Graphing f(x) = a(x - h)² + k

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Exercises 1 Let's start by reviewing the definition of even and odd functions.A function y=f(x) is even if and only if f(-x)=f(x) for each x in the domain of f. The graph of an even function is symmetric about the y-axis.A function y=f(x) is odd if and only if f(-x)=-f(x) for each x in the domain of f. The graph of an odd function is symmetric about the origin. This means that it looks the same after a reflection in the x-axis followed by a reflection in the y-axis. Recall that a quadratic function of the form f(x)=ax2, where a is any non-zero real number, has its vertex on the origin and is symmetric about the y-axis. Therefore, it is an even function. We can verify this by using the definition given above. f(x)=ax2x=-xf(-x)=a(-x)2(-a)2=a2f(-x)=ax2ax2=f(x)f(-x)=f(x) Since f(-x)=f(x) for each and every x, we know it is an even function. This is because squaring the variable eliminates the negative sign. Conversely, this will not happen if the exponent is an odd number. For example, let's consider g(x)=x3. g(x)=x3x=-xg(-x)=(-x)3(-a)3=-a3g(-x)=-x3x3=g(x)g(-x)=-g(x) Since g(-x)=-g(x) for each and every x, we know that g(x)=x3 is an odd function. Finally, let's draw both functions together and compare them. For f(x), we can choose any non-zero value of a. Let's choose a=41​. We will graph the even function f(x)=41​x2 and the odd function g(x)=x3.window.JXQ = window.JXQ || []; window.JXQtable = window.JXQtable || {}; if(!window.JXQtable["Solution20725_1_1128130442_l"]) { window.JXQ.push( function () { var code = function (gid) { var images = []; try { var b = mlg.board([-4.5, 3.5, 4.5, -3.5], {desktopSize:'medium', style:'usa'}); b.xaxis(1,0); b.yaxis(1,0); var t = 750; var f1 = b.func("x^3",{strokeColor:'blue',xMin:0,firstArrow:false}); var f3 = b.func("x^3",{opacity:0,dash:3, strokeColor:'blue',xMax:0,lastArrow:false}); var f2 = b.func("-x^3",{opacity:0,dash:3, strokeColor:'blue',xMax:0,lastArrow:false}); var f4 = b.func("x^3",{strokeColor:'blue'}); var l1 = b.legend(f4,[1.5,2],"g(x)=x^3"); var e1 = b.ellipse([0,1],1.3,0.5,{opacity:0, startAngle:0,endAngle:-180,strokeColor:"deepskyblue",strokeWidth:1.2,dash:3,lastArrow:true});var e2 = b.ellipse([-1.5,0],0.5,1.3,{opacity:0, startAngle:90,endAngle:270,strokeColor:"deepskyblue",strokeWidth:1.2,dash:3,lastArrow:true});var f12 = b.func("0.25*x^2",{strokeColor:'red',xMin:0,firstArrow:false}); var f22 = b.func("0.25*x^2",{opacity:0,dash:3, strokeColor:'red',xMax:0,lastArrow:false}); var f32 = b.func("0.25*x^2",{strokeColor:'red'}); var l12 = b.legend(f32,[-2.8,2],"f(x)=\\frac{1}{4}x^2"); var e12 = b.ellipse([0,1],2.3,0.5,{opacity:0, startAngle:0,endAngle:-180,strokeColor:"#FF8C00",strokeWidth:1.2,dash:3,lastArrow:true});mlg.af("evenFunction",function() { $('#btn12').css({'background-color':'#FFB6A6'});b.hide(f1,t); b.hide(f4,t); b.hide(l1,t); b.hide(f32,t); b.hide(l12,t);b.show(f12,t);setTimeout( function() { b.show(e12,t); }, 2*t); setTimeout( function() { b.show(f22,t); }, 4*t); setTimeout( function() { b.hide(e12,t); }, 6*t); setTimeout( function() { b.hide(f22,t); }, 8*t); setTimeout( function() { b.show(f32,t); }, 8*t); setTimeout( function() { b.show(l12,t); }, 8*t); setTimeout( function() { b.show(f4,t); }, 10*t); setTimeout( function() { b.show(l1,t); }, 10*t); setTimeout( function() {$('#btn12').css({'background-color':'#d8dce3'}); }, 11*t); });mlg.af("oddFunction",function() { $('#btn1').css({'background-color':'#6ac5fe'}); b.hide(f12,t); b.hide(f4,t); b.hide(l1,t); b.hide(f32,t); b.hide(l12,t);b.show(f1,t);setTimeout( function() { b.show(e1,t); }, 2*t);setTimeout( function() { b.show(f2,t); }, 4*t); setTimeout( function() { b.hide(e1,t); }, 6*t); setTimeout( function() { b.show(e2,t); }, 8*t); setTimeout( function() { b.hide(f2,t); }, 10*t); setTimeout( function() { b.show(f3,t); }, 10*t); setTimeout( function() { b.hide(e2,t); }, 12*t); setTimeout( function() { b.hide(f3,t); }, 12*t); setTimeout( function() { b.show(f4,t); }, 12*t); setTimeout( function() { b.show(l1,t); }, 12*t); setTimeout( function() { b.show(f32,t); }, 14*t); setTimeout( function() { b.show(l12,t); }, 14*t);setTimeout( function() {$('#btn1').css({'background-color':'#d8dce3'}); }, 15*t); });} catch( e ) { mw.hook("ml.error.report").fire(e, { tags: { module: "jsxg", graph: "Solution20725_1_1128130442_l" } }); console.error(e); } if(b) {b.board.update()} }; mw.loader.using('ext.MLJSXGraph', function() { window.ml.jsxgraph.buildRealGraph("Solution20725_1_1128130442_l", "Solution20725_1_1128130442_p", 1, code); }); } ); } window.JXQtable["Solution20725_1_1128130442_l"] = true;Show y-axis symmetrywindow.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) {$('#btn12').on('touchstart mousedown', function () { try { mlg.cf("evenFunction"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );Show origin symmetrywindow.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) {$('#btn1').on('touchstart mousedown', function () { try { mlg.cf("oddFunction"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );Notice that this is just an example, as there are infinitely many odd and even functions.
Exercises 2 Let's start by recalling the vertex form of a quadratic function.Vertex Form​f(x)=a(x−h)2+k​ In this format, the vertex is located at (h,k). Since we want our function to have a vertex at (1,2) we can wirte it in vertex form using h=1 and k=2. Note that a can take any value as it does not affect the location of the vertex. For simplicity, we will use a=1. f(x)=1(x−1)2+2⇔f(x)=(x−1)2+2​ Notice that we could have chosen any nonzero value for a. Therefore, there are infinitely many quadratic functions whose vertex is located at (1,2). This is just one example.
Exercises 3 Let's star by reviewing the vertex form of a quadratic function.Vertex Form​f(x)=a(x−h)2+k​ In this format, the vertex is located at (h,k). This form is a transformation of the simpler function y=ax2, which has its vertex at the origin. The parameter h performs a horizontal translation.If h>0 the function is translated to the right by h units. If h<0 the function is translated to the left by ∣h∣ units.On the other hand, the parameter k performs a vertical translation.If k>0 the function is translated up by k units. If k<0 the function is translated down by ∣k∣ units.This is illustrated in the graph below.
Exercises 4 We are given four quadratic functions to compare and find the one which doesn't belong. Let's start by labeling them. ​f1​(x)=8(x+4)2f2​(x)=(x−2)2+4f3​(x)=2(x+0)2f4​(x)=3(x+1)2+1​ All of these functions are written in vertex form. Therefore, it will be useful to recall how to identify the vertex from this notation.Vertex Form​f(x)=a(x−h)2+k​ In this format, the vertex is located at (h,k). Let's identify the vertex of each of the given functions.FunctionVertex f1​(x)=8(x+4)2⇕f1​(x)=8(x−(-4))2+0(-4,0) f2​(x)=(x−2)2+4⇕f2​(x)=1(x−2)2+4(2,4) f3​(x)=2(x+0)2⇕f3​(x)=2(x−(-0))2+0(0,0) f4​(x)=3(x+1)2+1⇕f4​(x)=3(x−(-1))2+1(-1,1) As we can see, function f3​(x)=2(x+0)2 is the only one whose vertex is the origin. Therefore, this is the odd one.
Exercises 5 Before we begin, let's recall two important definitions.A function f is an even function when f(-x)=f(x) for all x in its domain. A function f is an odd function when f(-x)=-f(x) for all x in its domain. Let's see how the graphs of these types of functions look.window.JXQ = window.JXQ || []; window.JXQtable = window.JXQtable || {}; if(!window.JXQtable["Solution20729_1_759350470_l"]) { window.JXQ.push( function () { var code = function (gid) { var images = []; try { /*board elements*/ var b = mlg.board([-11,11,11,-11],{desktopSize:'large',"style":"usa"}); var b1 = b.subBoard([0,0],[-1.75,1.75,1.75,-1.75]); var b2 = b.subBoard(b1,[-1.75,1.75,1.75,-1.75],{padding:0.1}); b.subCrop([b1,b2],0); var bb = b.board.getBoundingBox(), xMax = bb[2]; var el = {};/*default display*/ b1.xaxis(2,0); b1.yaxis(2,0); b2.xaxis(2,0); b2.yaxis(2,0); el.txt0 = b.txt(0.33*xMax,0, "\\begin{gathered}\\text{See the graph of even}\\\\\\text{and odd functions}\\end{gathered}", { mathMode:true, flag:true, fontSize:1.2, layer:100 } );/*Reset*/ mlg.af("showC",function() { b.remove(el); el.txt0 = b.txt(0.33*xMax,0, "\\begin{gathered}\\text{See the graph of even}\\\\\\text{and odd functions}\\end{gathered}", { mathMode:true, flag:true, fontSize:1.2, layer:100 } ); });/*A even functions: blue #628aff and #62d8ff*/ mlg.af("showA",function() { b.remove(el); el.p1 = b1.point(1.2,0.44,{size:0.25,"name":"(x,y)","label":{"position":160,"distance":1.5,"fontSize":0.75}}); el.p2 = b1.point(-1.2,0.44,{size:0.25,"name":"(\\text{-}x,y)","label":{"position":20,"distance":1.6,"fontSize":0.75}}); el.p3 = b2.point(0.5,-0.4375,{size:0.25,"name":"(x,y)","label":{"position":20,"distance":1.5,"fontSize":0.75}}); el.p4 = b2.point(-0.5,-0.4375,{size:0.25,"name":"(\\text{-}x,y)","label":{"position":160,"distance":1.6,"fontSize":0.75}}); el.f1 = b1.func('x^2-1',{strokeWidth:1.8,strokeColor:'#628aff',xMax:1.7,xMin:-1.7}); el.f2 = b2.func('x^4-2*x^2',{strokeWidth:1.8,strokeColor:'#628aff',xMax:1.7,xMin:-1.7}); });/*B odd functions: green #00c997 and #62ffd8*/ mlg.af("showB",function() { b.remove(el); el.p1 = b1.point(1,1,{size:0.25,"name":"(x,y)","label":{"position":340,"distance":1.75,"fontSize":0.75}}); el.p2 = b1.point(-1,-1,{size:0.25,"name":"(\\text{-}x,\\text{-}y)","label":{"position":160,"distance":1.75,"fontSize":0.75}}); el.p3 = b2.point(0.5,0.5,{size:0.25,"name":"(x,y)","label":{"position":340,"distance":1.75,"fontSize":0.75}}); el.p4 = b2.point(-0.5,-0.5,{size:0.25,"name":"(\\text{-}x,\\text{-}y)","label":{"position":160,"distance":1.75,"fontSize":0.75}}); el.f1 = b1.func('x^3',{strokeWidth:1.8,strokeColor:'#00c997'}); el.f2 = b2.func('x',{strokeWidth:1.8,strokeColor:'#00c997'}); });} catch( e ) { mw.hook("ml.error.report").fire(e, { tags: { module: "jsxg", graph: "Solution20729_1_759350470_l" } }); console.error(e); } if(b) {b.board.update()} }; mw.loader.using('ext.MLJSXGraph', function() { window.ml.jsxgraph.buildRealGraph("Solution20729_1_759350470_l", "Solution20729_1_759350470_p", 1, code); }); } ); } window.JXQtable["Solution20729_1_759350470_l"] = true;Even functionswindow.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) {$('#jsxbtn2091719257_1117097061').on('touchstart mousedown', function () { try { mlg.cf("showA"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );Odd functionswindow.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) {$('#jsxbtn594860332_998356919').on('touchstart mousedown', function () { try { mlg.cf("showB"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );Resetwindow.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) {$('#jsxbtn1001813895_1058884887').on('touchstart mousedown', function () { try { mlg.cf("showC"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );Consider the graphs above. We can see that for even functions, if (x,y) is on the graph, then (-x,y) is also on the graph. Meanwhile, for an odd function, if (x,y) is on the graph, then (-x,-y) is also on the graph. Now, consider the given function. f(x)=4x+3​ Let's calculate f(-x). f(-x)=4(-x)+3a(-b)=-a⋅bf(-x)=-4x+3 Next, let's calculate -f(x). -f(x)=-(4x+3)Distribute -1-f(x)=-4x−3 Finally, let's think about what these results tell us.f(x)f(-x)-f(x) 4x+3-4x+3-4x−3 Since f(x)​=f(-x) and f(-x)​=-f(x), the function f is neither an even nor an odd function.
Exercises 6 Before we begin, let's recall two important definitions.A function f is an even function when f(-x)=f(x) for all x in its domain. A function f is an odd function when f(-x)=-f(x) for all x in its domain. Let's see how the graphs of these types of functions look.window.JXQ = window.JXQ || []; window.JXQtable = window.JXQtable || {}; if(!window.JXQtable["Solution20730_1_1390419149_l"]) { window.JXQ.push( function () { var code = function (gid) { var images = []; try { /*board elements*/ var b = mlg.board([-11,11,11,-11],{desktopSize:'large',"style":"usa"}); var b1 = b.subBoard([0,0],[-1.75,1.75,1.75,-1.75]); var b2 = b.subBoard(b1,[-1.75,1.75,1.75,-1.75],{padding:0.1}); b.subCrop([b1,b2],0); var bb = b.board.getBoundingBox(), xMax = bb[2]; var el = {};/*default display*/ b1.xaxis(2,0); b1.yaxis(2,0); b2.xaxis(2,0); b2.yaxis(2,0); el.txt0 = b.txt(0.33*xMax,0, "\\begin{gathered}\\text{See the graph of even}\\\\\\text{and odd functions}\\end{gathered}", { mathMode:true, flag:true, fontSize:1.2, layer:100 } );/*Reset*/ mlg.af("showC",function() { b.remove(el); el.txt0 = b.txt(0.33*xMax,0, "\\begin{gathered}\\text{See the graph of even}\\\\\\text{and odd functions}\\end{gathered}", { mathMode:true, flag:true, fontSize:1.2, layer:100 } ); });/*A even functions: blue #628aff and #62d8ff*/ mlg.af("showA",function() { b.remove(el); el.p1 = b1.point(1.2,0.44,{size:0.25,"name":"(x,y)","label":{"position":160,"distance":1.5,"fontSize":0.75}}); el.p2 = b1.point(-1.2,0.44,{size:0.25,"name":"(\\text{-}x,y)","label":{"position":20,"distance":1.6,"fontSize":0.75}}); el.p3 = b2.point(0.5,-0.4375,{size:0.25,"name":"(x,y)","label":{"position":20,"distance":1.5,"fontSize":0.75}}); el.p4 = b2.point(-0.5,-0.4375,{size:0.25,"name":"(\\text{-}x,y)","label":{"position":160,"distance":1.6,"fontSize":0.75}}); el.f1 = b1.func('x^2-1',{strokeWidth:1.8,strokeColor:'#628aff',xMax:1.7,xMin:-1.7}); el.f2 = b2.func('x^4-2*x^2',{strokeWidth:1.8,strokeColor:'#628aff',xMax:1.7,xMin:-1.7}); });/*B odd functions: green #00c997 and #62ffd8*/ mlg.af("showB",function() { b.remove(el); el.p1 = b1.point(1,1,{size:0.25,"name":"(x,y)","label":{"position":340,"distance":1.75,"fontSize":0.75}}); el.p2 = b1.point(-1,-1,{size:0.25,"name":"(\\text{-}x,\\text{-}y)","label":{"position":160,"distance":1.75,"fontSize":0.75}}); el.p3 = b2.point(0.5,0.5,{size:0.25,"name":"(x,y)","label":{"position":340,"distance":1.75,"fontSize":0.75}}); el.p4 = b2.point(-0.5,-0.5,{size:0.25,"name":"(\\text{-}x,\\text{-}y)","label":{"position":160,"distance":1.75,"fontSize":0.75}}); el.f1 = b1.func('x^3',{strokeWidth:1.8,strokeColor:'#00c997'}); el.f2 = b2.func('x',{strokeWidth:1.8,strokeColor:'#00c997'}); });} catch( e ) { mw.hook("ml.error.report").fire(e, { tags: { module: "jsxg", graph: "Solution20730_1_1390419149_l" } }); console.error(e); } if(b) {b.board.update()} }; mw.loader.using('ext.MLJSXGraph', function() { window.ml.jsxgraph.buildRealGraph("Solution20730_1_1390419149_l", "Solution20730_1_1390419149_p", 1, code); }); } ); } window.JXQtable["Solution20730_1_1390419149_l"] = true;Even functionswindow.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) {$('#jsxbtn130409524_1937692575').on('touchstart mousedown', function () { try { mlg.cf("showA"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );Odd functionswindow.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) {$('#jsxbtn483821698_706426761').on('touchstart mousedown', function () { try { mlg.cf("showB"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );Resetwindow.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) {$('#jsxbtn683540790_1708375496').on('touchstart mousedown', function () { try { mlg.cf("showC"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );Consider the graphs above. We can see that for even functions, if (x,y) is on the graph, then (-x,y) is also on the graph. Meanwhile, for an odd function, if (x,y) is on the graph, then (-x,-y) is also on the graph. Now, consider the given function. g(x)=3x2​ Let's calculate g(-x). g(-x)=3(-x)2(-a)2=a2g(-x)=3x2 Next, let's calculate -g(x). -g(x)=-(3x2)Distribute -1-g(x)=-3x2 Finally, let's think about what these results tell us.g(x)g(-x)-g(x) 3x23x2-3x2 We can see above that g(x)=g(-x). Therefore, g is an even function.
Exercises 7 Before we begin, let's recall two important definitions.A function f is an even function when f(-x)=f(x) for all x in its domain. A function f is an odd function when f(-x)=-f(x) for all x in its domain. Let's see how the graphs of these types of functions look.window.JXQ = window.JXQ || []; window.JXQtable = window.JXQtable || {}; if(!window.JXQtable["Solution20731_1_765760489_l"]) { window.JXQ.push( function () { var code = function (gid) { var images = []; try { /*board elements*/ var b = mlg.board([-11,11,11,-11],{desktopSize:'large',"style":"usa"}); var b1 = b.subBoard([0,0],[-1.75,1.75,1.75,-1.75]); var b2 = b.subBoard(b1,[-1.75,1.75,1.75,-1.75],{padding:0.1}); b.subCrop([b1,b2],0); var bb = b.board.getBoundingBox(), xMax = bb[2]; var el = {};/*default display*/ b1.xaxis(2,0); b1.yaxis(2,0); b2.xaxis(2,0); b2.yaxis(2,0); el.txt0 = b.txt(0.33*xMax,0, "\\begin{gathered}\\text{See the graph of even}\\\\\\text{and odd functions}\\end{gathered}", { mathMode:true, flag:true, fontSize:1.2, layer:100 } );/*Reset*/ mlg.af("showC",function() { b.remove(el); el.txt0 = b.txt(0.33*xMax,0, "\\begin{gathered}\\text{See the graph of even}\\\\\\text{and odd functions}\\end{gathered}", { mathMode:true, flag:true, fontSize:1.2, layer:100 } ); });/*A even functions: blue #628aff and #62d8ff*/ mlg.af("showA",function() { b.remove(el); el.p1 = b1.point(1.2,0.44,{size:0.25,"name":"(x,y)","label":{"position":160,"distance":1.5,"fontSize":0.75}}); el.p2 = b1.point(-1.2,0.44,{size:0.25,"name":"(\\text{-}x,y)","label":{"position":20,"distance":1.6,"fontSize":0.75}}); el.p3 = b2.point(0.5,-0.4375,{size:0.25,"name":"(x,y)","label":{"position":20,"distance":1.5,"fontSize":0.75}}); el.p4 = b2.point(-0.5,-0.4375,{size:0.25,"name":"(\\text{-}x,y)","label":{"position":160,"distance":1.6,"fontSize":0.75}}); el.f1 = b1.func('x^2-1',{strokeWidth:1.8,strokeColor:'#628aff',xMax:1.7,xMin:-1.7}); el.f2 = b2.func('x^4-2*x^2',{strokeWidth:1.8,strokeColor:'#628aff',xMax:1.7,xMin:-1.7}); });/*B odd functions: green #00c997 and #62ffd8*/ mlg.af("showB",function() { b.remove(el); el.p1 = b1.point(1,1,{size:0.25,"name":"(x,y)","label":{"position":340,"distance":1.75,"fontSize":0.75}}); el.p2 = b1.point(-1,-1,{size:0.25,"name":"(\\text{-}x,\\text{-}y)","label":{"position":160,"distance":1.75,"fontSize":0.75}}); el.p3 = b2.point(0.5,0.5,{size:0.25,"name":"(x,y)","label":{"position":340,"distance":1.75,"fontSize":0.75}}); el.p4 = b2.point(-0.5,-0.5,{size:0.25,"name":"(\\text{-}x,\\text{-}y)","label":{"position":160,"distance":1.75,"fontSize":0.75}}); el.f1 = b1.func('x^3',{strokeWidth:1.8,strokeColor:'#00c997'}); el.f2 = b2.func('x',{strokeWidth:1.8,strokeColor:'#00c997'}); });} catch( e ) { mw.hook("ml.error.report").fire(e, { tags: { module: "jsxg", graph: "Solution20731_1_765760489_l" } }); console.error(e); } if(b) {b.board.update()} }; mw.loader.using('ext.MLJSXGraph', function() { window.ml.jsxgraph.buildRealGraph("Solution20731_1_765760489_l", "Solution20731_1_765760489_p", 1, code); }); } ); } window.JXQtable["Solution20731_1_765760489_l"] = true;Even functionswindow.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) {$('#jsxbtn1383706121_644227348').on('touchstart mousedown', function () { try { mlg.cf("showA"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );Odd functionswindow.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) {$('#jsxbtn1802302137_2004657727').on('touchstart mousedown', function () { try { mlg.cf("showB"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );Resetwindow.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) {$('#jsxbtn808829924_515336364').on('touchstart mousedown', function () { try { mlg.cf("showC"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );Consider the graphs above. We can see that for even functions, if (x,y) is on the graph, then (-x,y) is also on the graph. Meanwhile, for an odd function, if (x,y) is on the graph, then (-x,-y) is also on the graph. Now, consider the given function. h(x)=5x+2​ Let's calculate h(-x). h(-x)=5-x+2a-m=am1​h(-x)=5x1​+21a=1h(-x)=5x1x​+2bmam​=(ba​)mh(-x)=(51​)x+2 Next, let's calculate -h(x). -h(x)=-(5x+2)Remove parentheses-h(x)=-5x−2 Finally, let's think about what these results tell us.h(x)h(-x)-h(x) 5x+2(51​)x+2-5x−2 Since h(x)​=h(-x) and h(-x)​=-h(x), the function h is neither an even nor an odd function.
Exercises 8 Before we begin, let's recall two important definitions.A function f is an even function when f(-x)=f(x) for all x in its domain. A function f is an odd function when f(-x)=-f(x) for all x in its domain. Let's see how the graphs of these types of functions look.window.JXQ = window.JXQ || []; window.JXQtable = window.JXQtable || {}; if(!window.JXQtable["Solution20732_1_2020172868_l"]) { window.JXQ.push( function () { var code = function (gid) { var images = []; try { /*board elements*/ var b = mlg.board([-11,11,11,-11],{desktopSize:'large',"style":"usa"}); var b1 = b.subBoard([0,0],[-1.75,1.75,1.75,-1.75]); var b2 = b.subBoard(b1,[-1.75,1.75,1.75,-1.75],{padding:0.1}); b.subCrop([b1,b2],0); var bb = b.board.getBoundingBox(), xMax = bb[2]; var el = {};/*default display*/ b1.xaxis(2,0); b1.yaxis(2,0); b2.xaxis(2,0); b2.yaxis(2,0); el.txt0 = b.txt(0.33*xMax,0, "\\begin{gathered}\\text{See the graph of even}\\\\\\text{and odd functions}\\end{gathered}", { mathMode:true, flag:true, fontSize:1.2, layer:100 } );/*Reset*/ mlg.af("showC",function() { b.remove(el); el.txt0 = b.txt(0.33*xMax,0, "\\begin{gathered}\\text{See the graph of even}\\\\\\text{and odd functions}\\end{gathered}", { mathMode:true, flag:true, fontSize:1.2, layer:100 } ); });/*A even functions: blue #628aff and #62d8ff*/ mlg.af("showA",function() { b.remove(el); el.p1 = b1.point(1.2,0.44,{size:0.25,"name":"(x,y)","label":{"position":160,"distance":1.5,"fontSize":0.75}}); el.p2 = b1.point(-1.2,0.44,{size:0.25,"name":"(\\text{-}x,y)","label":{"position":20,"distance":1.6,"fontSize":0.75}}); el.p3 = b2.point(0.5,-0.4375,{size:0.25,"name":"(x,y)","label":{"position":20,"distance":1.5,"fontSize":0.75}}); el.p4 = b2.point(-0.5,-0.4375,{size:0.25,"name":"(\\text{-}x,y)","label":{"position":160,"distance":1.6,"fontSize":0.75}}); el.f1 = b1.func('x^2-1',{strokeWidth:1.8,strokeColor:'#628aff',xMax:1.7,xMin:-1.7}); el.f2 = b2.func('x^4-2*x^2',{strokeWidth:1.8,strokeColor:'#628aff',xMax:1.7,xMin:-1.7}); });/*B odd functions: green #00c997 and #62ffd8*/ mlg.af("showB",function() { b.remove(el); el.p1 = b1.point(1,1,{size:0.25,"name":"(x,y)","label":{"position":340,"distance":1.75,"fontSize":0.75}}); el.p2 = b1.point(-1,-1,{size:0.25,"name":"(\\text{-}x,\\text{-}y)","label":{"position":160,"distance":1.75,"fontSize":0.75}}); el.p3 = b2.point(0.5,0.5,{size:0.25,"name":"(x,y)","label":{"position":340,"distance":1.75,"fontSize":0.75}}); el.p4 = b2.point(-0.5,-0.5,{size:0.25,"name":"(\\text{-}x,\\text{-}y)","label":{"position":160,"distance":1.75,"fontSize":0.75}}); el.f1 = b1.func('x^3',{strokeWidth:1.8,strokeColor:'#00c997'}); el.f2 = b2.func('x',{strokeWidth:1.8,strokeColor:'#00c997'}); });} catch( e ) { mw.hook("ml.error.report").fire(e, { tags: { module: "jsxg", graph: "Solution20732_1_2020172868_l" } }); console.error(e); } if(b) {b.board.update()} }; mw.loader.using('ext.MLJSXGraph', function() { window.ml.jsxgraph.buildRealGraph("Solution20732_1_2020172868_l", "Solution20732_1_2020172868_p", 1, code); }); } ); } window.JXQtable["Solution20732_1_2020172868_l"] = true;Even functionswindow.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) {$('#jsxbtn1313188204_985448469').on('touchstart mousedown', function () { try { mlg.cf("showA"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );Odd functionswindow.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) {$('#jsxbtn1772013092_878451154').on('touchstart mousedown', function () { try { mlg.cf("showB"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );Resetwindow.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) {$('#jsxbtn1131005996_1855773130').on('touchstart mousedown', function () { try { mlg.cf("showC"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );Consider the graphs above. We can see that for even functions, if (x,y) is on the graph, then (-x,y) is also on the graph. Meanwhile, for an odd function, if (x,y) is on the graph, then (-x,-y) is also on the graph. Now, consider the given function. m(x)=2x2−7x​ Let's calculate m(-x). m(-x)=2(-x)2−7(-x) Simplify right-hand side (-a)2=a2m(-x)=2x2−7(-x)-a(-b)=a⋅b m(-x)=2x2+7x Next, let's calculate -m(x). -m(x)=-(2x2−7x)Distribute -1-m(x)=-2x2+7x Finally, let's think about what these results tell us.m(x)m(-x)-m(x) 2x2−7x2x2+7x-2x2+7x Since f(x)​=f(-x) and f(-x)​=-f(x), the function f is neither an even nor an odd function.
Exercises 9 Before we begin, let's recall two important definitions.A function f is an even function when f(-x)=f(x) for all x in its domain. A function f is an odd function when f(-x)=-f(x) for all x in its domain. Let's see how the graphs of these types of functions look.window.JXQ = window.JXQ || []; window.JXQtable = window.JXQtable || {}; if(!window.JXQtable["Solution20733_1_19388164_l"]) { window.JXQ.push( function () { var code = function (gid) { var images = []; try { /*board elements*/ var b = mlg.board([-11,11,11,-11],{desktopSize:'large',"style":"usa"}); var b1 = b.subBoard([0,0],[-1.75,1.75,1.75,-1.75]); var b2 = b.subBoard(b1,[-1.75,1.75,1.75,-1.75],{padding:0.1}); b.subCrop([b1,b2],0); var bb = b.board.getBoundingBox(), xMax = bb[2]; var el = {};/*default display*/ b1.xaxis(2,0); b1.yaxis(2,0); b2.xaxis(2,0); b2.yaxis(2,0); el.txt0 = b.txt(0.33*xMax,0, "\\begin{gathered}\\text{See the graph of even}\\\\\\text{and odd functions}\\end{gathered}", { mathMode:true, flag:true, fontSize:1.2, layer:100 } );/*Reset*/ mlg.af("showC",function() { b.remove(el); el.txt0 = b.txt(0.33*xMax,0, "\\begin{gathered}\\text{See the graph of even}\\\\\\text{and odd functions}\\end{gathered}", { mathMode:true, flag:true, fontSize:1.2, layer:100 } ); });/*A even functions: blue #628aff and #62d8ff*/ mlg.af("showA",function() { b.remove(el); el.p1 = b1.point(1.2,0.44,{size:0.25,"name":"(x,y)","label":{"position":160,"distance":1.5,"fontSize":0.75}}); el.p2 = b1.point(-1.2,0.44,{size:0.25,"name":"(\\text{-}x,y)","label":{"position":20,"distance":1.6,"fontSize":0.75}}); el.p3 = b2.point(0.5,-0.4375,{size:0.25,"name":"(x,y)","label":{"position":20,"distance":1.5,"fontSize":0.75}}); el.p4 = b2.point(-0.5,-0.4375,{size:0.25,"name":"(\\text{-}x,y)","label":{"position":160,"distance":1.6,"fontSize":0.75}}); el.f1 = b1.func('x^2-1',{strokeWidth:1.8,strokeColor:'#628aff',xMax:1.7,xMin:-1.7}); el.f2 = b2.func('x^4-2*x^2',{strokeWidth:1.8,strokeColor:'#628aff',xMax:1.7,xMin:-1.7}); });/*B odd functions: green #00c997 and #62ffd8*/ mlg.af("showB",function() { b.remove(el); el.p1 = b1.point(1,1,{size:0.25,"name":"(x,y)","label":{"position":340,"distance":1.75,"fontSize":0.75}}); el.p2 = b1.point(-1,-1,{size:0.25,"name":"(\\text{-}x,\\text{-}y)","label":{"position":160,"distance":1.75,"fontSize":0.75}}); el.p3 = b2.point(0.5,0.5,{size:0.25,"name":"(x,y)","label":{"position":340,"distance":1.75,"fontSize":0.75}}); el.p4 = b2.point(-0.5,-0.5,{size:0.25,"name":"(\\text{-}x,\\text{-}y)","label":{"position":160,"distance":1.75,"fontSize":0.75}}); el.f1 = b1.func('x^3',{strokeWidth:1.8,strokeColor:'#00c997'}); el.f2 = b2.func('x',{strokeWidth:1.8,strokeColor:'#00c997'}); });} catch( e ) { mw.hook("ml.error.report").fire(e, { tags: { module: "jsxg", graph: "Solution20733_1_19388164_l" } }); console.error(e); } if(b) {b.board.update()} }; mw.loader.using('ext.MLJSXGraph', function() { window.ml.jsxgraph.buildRealGraph("Solution20733_1_19388164_l", "Solution20733_1_19388164_p", 1, code); }); } ); } window.JXQtable["Solution20733_1_19388164_l"] = true;Even functionswindow.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) {$('#jsxbtn33319424_438705368').on('touchstart mousedown', function () { try { mlg.cf("showA"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );Odd functionswindow.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) {$('#jsxbtn754820621_941712779').on('touchstart mousedown', function () { try { mlg.cf("showB"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );Resetwindow.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) {$('#jsxbtn2018390663_1583115664').on('touchstart mousedown', function () { try { mlg.cf("showC"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );Consider the graphs above. We can see that for even functions, if (x,y) is on the graph, then (-x,y) is also on the graph. Meanwhile, for an odd function, if (x,y) is on the graph, then (-x,-y) is also on the graph. Now, consider the given function. p(x)=-x2+8​ Let's calculate p(-x). p(-x)=-(-x)2+8(-a)2=a2p(-x)=-x2+8 Next, let's calculate -p(x). -p(x)=-(-x2+8)Distribute -1-p(x)=x2−8 Finally, let's think about what these results tell us.p(x)p(-x)-p(x) -x2+8-x2+8x2−8 We can see above that p(x)=p(-x). Therefore, p is an even function.
Exercises 10 Before we begin, let's recall two important definitions.A function f is an even function when f(-x)=f(x) for all x in its domain. A function f is an odd function when f(-x)=-f(x) for all x in its domain. Let's see how the graphs of these types of functions look.window.JXQ = window.JXQ || []; window.JXQtable = window.JXQtable || {}; if(!window.JXQtable["Solution20734_1_563353186_l"]) { window.JXQ.push( function () { var code = function (gid) { var images = []; try { /*board elements*/ var b = mlg.board([-11,11,11,-11],{desktopSize:'large',"style":"usa"}); var b1 = b.subBoard([0,0],[-1.75,1.75,1.75,-1.75]); var b2 = b.subBoard(b1,[-1.75,1.75,1.75,-1.75],{padding:0.1}); b.subCrop([b1,b2],0); var bb = b.board.getBoundingBox(), xMax = bb[2]; var el = {};/*default display*/ b1.xaxis(2,0); b1.yaxis(2,0); b2.xaxis(2,0); b2.yaxis(2,0); el.txt0 = b.txt(0.33*xMax,0, "\\begin{gathered}\\text{See the graph of even}\\\\\\text{and odd functions}\\end{gathered}", { mathMode:true, flag:true, fontSize:1.2, layer:100 } );/*Reset*/ mlg.af("showC",function() { b.remove(el); el.txt0 = b.txt(0.33*xMax,0, "\\begin{gathered}\\text{See the graph of even}\\\\\\text{and odd functions}\\end{gathered}", { mathMode:true, flag:true, fontSize:1.2, layer:100 } ); });/*A even functions: blue #628aff and #62d8ff*/ mlg.af("showA",function() { b.remove(el); el.p1 = b1.point(1.2,0.44,{size:0.25,"name":"(x,y)","label":{"position":160,"distance":1.5,"fontSize":0.75}}); el.p2 = b1.point(-1.2,0.44,{size:0.25,"name":"(\\text{-}x,y)","label":{"position":20,"distance":1.6,"fontSize":0.75}}); el.p3 = b2.point(0.5,-0.4375,{size:0.25,"name":"(x,y)","label":{"position":20,"distance":1.5,"fontSize":0.75}}); el.p4 = b2.point(-0.5,-0.4375,{size:0.25,"name":"(\\text{-}x,y)","label":{"position":160,"distance":1.6,"fontSize":0.75}}); el.f1 = b1.func('x^2-1',{strokeWidth:1.8,strokeColor:'#628aff',xMax:1.7,xMin:-1.7}); el.f2 = b2.func('x^4-2*x^2',{strokeWidth:1.8,strokeColor:'#628aff',xMax:1.7,xMin:-1.7}); });/*B odd functions: green #00c997 and #62ffd8*/ mlg.af("showB",function() { b.remove(el); el.p1 = b1.point(1,1,{size:0.25,"name":"(x,y)","label":{"position":340,"distance":1.75,"fontSize":0.75}}); el.p2 = b1.point(-1,-1,{size:0.25,"name":"(\\text{-}x,\\text{-}y)","label":{"position":160,"distance":1.75,"fontSize":0.75}}); el.p3 = b2.point(0.5,0.5,{size:0.25,"name":"(x,y)","label":{"position":340,"distance":1.75,"fontSize":0.75}}); el.p4 = b2.point(-0.5,-0.5,{size:0.25,"name":"(\\text{-}x,\\text{-}y)","label":{"position":160,"distance":1.75,"fontSize":0.75}}); el.f1 = b1.func('x^3',{strokeWidth:1.8,strokeColor:'#00c997'}); el.f2 = b2.func('x',{strokeWidth:1.8,strokeColor:'#00c997'}); });} catch( e ) { mw.hook("ml.error.report").fire(e, { tags: { module: "jsxg", graph: "Solution20734_1_563353186_l" } }); console.error(e); } if(b) {b.board.update()} }; mw.loader.using('ext.MLJSXGraph', function() { window.ml.jsxgraph.buildRealGraph("Solution20734_1_563353186_l", "Solution20734_1_563353186_p", 1, code); }); } ); } window.JXQtable["Solution20734_1_563353186_l"] = true;Even functionswindow.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) {$('#jsxbtn1355882489_2103328826').on('touchstart mousedown', function () { try { mlg.cf("showA"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );Odd functionswindow.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) {$('#jsxbtn1834897347_300693291').on('touchstart mousedown', function () { try { mlg.cf("showB"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );Resetwindow.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) {$('#jsxbtn1952433274_887070668').on('touchstart mousedown', function () { try { mlg.cf("showC"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );Consider the graphs above. We can see that for even functions, if (x,y) is on the graph, then (-x,y) is also on the graph. Meanwhile, for an odd function, if (x,y) is on the graph, then (-x,-y) is also on the graph. Now, consider the given function. f(x)=-21​x​ Let's calculate f(-x). f(-x)=-21​(-x)-a(-b)=a⋅bf(-x)=21​x Next, let's calculate -f(x). -f(x)=-(-21​x)-(-a)=a-f(x)=21​x Finally, let's think about what these results tell us.f(x)f(-x)-f(x) -21​x21​x21​x We can see above that -f(x)=f(-x). Therefore, f is an odd function.
Exercises 11 Before we begin, let's recall two important definitions.A function f is an even function when f(-x)=f(x) for all x in its domain. A function f is an odd function when f(-x)=-f(x) for all x in its domain. Let's see how the graphs of these types of functions look.window.JXQ = window.JXQ || []; window.JXQtable = window.JXQtable || {}; if(!window.JXQtable["Solution20735_1_1539489769_l"]) { window.JXQ.push( function () { var code = function (gid) { var images = []; try { /*board elements*/ var b = mlg.board([-11,11,11,-11],{desktopSize:'large',"style":"usa"}); var b1 = b.subBoard([0,0],[-1.75,1.75,1.75,-1.75]); var b2 = b.subBoard(b1,[-1.75,1.75,1.75,-1.75],{padding:0.1}); b.subCrop([b1,b2],0); var bb = b.board.getBoundingBox(), xMax = bb[2]; var el = {};/*default display*/ b1.xaxis(2,0); b1.yaxis(2,0); b2.xaxis(2,0); b2.yaxis(2,0); el.txt0 = b.txt(0.33*xMax,0, "\\begin{gathered}\\text{See the graph of even}\\\\\\text{and odd functions}\\end{gathered}", { mathMode:true, flag:true, fontSize:1.2, layer:100 } );/*Reset*/ mlg.af("showC",function() { b.remove(el); el.txt0 = b.txt(0.33*xMax,0, "\\begin{gathered}\\text{See the graph of even}\\\\\\text{and odd functions}\\end{gathered}", { mathMode:true, flag:true, fontSize:1.2, layer:100 } ); });/*A even functions: blue #628aff and #62d8ff*/ mlg.af("showA",function() { b.remove(el); el.p1 = b1.point(1.2,0.44,{size:0.25,"name":"(x,y)","label":{"position":160,"distance":1.5,"fontSize":0.75}}); el.p2 = b1.point(-1.2,0.44,{size:0.25,"name":"(\\text{-}x,y)","label":{"position":20,"distance":1.6,"fontSize":0.75}}); el.p3 = b2.point(0.5,-0.4375,{size:0.25,"name":"(x,y)","label":{"position":20,"distance":1.5,"fontSize":0.75}}); el.p4 = b2.point(-0.5,-0.4375,{size:0.25,"name":"(\\text{-}x,y)","label":{"position":160,"distance":1.6,"fontSize":0.75}}); el.f1 = b1.func('x^2-1',{strokeWidth:1.8,strokeColor:'#628aff',xMax:1.7,xMin:-1.7}); el.f2 = b2.func('x^4-2*x^2',{strokeWidth:1.8,strokeColor:'#628aff',xMax:1.7,xMin:-1.7}); });/*B odd functions: green #00c997 and #62ffd8*/ mlg.af("showB",function() { b.remove(el); el.p1 = b1.point(1,1,{size:0.25,"name":"(x,y)","label":{"position":340,"distance":1.75,"fontSize":0.75}}); el.p2 = b1.point(-1,-1,{size:0.25,"name":"(\\text{-}x,\\text{-}y)","label":{"position":160,"distance":1.75,"fontSize":0.75}}); el.p3 = b2.point(0.5,0.5,{size:0.25,"name":"(x,y)","label":{"position":340,"distance":1.75,"fontSize":0.75}}); el.p4 = b2.point(-0.5,-0.5,{size:0.25,"name":"(\\text{-}x,\\text{-}y)","label":{"position":160,"distance":1.75,"fontSize":0.75}}); el.f1 = b1.func('x^3',{strokeWidth:1.8,strokeColor:'#00c997'}); el.f2 = b2.func('x',{strokeWidth:1.8,strokeColor:'#00c997'}); });} catch( e ) { mw.hook("ml.error.report").fire(e, { tags: { module: "jsxg", graph: "Solution20735_1_1539489769_l" } }); console.error(e); } if(b) {b.board.update()} }; mw.loader.using('ext.MLJSXGraph', function() { window.ml.jsxgraph.buildRealGraph("Solution20735_1_1539489769_l", "Solution20735_1_1539489769_p", 1, code); }); } ); } window.JXQtable["Solution20735_1_1539489769_l"] = true;Even functionswindow.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) {$('#jsxbtn598464028_772960888').on('touchstart mousedown', function () { try { mlg.cf("showA"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );Odd functionswindow.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) {$('#jsxbtn1580839272_2050440454').on('touchstart mousedown', function () { try { mlg.cf("showB"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );Resetwindow.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) {$('#jsxbtn298850605_2133282529').on('touchstart mousedown', function () { try { mlg.cf("showC"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );Consider the graphs above. We can see that for even functions, if (x,y) is on the graph, then (-x,y) is also on the graph. Meanwhile, for an odd function, if (x,y) is on the graph, then (-x,-y) is also on the graph. Now, consider the given function. n(x)=2x2−7x+3​ Let's calculate n(-x). n(-x)=2(-x)2−7(-x)+3 Simplify right-hand side (-a)2=a2n(-x)=2x2−7(-x)+3-a(-b)=a⋅b n(-x)=2x2+7x+3 Next, let's calculate -n(x). -n(x)=-(2x2−7x+3)Distribute -1-n(x)=-2x2+7x−3 Finally, let's think about what these results tell us.n(x)n(-x)-n(x) 2x2−7x+32x2+7x+3-2x2+7x−3 Since n(x)​=n(-x) and n(-x)​=-n(x), the function n is neither an even nor an odd function.
Exercises 12 Before we begin, let's recall two important definitions.A function f is an even function when f(-x)=f(x) for all x in its domain. A function f is an odd function when f(-x)=-f(x) for all x in its domain. Let's see how the graphs of these types of functions look.window.JXQ = window.JXQ || []; window.JXQtable = window.JXQtable || {}; if(!window.JXQtable["Solution20736_1_471284253_l"]) { window.JXQ.push( function () { var code = function (gid) { var images = []; try { /*board elements*/ var b = mlg.board([-11,11,11,-11],{desktopSize:'large',"style":"usa"}); var b1 = b.subBoard([0,0],[-1.75,1.75,1.75,-1.75]); var b2 = b.subBoard(b1,[-1.75,1.75,1.75,-1.75],{padding:0.1}); b.subCrop([b1,b2],0); var bb = b.board.getBoundingBox(), xMax = bb[2]; var el = {};/*default display*/ b1.xaxis(2,0); b1.yaxis(2,0); b2.xaxis(2,0); b2.yaxis(2,0); el.txt0 = b.txt(0.33*xMax,0, "\\begin{gathered}\\text{See the graph of even}\\\\\\text{and odd functions}\\end{gathered}", { mathMode:true, flag:true, fontSize:1.2, layer:100 } );/*Reset*/ mlg.af("showC",function() { b.remove(el); el.txt0 = b.txt(0.33*xMax,0, "\\begin{gathered}\\text{See the graph of even}\\\\\\text{and odd functions}\\end{gathered}", { mathMode:true, flag:true, fontSize:1.2, layer:100 } ); });/*A even functions: blue #628aff and #62d8ff*/ mlg.af("showA",function() { b.remove(el); el.p1 = b1.point(1.2,0.44,{size:0.25,"name":"(x,y)","label":{"position":160,"distance":1.5,"fontSize":0.75}}); el.p2 = b1.point(-1.2,0.44,{size:0.25,"name":"(\\text{-}x,y)","label":{"position":20,"distance":1.6,"fontSize":0.75}}); el.p3 = b2.point(0.5,-0.4375,{size:0.25,"name":"(x,y)","label":{"position":20,"distance":1.5,"fontSize":0.75}}); el.p4 = b2.point(-0.5,-0.4375,{size:0.25,"name":"(\\text{-}x,y)","label":{"position":160,"distance":1.6,"fontSize":0.75}}); el.f1 = b1.func('x^2-1',{strokeWidth:1.8,strokeColor:'#628aff',xMax:1.7,xMin:-1.7}); el.f2 = b2.func('x^4-2*x^2',{strokeWidth:1.8,strokeColor:'#628aff',xMax:1.7,xMin:-1.7}); });/*B odd functions: green #00c997 and #62ffd8*/ mlg.af("showB",function() { b.remove(el); el.p1 = b1.point(1,1,{size:0.25,"name":"(x,y)","label":{"position":340,"distance":1.75,"fontSize":0.75}}); el.p2 = b1.point(-1,-1,{size:0.25,"name":"(\\text{-}x,\\text{-}y)","label":{"position":160,"distance":1.75,"fontSize":0.75}}); el.p3 = b2.point(0.5,0.5,{size:0.25,"name":"(x,y)","label":{"position":340,"distance":1.75,"fontSize":0.75}}); el.p4 = b2.point(-0.5,-0.5,{size:0.25,"name":"(\\text{-}x,\\text{-}y)","label":{"position":160,"distance":1.75,"fontSize":0.75}}); el.f1 = b1.func('x^3',{strokeWidth:1.8,strokeColor:'#00c997'}); el.f2 = b2.func('x',{strokeWidth:1.8,strokeColor:'#00c997'}); });} catch( e ) { mw.hook("ml.error.report").fire(e, { tags: { module: "jsxg", graph: "Solution20736_1_471284253_l" } }); console.error(e); } if(b) {b.board.update()} }; mw.loader.using('ext.MLJSXGraph', function() { window.ml.jsxgraph.buildRealGraph("Solution20736_1_471284253_l", "Solution20736_1_471284253_p", 1, code); }); } ); } window.JXQtable["Solution20736_1_471284253_l"] = true;Even functionswindow.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) {$('#jsxbtn2147309102_2099464284').on('touchstart mousedown', function () { try { mlg.cf("showA"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );Odd functionswindow.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) {$('#jsxbtn1187670062_1465395534').on('touchstart mousedown', function () { try { mlg.cf("showB"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );Resetwindow.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) {$('#jsxbtn618622385_964202916').on('touchstart mousedown', function () { try { mlg.cf("showC"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );Consider the graphs above. We can see that for even functions, if (x,y) is on the graph, then (-x,y) is also on the graph. Meanwhile, for an odd function, if (x,y) is on the graph, then (-x,-y) is also on the graph. Now, consider the given function. r(x)=-6x2+5​ Let's calculate r(-x). r(-x)=-6(-x)2+5(-a)2=a2r(-x)=-6x2+5 Next, let's calculate -r(x). -r(x)=-(-6x2+5)Distribute -1-r(x)=6x2−5 Finally, let's think about what these results tell us.r(x)r(-x)-r(x) -6x2+5-6x2+56x2−5 We can see above that r(x)=r(-x). Therefore, r is an even function.
Exercises 13 Let's start by recalling the definitions of even functions and odd functions.Even FunctionOdd Function y=f(x) is even when f(-x)=f(x) for each x in the domain of f. The graph of an even function is symmetric about the y-axis.y=f(x) is odd when f(-x)=-f(x) for each x in the domain of f. The graph of an odd function is symmetric about the origin — it looks the same after a rotation of 180∘. With these definitions in mind, let's consider the given graph.We see that the graph is symmetric about the y-axis. Therefore, the function represented by the graph is even.
Exercises 14 Let's start by recalling the definitions of even functions and odd functions.Even FunctionOdd Function y=f(x) is even when f(-x)=f(x) for each x in the domain of f. The graph of an even function is symmetric about the y-axis.y=f(x) is odd when f(-x)=-f(x) for each x in the domain of f. The graph of an odd function is symmetric about the origin — it looks the same after a rotation of 180∘. With these definitions in mind, let's consider the given graph.We see that the graph is not symmetric about the y-axis, so it does not belong to an even function. We also see that the graph is not symmetric about the origin, so it does not belong to an odd function. Therefore, the function represented by the graph is neither even nor odd.
Exercises 15 Let's start by recalling the definitions of even functions and odd functions.Even FunctionOdd Function y=f(x) is even when f(-x)=f(x) for each x in the domain of f. The graph of an even function is symmetric about the y-axis.y=f(x) is odd when f(-x)=-f(x) for each x in the domain of f. The graph of an odd function is symmetric about the origin — it looks the same after a rotation of 180∘. With these definitions in mind, let's consider the given graph.We see that the graph is not symmetric about the y-axis, so it does not belong to an even function. We also see that the graph is not symmetric about the origin, so it does not belong to an odd function. Therefore, the function represented by the graph is neither even nor odd.
Exercises 16 Let's start by recalling the definitions of even functions and odd functions.Even FunctionOdd Function y=f(x) is even when f(-x)=f(x) for each x in the domain of f. The graph of an even function is symmetric about the y-axis.y=f(x) is odd when f(-x)=-f(x) for each x in the domain of f. The graph of an odd function is symmetric about the origin — it looks the same after a rotation of 180∘. With these definitions in mind, let's consider the given graph.We see that the graph is symmetric about the y-axis. Therefore, the function represented by the graph is even.
Exercises 17 Let's start by recalling the definitions of even functions and odd functions.Even FunctionOdd Function y=f(x) is even when f(-x)=f(x) for each x in the domain of f. The graph of an even function is symmetric about the y-axis.y=f(x) is odd when f(-x)=-f(x) for each x in the domain of f. The graph of an odd function is symmetric about the origin — it looks the same after a rotation of 180∘. With these definitions in mind, let's consider the given graph.If we reflect the graph in the x-axis and the y-axis, it will be mapped onto itself. Therefore, the function represented by the graph is odd.
Exercises 18 Let's start by recalling the definitions of even functions and odd functions.Even FunctionOdd Function y=f(x) is even when f(-x)=f(x) for each x in the domain of f. The graph of an even function is symmetric about the y-axis.y=f(x) is odd when f(-x)=-f(x) for each x in the domain of f. The graph of an odd function is symmetric about the origin — it looks the same after a rotation of 180∘. With these definitions in mind, let's consider the given graph.We see that the graph is not symmetric about the y-axis, so it does not belong to an even function. We also see that the graph is not symmetric about the origin, so it does not belong to an odd function. Therefore, the function represented by the graph is neither even nor odd.
Exercises 19 We want to identify the vertex and the axis of symmetry of the graph of the given quadratic function. To do so, we will first express it in vertex form, f(x)=a(x−h)2+k, where a, h, and k are either positive or negative numbers. f(x)=3(x+1)2⇔f(x)=3(x−(-1))2+0​ It is important to note that we do not need to graph the parabola to identify the desired information. Let's compare the general formula for the vertex form to our function. General formula: f(x)=Equation: f(x)=​ -a(x−-(h)2+k 3(x−(-1))2+0​ We can see that a=3, h=-1, and k=0.Vertex The vertex of a quadratic function written in vertex form is the point (h,k). For this exercise, we have h=-1 and k=0. Therefore, the vertex of the given equation is (-1,0).Axis of Symmetry The axis of symmetry of a quadratic function written in vertex form is the vertical line with equation x=h. As we have already noticed, for our function, this is h=-1. Thus, the axis of symmetry is the line x=-1.
Exercises 20 We want to identify the vertex and the axis of symmetry of the graph of the given quadratic function. Note that the function is already expressed in vertex form, f(x)=a(x−h)2+k, where a, h, and k are either positive or negative numbers. f(x)=41​(x−6)2⇔f(x)=41​(x−6)2+0​ It is important to note that we do not need to graph the parabola to identify the desired information. Let's compare the general formula for the vertex form to our function. General formula: f(x)=Equation: f(x)=​ a (x−h)2+k 41​(x−6)2+0​ We can see that a=41​, h=6, and k=0.Vertex The vertex of a quadratic function written in vertex form is the point (h,k). For this exercise, we have h=6 and k=0. Therefore, the vertex of the given equation is (6,0).Axis of Symmetry The axis of symmetry of a quadratic function written in vertex form is the vertical line with equation x=h. As we have already noticed, for our function, this is h=6. Thus, the axis of symmetry is the line x=6.
Exercises 21 We want to identify the vertex and the axis of symmetry of the graph of the given quadratic function. Note that the function is already expressed in vertex form, y=a(x−h)2+k, where a, h, and k are either positive or negative numbers. y=-81​(x−4)2⇔y=-81​(x−4)2+0​ It is important to note that we do not need to graph the parabola to identify the desired information. Let's compare the general formula for the vertex form to our function. General formula: y=Equation: y=​ a (x−h)2+k -81​(x−4)2+0​ We can see that a=-81​, h=4, and k=0.Vertex The vertex of a quadratic function written in vertex form is the point (h,k). For this exercise, we have h=4 and k=0. Therefore, the vertex of the given equation is (4,0).Axis of Symmetry The axis of symmetry of a quadratic function written in vertex form is the vertical line with equation x=h. As we have already noticed, for our function, this is h=4. Thus, the axis of symmetry is the line x=4.
Exercises 22 We want to identify the vertex and the axis of symmetry of the graph of the given quadratic function. To do so, we will first express it in vertex form, y=a(x−h)2+k, where a, h, and k are either positive or negative numbers. y=-5(x+9)2⇔y=-5(x−(-9))2+0​ It is important to note that we do not need to graph the parabola to identify the desired information. Let's compare the general formula for the vertex form to our function. General formula: f(x)=Equation: f(x)=​ -a(x−-(h)2+k -5(x−(-9))2+0​ We can see that a=-5, h=-9, and k=0.Vertex The vertex of a quadratic function written in vertex form is the point (h,k). For this exercise, we have h=-9 and k=0. Therefore, the vertex of the given equation is (-9,0).Axis of Symmetry The axis of symmetry of a quadratic function written in vertex form is the vertical line with equation x=h. As we have already noticed, for our function, this is h=-9. Thus, the axis of symmetry is the line x=-9.
Exercises 23 To graph the parabola, we first need to identify the axis of symmetry and the vertex. f(x)=a(x−h)2​ In this form, the axis of symmetry of the parabola is the vertical line x=h and the vertex is the point (h,0). Now consider the given function. g(x)=2(x+3)2⇔g(x)=2(x−(-3))2​ We can see that a=2 and h=-3. Therefore, the vertex is (-3,0), and the axis of symmetry is x=-3. To graph the function we need to find two more points on the graph. Let's choose two x-values less than the x-coordinate of the vertex and make a table of values.x2(x+3)2g(x)=2(x+3)2 -42(-4+3)22 -52(-5+3)28 Let's now plot the vertex and draw the axis of symmetry on a coordinate plane. We will also plot and reflect the obtained points across the axis of symmetry.Let's draw a smooth curve that connects the five points. We will also draw the parent function f(x)=x2.From the graph above, we can note the following.Both graphs open up. The graph of g(x)=2(x+3)2 is narrower than the graph of f(x)=x2. The axis of symmetry of the parent function is the y-axis. The axis of symmetry of the given function is x=-3. The vertex of the given function, (-3,0), is to the left of the vertex of the parent function, (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 2 and a horizontal translation left 3 units of the graph of f.
Exercises 24 To graph the parabola, we first need to identify the axis of symmetry and the vertex. f(x)=a(x−h)2​ In this form, the axis of symmetry of the parabola is the vertical line x=h and the vertex is the point (h,0). Now consider the given function. p(x)=3(x−1)2​ We can see that a=3 and h=1. Therefore, the vertex is (1,0), and the axis of symmetry is x=1. To graph the function we need to find two more points on the graph. Let's choose two x-values less than the x-coordinate of the vertex and make a table of values.x3(x−1)2p(x)=3(x−1)2 03(0−1)23 -13(-1−1)212 Let's now plot the vertex and draw the axis of symmetry on a coordinate plane. We will also plot and reflect the obtained points across the axis of symmetry.Let's draw a smooth curve that connects the five points. We will also draw the parent function f(x)=x2.From the graph above, we can note the following.Both graphs open up. The graph of p(x)=3(x−1)2 is narrower than the graph of f(x)=x2. The axis of symmetry of the parent function is the y-axis. The axis of symmetry of the given function is x=1. The vertex of the given function, (1,0), is to the right of the vertex of the parent function, (0,0). From the graph and the observations above, we can conclude that the graph of p is a vertical stretch by a factor of 3 and a horizontal translation right 1 unit of the graph of f.
Exercises 25 To graph the parabola, we first need to identify the axis of symmetry and the vertex. f(x)=a(x−h)2​ In this form, the axis of symmetry of the parabola is the vertical line x=h and the vertex is the point (h,0). Now consider the given function. r(x)=41​(x+10)2⇔r(x)=41​(x−(-10))2​ We can see that a=41​ and h=-10. Therefore, the vertex is (-10,0) and the axis of symmetry is x=-10. To graph the function we need to find two more points on the graph. Let's choose two x-values less than the x-coordinate of the vertex and make a table of values.x41​(x+10)2r(x)=41​(x+10)2 -1241​(-12+10)21 -1441​(-14+10)24 Let's now plot the vertex and draw the axis of symmetry on a coordinate plane. We will also plot and reflect the obtained points across the axis of symmetry.Let's draw a smooth curve that connects the five points. We will also draw the parent function f(x)=x2.From the graph above, we can note the following.Both graphs open up. The graph of r(x)=41​(x+10)2 is wider than the graph of f(x)=x2. The axis of symmetry of the parent function is the y-axis. The axis of symmetry of the given function is x=-10. The vertex of the given function, (-10,0), is to the left of the vertex of the parent function, (0,0). From the graph and the observations above, we can conclude that the graph of r is a vertical shrink by a factor of 41​ and a horizontal translation left 10 units of the graph of f.
Exercises 26 To graph the parabola, we first need to identify the axis of symmetry and the vertex. f(x)=a(x−h)2​ In this form, the axis of symmetry of the parabola is the vertical line x=h and the vertex is the point (h,0). Now consider the given function. n(x)=31​(x−6)2​ We can see that a=31​ and h=6. Therefore, the vertex is (6,0), and the axis of symmetry is x=6. To graph the function we need to find two more points on the graph. Let's choose two x-values less than the x-coordinate of the vertex and make a table of values.x31​(x−6)2n(x)=31​(x−6)2 031​(0−6)212 331​(3−6)23 Let's now plot the vertex and draw the axis of symmetry on a coordinate plane. We will also plot and reflect the obtained points across the axis of symmetry.Let's draw a smooth curve that connects the five points. We will also draw the parent function f(x)=x2.From the graph above, we can note the following.Both graphs open up. The graph of n(x)=31​(x−6)2 is wider than the graph of f(x)=x2. The axis of symmetry of the parent function is the y-axis. The axis of symmetry of the given function is x=6. The vertex of the given function, (6,0), is to the right of the vertex of the parent function, (0,0). From the graph and the observations above, we can conclude that the graph of n is a vertical shrink by a factor of 31​ and a horizontal translation right 6 units of the graph of f.
Exercises 27 To graph the parabola, we first need to identify the axis of symmetry and the vertex. f(x)=a(x−h)2​ In this form, the axis of symmetry of the parabola is the vertical line x=h and the vertex is the point (h,0). Now consider the given function. d(x)=51​(x−5)2​ We can see that a=51​ and h=5. Therefore, the vertex is (5,0), and the axis of symmetry is x=5. To graph the function we need to find two more points on the graph. Let's choose two x-values less than the x-coordinate of the vertex and make a table of values.x51​(x−5)2d(x)=51​(x−5)2 -551​(-5−5)220 051​(0−5)25 Let's now plot the vertex and draw the axis of symmetry on a coordinate plane. We will also plot and reflect the obtained points across the axis of symmetry.Let's draw a smooth curve that connects the five points. We will also draw the parent function f(x)=x2.From the graph above, we can note the following.Both graphs open up. The graph of d(x)=51​(x−5)2 is wider than the graph of f(x)=x2. The axis of symmetry of the parent function is the y-axis. The axis of symmetry of the given function is x=5. The vertex of the given function, (5,0), is to the right of the vertex of the parent function, (0,0). From the graph and the observations above, we can conclude that the graph of d is a vertical shrink by a factor of 51​ and a horizontal translation right 5 units of the graph of f.
Exercises 28 To graph the parabola, we first need to identify the axis of symmetry and the vertex. f(x)=a(x−h)2​ In this form, the axis of symmetry of the parabola is the vertical line x=h and the vertex is the point (h,0). Now consider the given function. q(x)=6(x+2)2⇔q(x)=6(x−(-2))2​ We can see that a=6 and h=-2. Therefore, the vertex is (-2,0), and the axis of symmetry is x=-2. To graph the function we need to find two more points on the graph. Let's choose two x-values less than the x-coordinate of the vertex and make a table of values.x6(x+2)2q(x)=6(x+2)2 -36(-3+2)26 -46(-4+2)224 Let's now plot the vertex and draw the axis of symmetry on a coordinate plane. We will also plot and reflect the obtained points across the axis of symmetry.Let's draw a smooth curve that connects the five points. We will also draw the parent function f(x)=x2.From the graph above, we can note the following.Both graphs open up. The graph of q(x)=6(x+2)2 is narrower than the graph of f(x)=x2. The axis of symmetry of the parent function is the y-axis. The axis of symmetry of the given function is x=-2. The vertex of the given function, (-2,0), is to the left of the vertex of the parent function, (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 6 and a horizontal translation left 2 units of the graph of f.
Exercises 29 To determine whether the function is even, odd, or neither, first let's recall how we determine this.RelationshipResulting Implication f(-x)=-f(x)Function is odd f(-x)=f(x)Function is evenUsing this information we can determine if the function f(x)=x2+3 is even, odd, or neither. f(x)=x2+3x=-xf(-x)=(-x)2+3(-a)2=a2f(-x)=x2+3 We can see that the function f(-x)=f(x), which implies the function is even, not odd.
Exercises 30 To determine the error, we will find the vertex by ourselves. To do so, we will first express the quadratic function in vertex form, y=a(x−h)2+k, where a, h, and k are either positive or negative constants. y=-(x+8)2⇔y=-1(x−(-8))2+0​ Let's compare the general formula for the vertex form with our equation. General formula: y=Equation: y=​ -a(x−-(h)2+k -1(x−(-8))2+0​ We can see that a=-1, h=-8, and k=0. The vertex of a quadratic function written in vertex form is the point (h,k). Therefore, the vertex of this parabola is (-8,0). Let's now examine the given solution.We see that the value of h was found correctly. However, the interpretation of the vertex form is not correct. Based on our operations at the beginning, we can correct the solution as follows.
Exercises 31 We want to identify the vertex and the axis of symmetry of the graph of the given quadratic function. To do so, we will first express it in vertex form, y=a(x−h)2+k, where a, h, and k are either positive or negative numbers. y=-6(x+4)2−3   ⇔   y=-6(x−(-4))2+(-3)​ It is important to note that we do not need to graph the parabola to identify the desired information. Let's compare the general formula for the vertex form with our equation. General formula: y=Equation: y=​ -a(x−-(h()2+--k -6(x−(-4))2+(-3)​ We can see that a=-6, h=-4, and k=-3.Vertex The vertex of a quadratic function written in vertex form is the point (h,k). For this exercise, we have h=-4 and k=-3. Therefore, the vertex of the given equation is (-4,-3).Axis of Symmetry The axis of symmetry of a quadratic function written in vertex form is the vertical line with equation x=h. As we have already noticed, for our function, this is h=-4. Thus, the axis of symmetry is the line x=-4.
Exercises 32 We want to identify the vertex and the axis of symmetry of the graph of the given quadratic function. Note that the function is already expressed in vertex form, f(x)=a(x−h)2+k, where a, h, and k are either positive or negative numbers. f(x)=3(x−3)2+6​ It is important to note that we do not need to graph the parabola to identify the desired information. Let's compare the general formula for the vertex form with our equation. General formula: y=Equation: y=​ a(x−h)2+k 3(x−3)2+6​ We can see that a=3, h=3, and k=6.Vertex The vertex of a quadratic function written in vertex form is the point (h,k). For this exercise, we have h=3 and k=6. Therefore, the vertex of the given equation is (3,6).Axis of Symmetry The axis of symmetry of a quadratic function written in vertex form is the vertical line with equation x=h. As we have already noticed, for our function, this is h=3. Thus, the axis of symmetry is the line x=3.
Exercises 33 We want to identify the vertex and the axis of symmetry of the graph of the given quadratic function. To do so, we will first express it in vertex form, y=a(x−h)2+k, where a, h, and k are either positive or negative numbers. y=-4(x+3)2+1   ⇔   y=-4(x−(-3))2+1​ It is important to note that we do not need to graph the parabola to identify the desired information. Let's compare the general formula for the vertex form with our equation. General formula: y=Equation: y=​ -a(x−-(h()2+--k -4(x−(-3))2+1​ We can see that a=-4, h=-3, and k=1.Vertex The vertex of a quadratic function written in vertex form is the point (h,k). For this exercise, we have h=-3 and k=1. Therefore, the vertex of the given equation is (-3,1).Axis of Symmetry The axis of symmetry of a quadratic function written in vertex form is the vertical line with equation x=h. As we have already noticed, for our function, this is h=-3. Thus, the axis of symmetry is the line x=-3.
Exercises 34 We want to identify the vertex and the axis of symmetry of the graph of the given quadratic function. To do so, we will first express it in vertex form, y=a(x−h)2+k, where a, h, and k are either positive or negative numbers. y=-(x−6)2−5⇔y=-(x−6)2+(-5)​ It is important to note that we do not need to graph the parabola to identify the desired information. Let's compare the general formula for the vertex form with our equation. General formula: y=Equation: y=​ -a(x−h)2+--k -1(x−6)2+(-5)​ We can see that a=-1, h=6, and k=-5.Vertex The vertex of a quadratic function written in vertex form is the point (h,k). For this exercise, we have h=6 and k=-5. Therefore, the vertex of the given equation is (6,-5).Axis of Symmetry The axis of symmetry of a quadratic function written in vertex form is the vertical line with equation x=h. As we have already noticed, for our function, this is h=6. Thus, the axis of symmetry is the line x=6.
Exercises 35 To match the function with its graph, we should first identify the vertex and then determine whether the parabola opens upward or downward. To do so, we will first express it in vertex form, y=a(x−h)2+k, where a, h, and k are either positive or negative numbers. y=-(x+1)2−3⇔y=-1(x−(-1))2+(-3)​ Let's compare the general formula for the vertex form with our equation. General formula: y=Equation: y=​ -a(x−-(h)2+--k -1(x−(-1))2+(-3)​ We can see that a=-1, h=-1, and k=-3. Since the vertex of a quadratic function written in vertex form is the point (h,k), the vertex of our function is (-1,-3). Let's now determine the direction of the parabola. Recall that if a>0, the parabola opens upwards. Conversely, if a<0, the parabola opens downwards.In the given function, we have a=-1, which is less than 0. Therefore, the parabola opens downward. As a result, the graph of the function is the graph given in choice C.
Exercises 36 To match the function with its graph, we should first identify the vertex and then determine whether the parabola opens upward or downward. Note that the equation is already expressed in vertex form, y=a(x−h)2+k, where a, h, and k are either positive or negative numbers. y=-21​(x−1)2+3​ Let's compare the general formula for the vertex form with our equation. General formula: y=Equation: y=​ -a(x−h)2+k -21​(x−1)2+3​ We can see that a=-21​, h=1, and k=3. Since the vertex of a quadratic function written in vertex form is the point (h,k), the vertex of our function is (1,3). Let's now determine the direction of the parabola. Recall that if a>0, the parabola opens upwards. Conversely, if a<0, the parabola opens downwards.In the given function, we have a=-21​, which is less than 0. Therefore, the parabola opens downward. As a result, the graph of the function is the graph given in choice A.
Exercises 37 To match the function with its graph, we should first identify the vertex and then determine whether the parabola opens upward or downward. Note that the equation is already expressed in vertex form, y=a(x−h)2+k, where a, h, and k are either positive or negative numbers. y=31​(x−1)2+3​ Let's compare the general formula for the vertex form with our equation. General formula: y=Equation: y=​ a(x−h)2+k 31​(x−1)2+3​ We can see that a=31​, h=1, and k=3. Since the vertex of a quadratic function written in vertex form is the point (h,k), the vertex of our function is (1,3). Let's now determine the direction of the parabola. Recall that if a>0, the parabola opens upwards. Conversely, if a<0, the parabola opens downwards.In the given function, we have a=31​, which is greater than 0. Therefore, the parabola opens upward. As a result, the graph of the function is the graph given in choice D.
Exercises 38 To match the function with its graph, we should first identify the vertex and then determine whether the parabola opens upward or downward. To do so, we will first express it in vertex form, y=a(x−h)2+k, where a, h, and k are either positive or negative numbers. y=2(x+1)2−3⇔y=2(x−(-1))2+(-3)​ Let's compare the general formula for the vertex form with our equation. General formula: y=Equation: y=​ a(x−-(h)2+--k 2(x−(-1))2+(-3)​ We can see that a=2, h=-1, and k=-3. Since the vertex of a quadratic function written in vertex form is the point (h,k), the vertex of our function is (-1,-3). Let's now determine the direction of the parabola.Recall that, if a>0, the parabola opens upwards. Conversely, if a<0, the parabola opens downwards.In the given function, we have a=2, which is greater than 0. Therefore, the parabola opens upward. As a result, the graph of the function is the graph given in choice B.
Exercises 39 To graph the given parabola, that is written in the vertex form, we first need to identify the axis of symmetry and the vertex. f(x)=a(x−h)2+k​ In this form, the axis of symmetry is the vertical line x=h and the vertex lies at (h,k). Now consider the given function. h(x)=1(x−2)2+4​ We can see that a=1, h=2 and k=4. Therefore, the vertex is (2,4), and the axis of symmetry is x=2. To graph the function we need to find two more points on the graph. Let's choose two x-values less than the x-coordinate of the vertex and make a table of values.x(x−2)2+4h(x)=(x−2)2+4 1(1−2)2+45 0(0−2)2+48 Next, we will plot the vertex and draw the axis of symmetry on a coordinate plane. We will also plot and reflect the obtained points across the axis of symmetry.Let's draw a smooth curve that connects the five 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. The axis of symmetry of the parent function is the y-axis. The axis of symmetry of h(x)=(x−2)2+4 is x=2. The vertex of the given function, (2,4), is above and to the right of the vertex of the parent function, (0,0). From the graph and the observations above, we can conclude that the graph of h is a translation right 2 units and up 4 units of the graph of f.
Exercises 40 To graph the given parabola, that is written in the vertex form, we first need to identify the axis of symmetry and the vertex. f(x)=a(x−h)2+k​ In this form, the axis of symmetry is the vertical line x=h and the vertex lies at (h,k). Now consider the given function. g(x)=(x+1)2−7⇔g(x)=1(x−(-1))2+(-7)​ We can see that a=1, h=-1 and k=-7. Therefore, the vertex is (-1,-7), and the axis of symmetry is x=-1. To graph the function we need to find two more points on the graph. Let's choose two x-values less than the x-coordinate of the vertex and make a table of values.x(x+1)2−7g(x)=(x+1)2−7 -2(-2+1)2−7-6 -3(-3+1)2−7-3 Next, we will plot the vertex and draw the axis of symmetry on a coordinate plane. We will also plot and reflect the obtained points across the axis of symmetry.Let's draw a smooth curve that connects the five 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. The axis of symmetry of the parent function is the y-axis. The axis of symmetry of g(x)=(x+1)2−7 is x=-1. The vertex of the given function, (−1,−7), is below and to the left of the vertex of the parent function, (0,0). From the graph and the observations above, we can conclude that the graph of g is a translation left 1 unit and down 7 units of the graph of f.
Exercises 41 To graph the given parabola, that is written in the vertex form, we first need to identify the axis of symmetry and the vertex. f(x)=a(x−h)2+k​ In this form, the axis of symmetry is the vertical line x=h and the vertex lies at (h,k). Now consider the given function. r(x)=4(x−1)2−5⇕r(x)=4(x−1)2+(-5)​ We can see that a=4, h=1 and k=-5. Therefore, the vertex is (1,-5), and the axis of symmetry is x=1. To graph the function we need to find two more points on the graph. Let's choose two x-values less than the x-coordinate of the vertex and make a table of values.x4(x−1)2−5r(x)=4(x−1)2−5 04(0−1)2−5-1 -14(-1−1)2−511 Next, we will plot the vertex and draw the axis of symmetry on a coordinate plane. We will also plot and reflect the obtained points across the axis of symmetry.Let's draw a smooth curve that connects the five 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. The graph of r(x)=4(x−1)2−5 is narrower than the graph of f(x)=x2. The axis of symmetry of the parent function is the y-axis. The axis of symmetry of the given function is x=1. The vertex of the given function, (1,-5), is below and to the right of the vertex of the parent function, (0,0). From the graph and the observations above, we can conclude that the graph of r is a vertical stretch by a factor of 4, followed by a translation right 1 unit and down 5 units of the graph of f.
Exercises 42 To graph the given parabola, that is written in the vertex form, we first need to identify the axis of symmetry and the vertex. f(x)=a(x−h)2+k​ In this form, the axis of symmetry is the vertical line x=h and the vertex lies at (h,k). Now consider the given function. n(x)=-(x+4)2+2⇕n(x)=-1(x−(-4))2+2​ We can see that a=-1, h=-4 and k=2. Therefore, the vertex is (-4,2), and the axis of symmetry is x=-4. To graph the function we need to find two more points on the graph. Let's choose two x-values less than the x-coordinate of the vertex and make a table of values.x-(x+4)2+2n(x)=-(x+4)2+2 -5-(-5+4)2+21 -6-(-6+4)2+2-2 Next, we will plot the vertex and draw the axis of symmetry on a coordinate plane. We will also plot and reflect the obtained points across the axis of symmetry.Let's draw a smooth curve that connects the five obtained points. We will also draw the parent function f(x)=x2.From the graph above, we can note the following.The graph of n(x)=-(x+4)2+2 opens down, while the graph of f(x)=x2 opens up. The axis of symmetry of the parent function is the y-axis. The axis of symmetry of the given function is x=-4. The vertex of the given function, (-4,2), is above and to the left of the vertex of the parent function, (0,0). From the graph and the observations above, we can conclude that the graph of n is a reflection in the x-axis, followed by a translation left 4 units and up 2 units of the graph of f.
Exercises 43 To graph the given parabola, that is written in the vertex form, we first need to identify the axis of symmetry and the vertex. f(x)=a(x−h)2+k​ In this form, the axis of symmetry is the vertical line x=h and the vertex lies at (h,k). Now consider the given function. g(x)=-31​(x+3)2−2⇕g(x)=-31​(x−(-3))2+(-2)​ We can see that a=-31​, h=-3 and k=-2. Therefore, the vertex is (-3,-2), and the axis of symmetry is x=-3. To graph the function we need to find two more points on the graph. Let's choose two x-values less than the x-coordinate of the vertex and make a table of values.x-31​(x+3)2−2g(x)=-31​(x+3)2−2 -6-31​(-6+3)2−2-5 -9-31​(-9+3)2−2-14 Next, we will plot the vertex and draw the axis of symmetry on a coordinate plane. We will also plot and reflect the obtained points across the axis of symmetry.Let's draw a smooth curve that connects the five obtained points. We will also draw the parent function f(x)=x2.From the graph above, we can note the following.The graph of g(x)=-31​(x+3)2−2 opens down, while the graph of f(x)=x2 opens up. The graph of g is wider than the graph of f(x)=x2. The axis of symmetry of the parent function is the y-axis. The axis of symmetry of the given function is x=-3. The vertex of the given function, (-3,-2), is below and to the left of the vertex of the parent function, (0,0). From the graph and the observations above, we can conclude that the graph of g is a reflection in the x-axis, vertical shrink by a factor of 31​, followed by a translation left 3 units and down 2 units of the graph of f.
Exercises 44 To graph the given parabola, that is written in the vertex form, we first need to identify the axis of symmetry and the vertex. f(x)=a(x−h)2+k​ In this form, the axis of symmetry is the vertical line x=h and the vertex lies at (h,k). Now consider the given function. r(x)=21​(x−2)2−4⇕r(x)=21​(x−2)2+(-4)​ We can see that a=21​, h=2 and k=-4. Therefore, the vertex is (2,-4), and the axis of symmetry is x=2. To graph the function we need to find two more points on the graph. Let's choose two x-values less than the x-coordinate of the vertex and make a table of values.x21​(x−2)2−4r(x)=21​(x−2)2−4 021​(0−2)2−4-2 -221​(-2−2)2−44 Next, we will plot the vertex and draw the axis of symmetry on a coordinate plane. We will also plot and reflect the obtained points across the axis of symmetry.Let's draw a smooth curve that connects the five 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. The graph of r(x)=21​(x−2)2−4 is wider than the graph of f(x)=x2. The axis of symmetry of the parent function is the y-axis. The axis of symmetry of the given function is x=2. The vertex of the given function, (2,-4), is below and to the right of the vertex of the parent function, (0,0). From the graph and the observations above, we can conclude that the graph of r is a vertical shrink by a factor of 21​, followed by a translation right 2 units and down 4 units of the graph of f.
Exercises 45 In order to find the function for g(x), substitute (x−1) for x into f(x)=(x−2)2+1. f(x)=(x−2)2+1x=(x−1)f(x−1)=((x−1)−2)2+1Subtract termf(x−1)=(x−3)2+1 In order to graph this quadratic function we must identify the values of h and k. g(x)=a(x−h)2+k⇔g(x)=(x−3)2+1​ We have identified the values of h=3 and k=1, which identifies the vertex (3,1). Let's begin graphing by plotting the vertex and the axis of symmetry. Because h=3, we will graph the line x=3.Now we must find and plot two more points on the graph. We will choose two x-values less than the x-coordinate of the vertex. Then we will find g(x) for each x-value. Let's use x=1 and x=2.x-valueg(x)=(x−3)2+1Simplify(x,g(x)) 1g(1)=(1−3)2+1g(x)=5(1,5) 2g(2)=(2−3)2+1g(x)=2(2,2) Plot these points and reflect them across the axis of symmetry.Finally, we will draw a smooth curve through the points to graph the function.We can see that the graph of g(x) is represented by option A.
Exercises 46 In order to find the function for r(x), we will substitute (x+2) for x into f(x)=(x−2)2+1. f(x)=(x−2)2+1x=(x+2)f(x+2)=((x+2)−2)2+1Subtract termf(x+2)=x2+1 In order to graph this quadratic function, we must identify the values of h and k. r(x)=a(x−h)2+k⇔r(x)=(x−0)2+1​ We have identified the values of h=0 and k=1, which identifies the vertex (0,1). Let's begin graphing by plotting the vertex and the axis of symmetry. Because h=0, we will graph the line x=0.Now we will find and plot two more points on the graph. We should choose two x-values less than the x-coordinate of the vertex. Then we will find r(x) for each x-value. Let's use x=-2 and x=-1.x-valuer(x)=x2+1Simplify(x,r(x)) -2r(-2)=(-2)2+1r(x)=5(-2,5) -1r(-1)=(-1)2+1r(x)=2(-1,2) Plot these points and reflect them across the axis of symmetry.Finally, we will draw a smooth curve through the points to graph the function.We can see that the graph of r(x) is represented by option C.
Exercises 47 Let's determine the function for h(x). h(x)=f(x)+2⇔h(x)=(x−2)2+1+2⇔h(x)=(x−2)2+3​ In order to graph this quadratic function, we must identify the values of h and k. h(x)=a(x−h)2+k⇔h(x)=(x−2)2+3​ We have identified the values of h=2 and k=3, which identifies the vertex (2,3). Let's begin graphing by plotting the vertex and the axis of symmetry. Because h=2, we will graph the line x=2.Now we will find and plot two more points on the graph. We will choose two x-values less than the x-coordinate of the vertex. Then we will find h(x) for each x-value. Let's use x=0 and x=1.x-valueh(x)=(x−2)2+3Simplify(x,h(x)) 0h(0)=(0−2)2+3h(x)=7(0,7) 1h(1)=(1−2)2+3h(x)=4(1,4) Plot these points and reflect them across the axis of symmetry.Finally, we will draw a smooth curve through the points to graph the function.We can see that the graph of h(x) is represented by option B.
Exercises 48 Let's determine the function for p(x). p(x)=f(x)−3⇔p(x)=(x−2)2+1−3⇔p(x)=(x−2)2−2​ In order to graph this quadratic function, we must identify the values of h and k. p(x)=a(x−h)2+k⇔p(x)=(x−2)2−2​ We have identified the values of h=2 and k=-2, which identifies the vertex (2,-2). Let's begin graphing by plotting the vertex and the axis of symmetry. Because h=2, we will graph the line x=2.Now we will find and plot two more points on the graph. We will choose two x-values less than the x-coordinate of the vertex. Then we will find p(x) for each x-value. Let's use x=0 and x=1.x-valuep(x)=(x−2)2−2Simplify(x,p(x)) 0p(0)=(0−2)2−2p(x)=2(0,2) 1p(1)=(1−2)2−2p(x)=-1(1,-1) Plot these points and reflect them across the axis of symmetry.Finally, we will draw a smooth curve through the points to graph the function.We can see that the graph of p(x) is represented by option D.
Exercises 49 We have to draw the graph of g, which is a transformation of the graph of f.Describing the Transformation We will first describe the given transformation g(x)=f(x+3). Let's look at the table.Transformations of f(x) Horizontal TranslationsTranslation right h units, h>0y=f(x−h)​ Translation left h units, h>0y=f(x+h)​ Now, using the table, let's highlight the transformation of f(x). g(x)=f(x+3)​ The function g is of the form y=f(x+h) where h=3. So the graph of g is a horizontal translation 3 units left of the graph of f.Graphing the Functions Let's make a table of values for the function f(x)=2(x−1)2+1.x2(x−1)2+1f(x)=2(x−1)2+1 -12(-1−1)2+19 02(0−1)2+13 12(1−1)2+11 22(2−1)2+13 32(3−1)2+19 To graph f, let's draw a smooth curve that connects the five obtained points.Finally, we will graph g as a horizontal translation left 3 units of a graph of f. To do so, let's substract 3 from the x-coordinates of the points on the graph of f.
Exercises 50 We have to draw the graph of g, which is a transformation of the graph of f.Describing the Transformation We will first describe the given transformation g(x)=21​f(x). Let's look at the table.Transformations of f(x) Vertical Stretch or ShrinkVertical stretch, a>1y=af(x)​ Vertical shrink, 0<a<1y=af(x)​ Now, using the table, let's highlight the transformation of f(x). g(x)=21​f(x)​ The function g is of the form y=af(x) where a=21​. Since 0<a<1, the graph of g is a vertical shrink by a factor of 21​ of the graph of f.Graphing the Functions Let's make a table of values for the function f(x)=-(x+1)2+2.x-(x+1)2+2f(x)=-(x+1)2+2 -3-(-3+1)2+2-2 -2-(-2+1)2+21 -1-(-1+1)2+22 0-(0+1)2+21 1-(1+1)2+2-2 To graph f, let's draw a smooth curve that connects the five obtained points.Finally, we will graph g as a vertical shrink by a factor of 21​ of the graph of f. To do it, let's multiply by 21​ each y-coordinate of the points on the graph of f.
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