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###### Communicate Your Answer

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###### Exercises

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Exercises 21 Let's analyze the given function. f(x)=2(0.5)x We see that the initial value of the function and constant multiplier are 2 and 0.5. Therefore, it intercepts the y-axis at (0,2). Since 0<0.5<1, it is a decreasing function. The function in choice C has these properties. The answer is C.Extra Graphing Exponential Functions When we draw an exponential function, y=abx, the value of a and b changes the shape of the graph. The graphs below show the possible graphs of the exponential functions. | |

Exercises 22 We see that there is a minus sign in front of the function. y=-2(0.5)x The graph of it is the graph of f(x)=2(0.5)x reflected across the x-axis. See the graph of the function f(x)=2(0.5)x. Therefore, the graph in B matches with the function y=-2(0.5)x.Extra Graphing Exponential Functions When we draw an exponential function, y=abx, the value of a and b changes the shape of the graph. The graphs below show the possible graphs of the exponential functions. | |

Exercises 23 Let's analyze the given function. f(x)=2(2)x We see that the initial value of the function and constant multiplier are 2 and 2. Therefore, it intercepts the y-axis at (0,2). Since 2>1, it is an increasing function. The function in A has these properties. The answer is A.Extra Graphing Exponential Functions When we draw an exponential function, y=abx, the value of a and b changes the shape of the graph. The graphs below show the possible graphs of the exponential functions. | |

Exercises 24 We see that there is a minus sign in front of the function. y=-2(2)x The graph of it is the graph of f(x)=2(2)x reflected across the x-axis. See the graph of the function f(x)=2(2)x. Therefore, the graph in D matches with the function y=-2(2)x.Extra Graphing Exponential Functions When we draw an exponential function, y=abx, the value of a and b changes the shape of the graph. The graphs below show the possible graphs of the exponential functions. | |

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Exercises 37 Looking at the graph, we see that the graph of f(x)=2x is shrunk vertically.The absolute value of a should be less than 1 due to the vertical compression. g(x)=a(2)x We see that the point (2,2) is on the graph of g(x)=a(2)x. Let's substitute it and solve for a. g(x)=a(2)xx=2, y=22=a(2)2 Solve for a Calculate power2=a(4)LHS/4=RHS/442=aba=b/2a/221=aRearrange equation a=21 The value of a is 21. | |

Exercises 38 Looking at the graph, we see that the graph of f(x)=0.25x is translated k units up.Notice that the point (0,4) is on the graph of g(x)=0.25x+k. Let's substitute it and solve for k. g(x)=0.25x+kx=0, y=44=0.250+k Solve for k a0=14=1+kLHS−1=RHS−13=kRearrange equation k=3 The value of k is 3. | |

Exercises 39 Looking at the graph, we can say that the graph of f(x)=-3x is translated h units to the right.Notice that the point (4,-1) is on the graph of g(x)=-3x−h. Let's substitute it and solve for h. g(x)=-3x−hx=4, y=-1-1=-34−h Solve for h LHS/(-1)=RHS/(-1)1=34−ham−n=anam1=3h34LHS⋅3h=RHS⋅3h3h=34Equate exponents h=4 The value of h is 4. | |

Exercises 40 Looking at the graph, we can say that the graph of f(x)=31(6)x is translated h units to the left.Notice that the point (-1,2) is on the graph of g(x)=31(6)x−h. Let's substitute it and solve for k. g(x)=31(6)x−hx=-1, y=22=31(6)-1−h Solve for h LHS⋅3=RHS⋅36=6-1−ha=a161=6-1−hEquate exponents1=-1−hLHS+1=RHS+12=-hLHS/(-1)=RHS/(-1)-2=hRearrange equation h=-2 The value of h is -2. | |

Exercises 41 Let's try to find the value of g(x) when x=-2. g(x)=6(0.5)xx=-2g(-2)=6(0.5)-2Calculate powerg(-2)=6⋅4Multiplyg(-2)=24 The function is equal to 24 when x=-2. In the given solution, the multiplication is performed before calculating the power of 0.5. Correct:Incorrect: 6 (0.5)-2=6 (4) ✓ 6 (0.5)-2=3-2× | |

Exercises 42 We want to find the domain and range of the given exponential function. y=-(0.5)x−1⇓y=-1(0.5)x−1 Looking at the function, we can see that its parent function has been reflected across the x-axis, and translated 1 unit down. Therefore, the asymptote of the function is the line y=-1.The domain of the function is all real numbers. The function cannot be greater than -1, so the range of the function is all real numbers less than -1, not 0. Domain:Range: -∞<x<∞ y<-1 | |

Exercises 43 To draw the graph of g(x), we will first make a table of values. We know that g(0)=8, and each term is 2.5 times the previous term.The ordered pairs (-1,3.2), (0,8), (1,20), and (2,50) all lie on the function. Now, we will plot and connect these points with a smooth curve.Now let's draw the functions f(x)=0.5(4)x and g(x) on the same axes and compare them. We see that both functions have the different y-intercepts when x=0. The value of g(x) is greater than the value of f(x) over the rest of the interval. Showing Our Work Graphing f(x) To graph the exponential function, f(x)=0.5(4)x, we will first make a table of values.x0.5(4)xf(x)=0.5(4)x -10.5(4)-10.125 00.5(4)00.5 10.5(4)12 20.5(4)28 30.5(4)332 Let's now plot and connect the points (-1,0.125), (0,0.5), (1,2), (2,8), and (3,32) with a smooth curve. | |

Exercises 44 To draw the graph of h(x), we will first make a table of values. We know that h(0)=32, and each term is 21 times the previous term.The ordered pairs (-1,64), (0,32), (1,16), and (2,8) all lie on the function. Now, we will plot and connect these points with a smooth curve.Now let's draw the functions f(x)=0.5(4)x and g(x) on the same axes and compare them. We see that both functions have the different y-intercepts when x=0. The value of h(x) is greater than the value of f(x) when 0<x<2. They have the same value when x=2. Showing Our Work Graphing f(x) To graph the exponential function, f(x)=0.5(4)x, we will first make a table of values.x0.5(4)xf(x)=0.5(4)x -10.5(4)-10.125 00.5(4)00.5 10.5(4)12 20.5(4)28 30.5(4)332 Let's now plot and connect the points (-1,0.125), (0,0.5), (1,2), (2,8), and (3,32) with a smooth curve. | |

Exercises 45 window.JXQ = window.JXQ || []; window.JXQtable = window.JXQtable || {}; if(!window.JXQtable["Solution19412_1_862612476_l"]) { window.JXQ.push( function () { var code = function (gid) { var images = []; try { var b = mlg.board([-4.5,7.1,4.5,-3.5],{desktopSize:'medium',style:'usa'}); var xax = b.xaxis(1,0); b.yaxis(1,0);var m = b.node(-1,0); var f = function(x){ return m.X()*m.X()*JXG.Math.pow(2,x); };var fNoM = function(x){ return JXG.Math.pow(2,x); };var fOpts = {doAdvancedPlot:false,numberPointsHigh:150,numberPointsLow:150,strokeWidth:2,firstArrow: {size: 7, type: 2},lastArrow: {size: 7, type: 2}};var f1 = b.board.create('functiongraph',[f,function(){ return -4 - m.X()*m.X(); },function(){ return 4 + m.X()*m.X(); }],fOpts);var f2 = b.func(fNoM,{opacity:0.3});var n1 = b.node(0,1); var n2 = b.node(0,function(){return f1.Y(0);});var a1 = b.measure(n2,n1,{label:{rotate:false,precision:1,distance:0.05},type:'flatArrow'});var n3 = b.node(1,2); var n4 = b.node(1,function(){return f1.Y(1);});var a2 = b.measure(n4,n3,{label:{rotate:false,precision:1,distance:0.05},type:'flatArrow'});var anim_state = 0; var animating = false; b.changeTick(xax,1,''); var l = b.legend(f1,[-4,-1],'f(x)',{anchorX:'left'});mlg.af('vertical.anim',function(){ if(!animating){ animating = true; var time = 2000; if(anim_state === 1){ m.moveTo([Math.sqrt(0.5),0],time *= 2); } if(anim_state === 2){ m.moveTo([1,0],time); } setTimeout(function(){ animating = false; anim_state = (anim_state + 1) % 3; if(anim_state === 1){ document.getElementById('vertstr').innerHTML = 'Shrink graph vertically'; }else if(anim_state === 2){ document.getElementById('vertstr').innerHTML = 'Reset graph'; } },time); } });m.coords.on('update',function(){ if(Math.abs(m.X()*m.X() - 1) < 0.05){ b.changeText(l,'f(x)'); }else if(m.X() > 0){ b.changeText(l,-(-(m.X()*m.X()).toFixed(1)) + 'f(x)'); } });} catch( e ) { mw.hook("ml.error.report").fire(e, { tags: { module: "jsxg", graph: "Solution19412_1_862612476_l" } }); } if(b) {b.board.update()} }; mw.loader.using('ext.MLJSXGraph', function() { window.ml.jsxgraph.buildRealGraph("Solution19412_1_862612476_l", "Solution19412_1_862612476_p", 1, code); }); } ); } window.JXQtable["Solution19412_1_862612476_l"] = true;Shrink graph verticallywindow.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#vertstr').on('touchstart mousedown', function () { try { event.preventDefault();mlg.cf("vertical.anim"); } catch(e) { mw.log.error(e); } }); })(jQuery); } ); | |

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