MATHHX B
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3.2 Grafer for lineære funktioner
Grafen for en lineær funktion kan tegnes på samme måde som grafen for enhver anden funktion, nemlig ved at lave et sildeben.
Øvelse 3.2.1
Lad \(f(x)=-2x+3\).
Løsning 3.2.1
-
a)
Kender man betydningen af \(a\) og \(b\) i forskriften \(f(x)=ax+b\), kan man tegne grafen med en hurtigere metode.
-
• Tallet \(b\) angiver, hvor grafen skærer y-aksen.
-
• Tallet \(a\) er det tal, man skal gå op, hver gang man går 1 ud af \(x\)-aksen. Hvis \(a\) er negativ, skal man gå ned i stedet for op.
Betydningen af \(a\) og \(b\) kan også ses på tegningen:
-
Eksempel 3.2.1
Lad os prøve at tegne funktionen \(f(x)=-2x+3\). Vi kan se, at \(b=3\), så funktionen skærer y-aksen i \(3\):
Da \(a=-2\), skal vi gå \(1\) ud og \(2\) ned for at finde det næste punkt på grafen.
Sådan fortsætter vi:
og vi kan forbinde punkterne med en linje:
Øvelse 3.2.2
Tegn med papir og blyant en skitse af nedenstående funktioner vha. metoden beskrevet ovenover:
-
a) \(f(x)=0{,}5x+2\)
-
b) \(f(x)=x+1\)
-
c) \(f(x)=-2x+1\)
-
d) \(f(x)=5\)
-
e) \(f(x)=0\)
Øvelse 3.2.3
Betragt graferne:
Bestem forskriften for følgende funktioner:
-
a) Bestem forskriften for \(f\).
-
b) Bestem forskriften for \(g\).
-
c) Bestem forskriften for \(h\).
-
d) Bestem forskriften for \(i\).
Løsning 3.2.3
-
a) \(f(x)=0{,}5x-1\)
-
b) \(g(x)=-x+1\)
-
c) \(h(x)=-1\)
-
d) \(i(x)=x\)
Forskrift ud fra to punkter på grafen
Det er klart, at hvis man har to punkter, så findes der netop én linje igennem punkterne. Vi vil nu præsentere en sætning, som kan bruges til at bestemme en forskriften for linjen. En sætning
er et matematisk resultat, som er særligt nyttigt.
-
Sætning 3.2.1
Lad \(f(x)=ax+b\) være en lineær funktion og antag, at \(f\) går igennem punkterne \(P(x_0,y_0)\) og
\(Q(x_1,y_1)\):
Da er konstanterne \(a\) og \(b\) givet ved:
\[a=\frac {y_1-y_0}{x_1-x_0}\qquad \textrm { og }\qquad b=y_0-ax_0\]
Notationen (symbolerne) i sætningen kræver lidt forklaring. Har man en funktion, er \(x\) en variabel, som kan være alt muligt. Nogle gange vil man dog tage udgangspunkt i en fast \(x\)-værdi, som man så kalder \(x_0\), så man
kan skelne den fra \(x\) (der jo ikke har en fast værdi). Tallet \(0\) i skrivemåden \(x_0\) kaldes et indeks, og har altså ikke nogen anden betydning end at signalere, at der er tale om en bestemt \(x\)-værdi. Har man
mere end én fast \(x\)-værdi, kan man kalde den næste for \(x_1\). Tilsvarende gælder for \(y_0\) og \(y_1\). Så når der står \(P(x_0,y_0)\) og \(Q(x_1,y_1)\), betyder det altså bare to punkter, kaldet \(P\) og \(Q\), med
koordinaterne \((x_0,y_0)\) og \((x_1,y_1)\), hvor \(x_0\), \(y_0\), \(x_1\) og \(y_1\) er nogle bestemte tal. Måske er det nemmest at forstå med et eksempel:
-
Eksempel 3.2.2
Vi vil bestemme forskriften for linjen gennem punkterne \(P(1,2)\) og \(Q(4,14)\).
Sammenligner vi med ?? får vi at
\(x_0=1\)
\(y_0=2\)
\(x_1=4\)
\(y_1=14\)
Vi regner først \(a\):
\(\seteqnumber{0}{3.}{0}\)
\begin{align*}
a & = \frac {y_1-y_0}{x_1-x_0} && (\text {Formlen for $a$ skrives op})\\ & =\frac {14-2}{4-1} && (\text {Værdierne for } x_0,\ x_1,\ y_0 \text { og } y_1 \text {
indsættes})\\ & =4
\end{align*}
og med den \(a\)-værdi i baglommen, kan vi så regne \(b\):
\(\seteqnumber{0}{3.}{0}\)
\begin{align*}
b & = y_0-ax_0 && (\text {Formlen for $b$ skrives op})\\ & =2-4\cdot 1 && (\text {Værdierne for } x_0,\ y_0 \text { og } a \text { indsættes})\\ & =-2
\end{align*}
Vi sætter de fundne \(a\) og \(b\)-værdier ind i formlen \(f(x)=ax+b\):
\(\seteqnumber{0}{3.}{0}\)
\begin{align*}
f(x) & = ax+b\\ & = 4x+(-2) && (\text {Værdierne for } a \text { og } b \text { indsættes})\\ & = 4x-2
\end{align*}
Altså er forskriften \(f(x)=4x-2\).
Som regel gider vi ikke give punkterne navne. Så i stedet for f.eks. \(P(1,2)\) og \(Q(4,14)\) vil vi bare skrive \((1,2)\) og \((4,14)\).
Øvelse 3.2.4
Bestem vha. ?? den lineær funktion, som går gennem punkterne
-
a) \((2,5)\) og \((4,11)\).
-
b) \((5,-3)\) og \((12,-10)\).
Løsning 3.2.4
-
a) \(f(x)=3x-1\).
-
b) \(f(x)=-x+2\).
Øvelse 3.2.5
Et sildeben for en lineær funktion \(f\) er givet ved:
\(\begin {array}{ | c | c | c | c | c |c |} \hline x & 4 & 8 \\ \hline f(x) & 130 & 70 \\ \hline \end {array}\)