import{c as M}from"./courbes-lqVoJQXx.js";import{E as F,aj as P,r as o,i as k,j as r,g as h,s as p,u as b,q as l,o as y,a as Q,l as L}from"./index-XCg2QAX4.js";import{r as S}from"./reperes-uV74d7Az.js";const w="Determine an affine function using the images of two numbers",T=!0,v="mathLive",j="b8b33",q="3F21-2";function U(){F.call(this),this.titre=w,this.interactifReady=T,this.interactifType=v,this.consigne="",this.sup=1,this.nbQuestions=2,this.nbCols=2,this.nbColsCorr=2,this.tailleDiaporama=3,this.video="",this.nouvelleVersion=function(){this.titre=w,this.listeQuestions=[],this.listeCorrections=[];let g;parseInt(this.sup)===1?g=[0,1]:parseInt(this.sup)===2?g=[1,2,3]:g=[3,4];const D=P(g,this.nbQuestions);for(let x=0,a,i,n,$,t,e,f,c,m,d,s,C=0;x<this.nbQuestions&&C<50;){switch(d="",s="",D[x]){case 0:t=0,e=o(-10,10,0),a=o(-5,-1),i=o(1,5),n=e,$=e,s=`We notice that $f(${a})=f(${i})=${e}$ therefore the line representing the function $f$ passes through two distinct points having the same ordinate.<br>`,s+=`It is therefore parallel to the abscissa axis. The function $f$ is a constant function and $f(x)=${e}$.`,p(this,x,`f(x)=${e}`),this.correctionDetaillee&&(f=b(l(a,n),"red"),c=b(l(i,$),"red"),m=S({xMin:-5,yMin:Math.min(-1,e-1),xMax:5,yMax:Math.max(e+1,2)}),s+=`<br><br>${y({xmin:-5,ymin:Math.min(-1,e-1),xmax:5,ymax:Math.max(e+1,2),pixelsParCm:20,scale:.7},m,M(u=>t*u+e,{mark:m,color:"blue"}),f,c)}`);break;case 1:t=o(-2,2,0),e=o(-5,5,0),a=0,n=e,i=o(-5,5,0),$=e+t*i,s=`Let $f(x)=ax+b$. We know that $f(0)=${n}=b$.<br>`,s+=`So $f(x)=ax${h(n)}$. Using the data $f(${i})=${$}$ we obtain: $a \\times ${r(i)}${h(e)} = ${$}$ hence $a \\times ${r(i)} = ${$}${h(-e)} = ${$-e}$ therefore $a=\\dfrac{${$-e}}{${i} } = ${t}$.<br>`,s+=`So $f(x)=${t}x${h(e)}$.`,p(this,x,`f(x)=${t}x${h(e)}`),this.correctionDetaillee&&(f=b(l(a,n),"red"),c=b(l(i,$),"red"),m=S({xMin:-5,yMin:Math.min(-5*t+e,5*t+e),xMax:5,yMax:Math.max(-5*t+e,5*t+e)}),s+=`<br><br>${y({xmin:-5,ymin:Math.min(-5*t+e,5*t+e),xmax:5,ymax:Math.max(-5*t+e,5*t+e),pixelsParCm:20,scale:.7},m,M(u=>t*u+e,{mark:m,color:"blue"}),f,c)}`);break;case 2:t=o(-5,5,0),e=o(-5,5,0),a=o(-5,5,[-1,0]),n=t*a+e,i=a+1,$=e+t*i,s=`Let $f(x)=ax+b$. We go from $${a}$ to $${i}$ by adding 1, so the slope $a$ of the line corresponds to $f(${i})-f(${a})=${$}-${r(n)}=`,n<0?s+=`${$}${h(-n)} = ${t}$.<br>`:s+=`${t}$.<br>`,s+=`So $f(x)=${t}x+b$.<br>Using the data $f(${i})=${$}$ we obtain: $${t} \\times ${r(i)}+b=${$}$ hence $${t*i}+b=${$} $ therefore $b=${$}${h(-t*i)} = ${e}$.<br>`,s+=`So $f(x)=${t}x${h(e)}$.`,p(this,x,`f(x)=${t}x${h(e)}`),this.correctionDetaillee&&(f=b(l(a,n),"red"),c=b(l(i,$),"red"),m=S({xMin:-5,yMin:Math.min(-5*t+e,5*t+e),xMax:5,yMax:Math.max(-5*t+e,5*t+e)}),s+=`<br><br>${y({xmin:-5,ymin:Math.min(-5*t+e,5*t+e),xmax:5,ymax:Math.max(-5*t+e,5*t+e),pixelsParCm:20,scale:.7},m,M(u=>t*u+e,{mark:m,color:"blue"}),f,c)}`);break;case 3:t=o(-5,5,0),e=o(-5,5,0),a=o(-5,5,0),n=t*a+e,i=o(-5,5,[0,a]),$=e+t*i,s=`Let $f(x)=ax+b$. Using the data from the statement, we obtain: $f(${a})=${n}=a \\times ${r(a)}+b$ and $f(${i})=${$}=a \\times ${r(i)}+b$<br>`,s+=`So on the one hand: $b=${n}+a\\times ${r(-a)}$ and on the other hand: $b=${$}+a\\times ${r(-i)}$.<br>`,s+=`By identification, we obtain: $${n}+a\\times ${r(-a)} = ${$}+a\\times ${r(-i)}$.<br>`,s+=`We deduce that $${n}${h(-$)}=a(${a}${h(-i)})$ is $${n-$} = ${a-i}a$.<br>`,s+=`So $a=\\dfrac{${n-$}}{${a-i}} = ${t}$.<br>`,s+=`So $b=${n}${h(t)}\\times ${r(-a)} = ${n}${h(-t*a)} = ${e}$.<br>`,s+=`So $f(x)=${t}x${h(e)}$.`,p(this,x,`f(x)=${t}x${h(e)}`),this.correctionDetaillee&&(f=b(l(a,n),"red"),c=b(l(i,$),"red"),m=S({xMin:-5,yMin:Math.min(-5*t+e,5*t+e),xMax:5,yMax:Math.max(-5*t+e,5*t+e)}),s+=`<br><br>${y({xmin:-5,ymin:Math.min(-5*t+e,5*t+e),xmax:5,ymax:Math.max(-5*t+e,5*t+e),pixelsParCm:20,scale:.7},m,M(u=>t*u+e,{mark:m,color:"blue"}),f,c)}`);break;case 4:a=o(-5,5,0),i=o(-5,5,[0,a]),n=o(-5,5),$=o(-5,5),t=k($-n,i-a),e=t.multiplieEntier(-a).ajouteEntier(n),s=`Let $f(x)=ax+b$. Using the data from the statement, we obtain: $f(${a})=${n}=a \\times ${r(a)}+b$ and $f(${i})=${$}=a \\times ${r(i)}+b$<br>`,s+=`So on the one hand: $b=${n}+a\\times ${r(-a)}$ and on the other hand: $b=${$}+a\\times ${r(-i)}$.<br>`,s+=`By identification, we obtain: $${n}+a\\times ${r(-a)} = ${$}+a\\times ${r(-i)}$.<br>`,s+=`We deduce that $${n}${h(-$)}=a(${a}${h(-i)})$ is $${n-$} = ${a-i}a$.<br>`,s+=`So $a=\\dfrac{${n-$}}{${a-i}} = ${t.texFractionSimplifiee}$.<br>`,s+=`So $b=${n}+${t.texFractionSimplifiee}\\times ${r(-a)} = ${k(n*t.denIrred,t.denIrred).texFraction}+${t.multiplieEntier(-a).texFractionSimplifiee} = ${e.texFractionSimplifiee}$.<br>`,s+=`So $f(x)=${t.texFractionSimplifiee}x${e.simplifie().texFractionSignee}$.`,p(this,x,`f(x)=${t.texFractionSimplifiee}x${e.simplifie().texFractionSignee}`),this.correctionDetaillee&&(f=b(l(a,n),"red"),c=b(l(i,$),"red"),t=t.n/t.d,e=e.n/e.d,m=S({xMin:-5,yMin:Math.round(Math.min(-5*t+e,5*t+e)),xMax:5,yMax:Math.round(Math.max(-5*t+e,5*t+e))}),s+=`<br><br>${y({xmin:-5,ymin:Math.round(Math.min(-5*t+e,5*t+e)),xmax:5,ymax:Math.round(Math.max(-5*t+e,5*t+e)),pixelsParCm:20,scale:.7},m,M(u=>t*u+e,{mark:m,color:"blue"}),f,c)}`);break}d=`The function $f$ is an affine function and we know that $f(${a})=${n}$ and $f(${i})=${$}$.<br>`,d+="Determine the algebraic form of the function $f$.",d+=Q(this,x),this.questionJamaisPosee(x,a,i,n,$,t,e,D[x])&&(this.listeQuestions.push(d),this.listeCorrections.push(s),x++),C++}L(this)},this.besoinFormulaireNumerique=["Difficulty level",3,`1: Easy
2: Difficult
3: Very difficult`]}export{U as default,T as interactifReady,v as interactifType,q as ref,w as titre,j as uuid};
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