MATHHX B
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2.6 Grafer i GeoGebra
Dette afsnit kan springes over, indtil du får brug for det.
Vi har set, hvordan vi bruger GeoGebra til at løse ligninger, reducerer osv. Men GeoGebra kan også bruges til at tegne grafer.
Øvelse 2.6.1
Tegn i GeoGebra grafen for :
Løsning 2.6.1
-
a)
Man kan zoome og "trække"i koordinatsystemet. Der er flere måder at gøre det på. Man kan højreklikke, man kan bruge to-fingre-zoom og kan man bruge værktøjerne her:
Klikker man på ”Flyt tegnefladen” kan man også trække i akserne, så enhederne kommer til at stå tættere eller mere spredt.
Øvelse 2.6.2
Tegn følgende grafer og zoom/træk, så de ser pæne ud
-
a) \(f(x)=x+1000\)
-
b) \(g(x)=1000x^2\)
-
c) \(h(x)=1000(x-1000)\)
Begrænsede funktioner
Man kan tegne begrænsede funktioner i Geogebra.
-
Eksempel 2.6.2
Vi vil tegne funktionen \(f(x)=x^2\) hvor \(-1\geq x \leq 2\). Altså funktionen \(f(x)=x^2\) begrænset til definitionsmængden \(\Dm (f)=[-1;2[\).
Vi taster først
\[\verb |x^2,-1| \leq \verb |x| < \verb |1|\]
Det giver:
hmmm.... ikke helt godt. GeoGebra kan ikke finde ud af at tilføje endepunkter, så det må vi selv gøre. Vi vælger ”Punkt”:
og klikker på endepunkterne:
Du skal muligvis kæmpe lidt med punkt \(B\), da punktet ikke ligger på grafen, og derfor bliver GeoGebra lidt sur, når den skal tegne det. Vi mangler nu bare at få punkterne til at se rigtige ud. Vi højreklikker på et af punkterne:
og fjerner flueben i ”vis navn”:
Hvorefter vi klikker ”Stil” og vælger den rigtige punktmarkering (vi husker, at punktet ikke skulle være med på grafen):
På samme måde fjerne vi navnet på punkt \(A\), og her er det endelige resultat:
Sehr gut.
Øvelse 2.6.3
Tegn i GeoGebra graferne for funktionerne
-
a) \(f(x)=2x-3\), hvor \(1<x\leq 4\)
-
b) \(g(x)=0{,}01x^2+100\) hvor \(x>20\).
Løsning 2.6.3
-
a)
-
b)
Øvelse 2.6.4
Det er fint at kunne tegne grafer i GeoGebra, men hvis man glemmer hvordan man gør i hånden, ja så har man også glemt, hvad en graf er, og så kan det hele være ligemeget.
Løsning 2.6.4
-
a) Man laver da et sildeben og plotter punkterne i et koordinatsystem. Hvis du er tvivl om hvad dette betyder, så prøv at gør det! (evt. med hjælp fra
mig). Tegn derefter grafen i GeoGebra for at tjekke, at du har tegnet korrekt.
Grafisk løsning af ligninger og uligheder
Man kan løse ligninger og uligheder vha. grafer.
-
Eksempel 2.6.3
Vi vil løse ligningen \(x+1=x^2-4x+5\).
Vi sætter \(f(x)\) lig med venstresiden af ligningen og \(g(x)\) lig med højresiden, dvs. \(f(x)=x+1\) og \(g(x)=x^2-4x+5\) og vi skal dermed finde ud af hvornår \(f(x)=g(x)\). Vi tegner graferne for begge funktioner i
GeoGebra:
Herefter bruger vi kommandoen:
\[\verb |Skaering(objekt,objekt)|\]
Vi bruger den på \(f\) og \(g\) så vi skriver:
\[\verb |Skaering(f,g)|\]
Vi får nu skæringspunkterne:
Løsningerne til ligningen er \(x\)-koordinaterne til de to skæringspunkterne. Ligningen har altså to løsninger:
\[x=1\qquad \text {og}\qquad x=4\]
I afsnittet om polynomier skal vi se, hvordan man løser den type ligninger uden at bruge GeoGebra.
Øvelse 2.6.5
Brug den grafiske metode til at løse ligningerne:
-
a) \(2x-7=5(x+1)\)
-
b) \(-x^2+1=2^x\)
Hvis der ikke er nogle skæringspunkter, så er der heller ingen løsninger til ligningen.
Øvelse 2.6.6
Brug den grafiske metode til at løse ligningen:
Øvelse 2.6.7
Nedenunder ses graferne for to funktioner \(f(x)\) og \(g(x)\).
Vi kan også løse uligheder grafisk.
-
Eksempel 2.6.4
Vi vil løse uligheden \(2x+1<-x+2\).
Vi tegner begge sider af uligheden ind i Geogebra:
Da \(2x+1\) skal være mindre end \(-x+2\), må løsningen være givet ved det område, hvor den grønne graf ligger under den røde. Vi finder skæringspunktet:
og løsningen må så være \(x<0{,}33\), fordi den grønne graf ligger under den røde når \(x<0{,}33\).
Øvelse 2.6.8
Brug den grafiske metode til at løse ulighederne:
-
a) \(-x+7 < x-4\)
-
b) \(x^2+x < 2x\)
-
c) \(x^2+x\geq 2x\)