). Exercises 24 Recall that the x-intercept is the x-coordinate of the point where y=0. Since we are considering the expression y=ax2, the x-intercept will be the value satisfying the equation shown below. 0=ax2​ Remember that a cannot be zero, otherwise we would have a constant zero function instead of a quadratic one. Therefore, the above equation is only satisfied when x=0. Another way to interpret this result is that the graph of any function of the form y=ax2 passes through the origin. Exercises 25 Let's start by plotting the points (-4,2) and (6,-3). We are told that the parabola opens upward, and therefore its vertex is the minimum point of the curve.We see above that the point (6,-3) is below (-4,2). Therefore, (-4,2) is not the minimum point of a parabola that opens upward and also passes through (6,-3). This means it is not the vertex. Exercises 26 Let's examine the graph below for the different values of a, greater than zero.window.JXQ = window.JXQ || []; window.JXQtable = window.JXQtable || {}; if(!window.JXQtable["Solution20560_1_1572907954_l"]) { window.JXQ.push( function () { var code = function (gid) { var images = []; try { var b = mlg.board([-5.5,16.5,5.5,-4.5],{desktopSize:'medium',style:'usa',yScale:3}); b.xaxis(1,0,'x'); b.yaxis(3,0,'y'); var p = b.node(0,1); var m = b.node(1,0); var f1 = b.func('x^2',{opacity:0.6}); b.legend(f1,[-3.2,9],'g(x)=x^2',{}); var graph = b.board.create('functiongraph', [function(x){ return p.Y()*JXG.Math.pow(x*m.X()*m.X(),2);},-10, 10],{strokeWidth:2,doAdvancedPlot:false,numberPointsLow:150,numberPointsHigh:150});var n1 = b.node(-1.5,2.25); var n2 = b.node(-1.5,function(){return graph.Y(-1.5);}); b.arrow(n1,n2,{strokeWidth:1.5});var n3 = b.node(1.5,2.25); var n4 = b.node(1.5,function(){return graph.Y(1.5);}); b.arrow(n3,n4,{strokeWidth:1.5});var l = b.txt(2.45,function(){return graph.Y(2.45);},'f(x)=(' + p.Y().toFixed(2) + ')x^2',{flag:true});(b.getId(l)).css({ "min-width":"28%", "text-align":"center", "padding":"2px", "background-color":"rgb(175, 175, 255)", "border":"1px solid blue" });mlg.af('villkorgraph645.k1', function(sliderValue){ p.moveTo([0,sliderValue]); if (sliderValue > 1) { b.changeText(l,'f(x)=('+ (sliderValue).toFixed(2) + ')x^2'); graph.setAttribute({strokeColor:'green'}); (b.getId(l)).css({ "min-width":"28%", "text-align":"center", "padding":"2px", "background-color":"rgb(51, 204, 51)", "border":"1px solid green"}); } else { if(sliderValue < 1 ){ b.changeText(l,'f(x)=('+ (sliderValue).toFixed(2) + ')x^2'); graph.setAttribute({strokeColor:'red'});(b.getId(l)).css({ "min-width":"28%", "text-align":"center", "padding":"2px", "background-color":"rgb(255,175,175)", "border":"1px solid red"}); } } });} catch( e ) { mw.hook("ml.error.report").fire(e, { tags: { module: "jsxg", graph: "Solution20560_1_1572907954_l" } }); console.error(e); } if(b) {b.board.update()} }; mw.loader.using('ext.MLJSXGraph', function() { window.ml.jsxgraph.buildRealGraph("Solution20560_1_1572907954_l", "Solution20560_1_1572907954_p", 1, code); }); } ); } window.JXQtable["Solution20560_1_1572907954_l"] = true;window.JXQ = window.JXQ || []; window.JXQtable = window.JXQtable || {}; if(!window.JXQtable["Solution20560_1-slider_857627496_l"]) { window.JXQ.push( function () { var code = function (gid) { var images = []; try { var b = mlg.board([0.15,0.5,2.5,-0.05],{desktopSize:'medium'}); var s = b.slider(1,null,null,{title:{label:'Choose ' + 'a'.italics() + '-value'},snapWidth:-1,precision:2}); var snapMode = false; s.slider.on('drag',function(){ if(Math.abs(s.slider.Value()) < 0.15){ if(!snapMode){ s.slider.setAttribute({snapWidth:0.15}); snapMode = true; } }else{ if(snapMode){ s.slider.setAttribute({snapWidth:-1}); snapMode = false; } } mlg.cf("villkorgraph645.k1", s.slider.Value()); });} catch( e ) { mw.hook("ml.error.report").fire(e, { tags: { module: "jsxg", graph: "Solution20560_1-slider_857627496_l" } }); console.error(e); } if(b) {b.board.update()} }; mw.loader.using('ext.MLJSXGraph', function() { window.ml.jsxgraph.buildRealGraph("Solution20560_1-slider_857627496_l", "Solution20560_1-slider_857627496_p", 1, code); }); } ); } window.JXQtable["Solution20560_1-slider_857627496_l"] = true;We see that the graph of f(x)=ax2 is narrower than the graph of g(x)=x2 only if a>1. For 0<a<1, the graph is wider. Therefore, the graph of f is sometimes narrower than the graph of the other function.
Exercises 27 Let's examine the graph below.window.JXQ = window.JXQ || []; window.JXQtable = window.JXQtable || {}; if(!window.JXQtable["Solution20561_1_1786169528_l"]) { window.JXQ.push( function () { var code = function (gid) { var images = []; try { var b = mlg.board([-5.5,16.5,5.5,-4.5],{desktopSize:'medium',style:'usa',yScale:3}); b.xaxis(1,0,'x'); b.yaxis(3,0,'y'); var p = b.node(0,1); var m = b.node(1,0); var f1 = b.func('x^2',{opacity:0.6}); b.legend(f1,[-3.2,9],'g(x)=x^2',{}); var graph = b.board.create('functiongraph', [function(x){ return p.Y()*JXG.Math.pow(x*m.X()*m.X(),2);},-10, 10],{strokeWidth:2,doAdvancedPlot:false,numberPointsLow:150,numberPointsHigh:150});var n1 = b.node(-1.5,2.25); var n2 = b.node(-1.5,function(){return graph.Y(-1.5);}); b.arrow(n1,n2,{strokeWidth:1.5});var n3 = b.node(1.5,2.25); var n4 = b.node(1.5,function(){return graph.Y(1.5);}); b.arrow(n3,n4,{strokeWidth:1.5});var l = b.txt(2.45,function(){return graph.Y(2.45);},'f(x)=(' + p.Y().toFixed(2) + ')x^2',{flag:true}); (b.getId(l)).css({ "min-width":"28%", "text-align":"center", "padding":"2px", "background-color":"rgb(175, 175, 255)", "border":"1px solid blue" });mlg.af('villkorgraph645.k1', function(sliderValue){ p.moveTo([0,sliderValue]); if (sliderValue > 1) { b.changeText(l,'f(x)=('+ (sliderValue).toFixed(2) + ')x^2'); graph.setAttribute({strokeColor:'green'});(b.getId(l)).css({ "min-width":"28%", "text-align":"center", "padding":"2px", "background-color":"rgb(51, 204, 51)", "border":"1px solid green"}); }else{ if (sliderValue===1) { graph.setAttribute({strokeColor:'blue'}); (b.getId(l)).css({ "min-width":"28%", "text-align":"center", "padding":"2px", "background-color":"rgb(175, 175, 255)", "border":"1px solid blue" }); } } });} catch( e ) { mw.hook("ml.error.report").fire(e, { tags: { module: "jsxg", graph: "Solution20561_1_1786169528_l" } }); console.error(e); } if(b) {b.board.update()} }; mw.loader.using('ext.MLJSXGraph', function() { window.ml.jsxgraph.buildRealGraph("Solution20561_1_1786169528_l", "Solution20561_1_1786169528_p", 1, code); }); } ); } window.JXQtable["Solution20561_1_1786169528_l"] = true;window.JXQ = window.JXQ || []; window.JXQtable = window.JXQtable || {}; if(!window.JXQtable["Solution20561_1-slider_556292055_l"]) { window.JXQ.push( function () { var code = function (gid) { var images = []; try { var b = mlg.board([1,0.2,2.5,-0.1],{desktopSize:'medium'}); var s = b.slider(1,null,null,{title:{label:'Choose ' + 'a'.italics() + '-value'},snapWidth:-1,precision:2}); var snapMode = false; s.slider.on('drag',function(){ if(Math.abs(s.slider.Value()) < 1.001){ if(!snapMode){ s.slider.setAttribute({snapWidth:0.001}); snapMode = true; } }else{ if(snapMode){ s.slider.setAttribute({snapWidth:-1}); snapMode = false; } } mlg.cf("villkorgraph645.k1", s.slider.Value()); });} catch( e ) { mw.hook("ml.error.report").fire(e, { tags: { module: "jsxg", graph: "Solution20561_1-slider_556292055_l" } }); console.error(e); } if(b) {b.board.update()} }; mw.loader.using('ext.MLJSXGraph', function() { window.ml.jsxgraph.buildRealGraph("Solution20561_1-slider_556292055_l", "Solution20561_1-slider_556292055_p", 1, code); }); } ); } window.JXQtable["Solution20561_1-slider_556292055_l"] = true;We see that the graph of f(x)=ax2 is narrower than the graph of g(x)=x2 when a>1. It is also true for the reflection in the x-axis of the graphs of the functions. In other words, it is true for all a<-1. Hence, the graph of f is always narrower. Exercises 28 Let's examine the graph below.window.JXQ = window.JXQ || []; window.JXQtable = window.JXQtable || {}; if(!window.JXQtable["Solution20562_1_709188695_l"]) { window.JXQ.push( function () { var code = function (gid) { var images = []; try { var b = mlg.board([-5.5,16.5,5.5,-4.5],{desktopSize:'medium',style:'usa',yScale:3}); b.xaxis(1,0,'x'); b.yaxis(3,0,'y'); var p = b.node(0,1); var m = b.node(1,0); var f1 = b.func('x^2',{opacity:0.6}); b.legend(f1,[-3.2,9],'g(x)=x^2',{}); var graph = b.board.create('functiongraph', [function(x){ return p.Y()*JXG.Math.pow(x*m.X()*m.X(),2);},-10, 10],{strokeWidth:2,doAdvancedPlot:false,numberPointsLow:150,numberPointsHigh:150});var n1 = b.node(-2.5,6.25); var n2 = b.node(-2.5,function(){return graph.Y(-2.5);}); b.arrow(n1,n2,{strokeWidth:1.5});var n3 = b.node(2.5,6.25); var n4 = b.node(2.5,function(){return graph.Y(2.5);}); b.arrow(n3,n4,{strokeWidth:1.5});var l = b.txt(2.45,function(){return graph.Y(2.45);},'f(x)=(' + p.Y().toFixed(2) + ')x^2',{flag:true});(b.getId(l)).css({ "min-width":"28%", "text-align":"center", "padding":"2px", "background-color":"rgb(175, 175, 255)", "border":"1px solid blue" });mlg.af('villkorgraph645.k1', function(sliderValue){ p.moveTo([0,sliderValue]); if (sliderValue < 1) { b.changeText(l,'f(x)=('+ (sliderValue).toFixed(2) + ')x^2'); graph.setAttribute({strokeColor:'red'}); (b.getId(l)).css({ "min-width":"28%", "text-align":"center", "padding":"2px", "background-color":"rgb(255,175,175)", "border":"1px solid red"}); }else{ if (sliderValue===1) { graph.setAttribute({strokeColor:'blue'});(b.getId(l)).css({ "min-width":"28%", "text-align":"center", "padding":"2px", "background-color":"rgb(175, 175, 255)", "border":"1px solid blue" }); } } });} catch( e ) { mw.hook("ml.error.report").fire(e, { tags: { module: "jsxg", graph: "Solution20562_1_709188695_l" } }); console.error(e); } if(b) {b.board.update()} }; mw.loader.using('ext.MLJSXGraph', function() { window.ml.jsxgraph.buildRealGraph("Solution20562_1_709188695_l", "Solution20562_1_709188695_p", 1, code); }); } ); } window.JXQtable["Solution20562_1_709188695_l"] = true;window.JXQ = window.JXQ || []; window.JXQtable = window.JXQtable || {}; if(!window.JXQtable["Solution20562_1-slider_302854787_l"]) { window.JXQ.push( function () { var code = function (gid) { var images = []; try { var b = mlg.board([0.01,0.1,1,-0.1],{desktopSize:'medium'}); var s = b.slider(1,null,null,{title:{label:'Choose ' + 'a'.italics() + '-value'},snapWidth:-1,precision:2}); var snapMode = false; s.slider.on('drag',function(){ if(Math.abs(s.slider.Value() - 1) < 0.001){ if(!snapMode){ s.slider.setAttribute({snapWidth:0.001}); snapMode = true; } }else{ if(snapMode){ s.slider.setAttribute({snapWidth:-1}); snapMode = false; } } mlg.cf("villkorgraph645.k1", s.slider.Value()); });} catch( e ) { mw.hook("ml.error.report").fire(e, { tags: { module: "jsxg", graph: "Solution20562_1-slider_302854787_l" } }); console.error(e); } if(b) {b.board.update()} }; mw.loader.using('ext.MLJSXGraph', function() { window.ml.jsxgraph.buildRealGraph("Solution20562_1-slider_302854787_l", "Solution20562_1-slider_302854787_p", 1, code); }); } ); } window.JXQtable["Solution20562_1-slider_302854787_l"] = true;We see that the graph of f(x)=ax2 is wider than the graph of g(x)=x2 when 0<a<1. It is also true for the reflection in the x-axis of the graphs of the functions. In other words, it is true for all a-values between -1 and 0. Hence, the graph of f is always wider.
Exercises 29 We want to determine if the graph of f(x)=ax2 is wider than the graph of g(x)=dx2 when ∣a∣>∣d∣. To do so, we will use two facts.The graph of f(x)=ax2 is narrower than the graph of h(x)=x2 when ∣a∣>1. See this exercise. The graph of f(x)=ax2 is wider than the graph of h(x)=x2 when 0<∣a∣<1. See this exercise.We will examine the graphs in three cases.When ∣a∣>∣d∣>1 As a result of the first fact, the greater the a-value, the narrower the graph of a function of the form f(x)=ax2. Hence, the graph of f will be narrower than the graph of g when ∣a∣>∣d∣>1.When ∣a∣>1>∣d∣ Combining two facts, the graph of the function f is apparently narrower than the graph of g since ∣a∣ is greater than 1, and ∣d∣ is less than 1.When 1>∣a∣>∣d∣ As a result of the second fact, the smaller the a-value, the wider the graph of a function of the form f(x)=ax2. Hence, the graph of g will be wider than the graph of f when 1>∣a∣>∣d∣.We see that in all cases the graph of f is narrower than the graph of g. The cases above are also true for the reflection in the x-axis of the graphs of the functions. Hence, the graph of the function f is never wider than the graph of g.
Exercises 30 Let's draw the shape in the coordinate plane.window.JXQ = window.JXQ || []; window.JXQtable = window.JXQtable || {}; if(!window.JXQtable["Solution20564_2_655353856_l"]) { window.JXQ.push( function () { var code = function (gid) { var images = []; try { /*board elements*/ var b=mlg.board([-8.5,6.5,8.5,-6.5],{desktopSize:"medium",style:"usa"}); var el = {};/*default display*/el.n1 = b.node(6,-4,{name:"(6,\\text{-}4)",label:{position:25,distance:2}}); el.n2 = b.node(-6,-4,{name:"(\\text{-}6,\\text{-}4)",label:{position:155,distance:2}}); el.n3 = b.node(0,4,{name:"(0,4)",label:{position:20,distance:2}}); el.segment1 = b.segment(el.n3,el.n2); el.segment2 = b.segment(el.n2,el.n1); el.segment3 = b.segment(el.n1,el.n3); el.sp1 = b.segmentPoints(el.segment1,9,{visible:false}); el.sp2 = b.segmentPoints(el.segment3,9,{visible:false}); el.segment4 = b.segment(el.sp1[1],el.sp2[1],{strokeColor:"blue",strokeWidth:1}); el.segment5 = b.segment(el.sp2[7],el.sp1[7],{strokeColor:"blue",strokeWidth:1});/*axes*/ b.xaxis(6,2); b.yaxis(6,2,{name:"",name2:"y"});/*A*/ mlg.af("showA",function() { b.remove(el); el.n1 = b.node(6,-4,{name:"(6,\\text{-}4)",label:{position:25,distance:2}}); el.n2 = b.node(-6,-4,{name:"(\\text{-}6,\\text{-}4)",label:{position:155,distance:2}}); el.n3 = b.node(0,4,{name:"(0,4)",label:{position:20,distance:2}}); el.segment1 = b.segment(el.n3,el.n2); el.segment2 = b.segment(el.n2,el.n1); el.segment3 = b.segment(el.n1,el.n3); el.sp1 = b.segmentPoints(el.segment1,9,{visible:false}); el.sp2 = b.segmentPoints(el.segment3,9,{visible:false}); el.segment4 = b.segment(el.sp1[1],el.sp2[1],{strokeColor:"blue",strokeWidth:1}); el.segment5 = b.segment(el.sp2[7],el.sp1[7],{strokeColor:"blue",strokeWidth:1}); el.segment6 = b.segment(el.sp1[2],el.sp2[2],{strokeColor:"blue",strokeWidth:1}); el.segment7 = b.segment(el.sp1[6],el.sp2[6],{strokeColor:"blue",strokeWidth:1}); el.segment8 = b.segment(el.sp1[3],el.sp2[3],{strokeColor:"blue",strokeWidth:1}); el.segment9 = b.segment(el.sp2[5],el.sp1[5],{strokeColor:"blue",strokeWidth:1}); el.segment10 = b.segment(el.sp1[4],el.sp2[4],{strokeColor:"blue",strokeWidth:1}); });/*B*/ mlg.af("showB",function() { b.remove(el);el.n1 = b.point(6,-4,{name:"(6,\\text{-}4)",label:{position:25,distance:2}}); el.n2 = b.point(-6,-4,{name:"(\\text{-}6,\\text{-}4)",label:{position:155,distance:2}}); el.n3 = b.node(0,4,{name:"(0,4)",label:{position:20,distance:2}}); el.segment1 = b.segment(el.n3,el.n2,{opacity:0.4}); el.segment2 = b.segment(el.n2,el.n1,{opacity:0.4}); el.segment3 = b.segment(el.n1,el.n3,{opacity:0.4}); el.sp1 = b.segmentPoints(el.segment1,9,{visible:false}); el.sp2 = b.segmentPoints(el.segment3,9,{visible:false}); el.segment4 = b.segment(el.sp1[1],el.sp2[1],{strokeColor:"blue",strokeWidth:1,opacity:0.4}); el.segment5 = b.segment(el.sp2[7],el.sp1[7],{strokeColor:"blue",strokeWidth:1,opacity:0.4}); el.segment6 = b.segment(el.sp1[2],el.sp2[2],{strokeColor:"blue",strokeWidth:1,opacity:0.4}); el.segment7 = b.segment(el.sp1[6],el.sp2[6],{strokeColor:"blue",strokeWidth:1,opacity:0.4}); el.segment8 = b.segment(el.sp1[3],el.sp2[3],{strokeColor:"blue",strokeWidth:1,opacity:0.4}); el.segment9 = b.segment(el.sp2[5],el.sp1[5],{strokeColor:"blue",strokeWidth:1,opacity:0.4}); el.segment10 = b.segment(el.sp1[4],el.sp2[4],{strokeColor:"blue",strokeWidth:1,opacity:0.4});el.func1 = b.func("-1/9*x^2",{strokeColor:"red"}); });/*C*/ mlg.af("showC",function() { b.remove(el); el.n1 = b.node(6,-4,{name:"(6,\\text{-}4)",label:{position:25,distance:2}}); el.n2 = b.node(-6,-4,{name:"(\\text{-}6,\\text{-}4)",label:{position:155,distance:2}}); el.n3 = b.node(0,4,{name:"(0,4)",label:{position:20,distance:2}}); el.segment1 = b.segment(el.n3,el.n2); el.segment2 = b.segment(el.n2,el.n1); el.segment3 = b.segment(el.n1,el.n3); el.sp1 = b.segmentPoints(el.segment1,9,{visible:false}); el.sp2 = b.segmentPoints(el.segment3,9,{visible:false}); el.segment4 = b.segment(el.sp1[1],el.sp2[1],{strokeColor:"blue",strokeWidth:1}); el.segment5 = b.segment(el.sp2[7],el.sp1[7],{strokeColor:"blue",strokeWidth:1}); });} catch( e ) { mw.hook("ml.error.report").fire(e, { tags: { module: "jsxg", graph: "Solution20564_2_655353856_l" } }); console.error(e); } if(b) {b.board.update()} }; mw.loader.using('ext.MLJSXGraph', function() { window.ml.jsxgraph.buildRealGraph("Solution20564_2_655353856_l", "Solution20564_2_655353856_p", 1, code); }); } ); } window.JXQtable["Solution20564_2_655353856_l"] = true;Draw Segmentswindow.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) {$('#jsxbtn1952780075_427795823').on('touchstart mousedown', function () { try { mlg.cf("showA"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );Shapewindow.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) {$('#jsxbtn1283423867_1493250506').on('touchstart mousedown', function () { try { mlg.cf("showB"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );Resetwindow.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) {$('#jsxbtn738356317_717568170').on('touchstart mousedown', function () { try { mlg.cf("showC"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );After we draw the segments, the shape constructed by the segments looks like a parabola that passes through the points (-6,-4) and (6,-4). Furthermore, the vertex of the parabola is (0,0). Hence, its function should have the form below. y=ax2​ Let's find the value of a. To do so, we will substitute the point (6,-4). y=ax2x=6, y=-4-4=a(6)2 Solve for a Calculate power-4=a(36)LHS/36=RHS/3636-4​=aba​=b/4a/4​9-1​=aPut minus sign in front of fraction-91​=aRearrange equation a=-91​ The graph of the function y=-91​x2 models the shape.
Exercises 31
Exercises 32 We are given an expression and asked to evaluate it for a given value. n=3​ This means that we should substitute 3 for n in the given expression and then evaluate. Let's do it! n2+5n=3(3)2+5Calculate power9+5Add terms14
Exercises 33 We are given an expression and asked to evaluate it for a given value. x=-2​ This means that we should substitute -2 for x in the given expression and then evaluate. Let's do it! 3x2−9x=-23(-2)2−9(-a)2=a23(2)2−9Calculate power3(4)−9Multiply12−9Subtract term3
Exercises 34 We are given an expression and asked to evaluate it for a given value. n=3​ This means that we should substitute 3 for n in the given expression and then evaluate. Let's do it! -4n2+11n=3-4(3)2+11Calculate power-4(9)+11Multiply-36+11Add terms-25
Exercises 35 We are given an expression and asked to evaluate it for some given values. n=3x=-2​ This means that we should substitute 3 for n and -2 for x in the given expression and then evaluate. Let's do it! n+2x2n=3, x=-23+2(-2)2(-a)2=a23+2(2)2Calculate power3+2(4)Multiply3+8Add terms11