Punktsymmetrie zum Ursprung

In diesem Artikel erkläre ich dir die Punktsymmetrie zum Ursprung. Als erstes veranschauliche ich sie graphisch, dann zeige ich dir einen kleinen Trick mit dem du sofort erkennen kannst, ob eine ganzrationale Funktion punktsymmetrisch ist und als letztes zeige ich dir, wie man mathematisch die Punktsymmetrie nachweist.

[one_third]Übersicht:[/one_third][two_third_last]Graphische Erklärung
Punktsymmetrie zum Ursprung (ganzrat. Funktionen)
Mathematischer Nachweis

Graphische Erklärung

punktsymmetrischer Graph

Dieser Graph ist punktsymmetrisch zum Ursprung. Warum?

punktsymmetrischer Graph

Wie du vielleicht siehst, dient der Ursprung als eine Art Drehzentrum. Wenn man also den Funktionsgraph um 180° (=Halbkreis) um den Ursprung dreht und der daraus resultierende Graph dann deckungsleich mit dem Ausgangsgraphen ist, dann gehört dieser Graph zu einer punktsymmetrischen Funktion.

Punktsymmetrie zum Ursprung bei ganzrationalen Funktion

Bei einer ganzrationalen Funktion genügt ein einziger Blick auf die Exponenten (Hochzahlen) um zu entscheiden, ob die vorliegende Funktion punktsymmetrisch ist, denn es gilt:

Punktsymmetrie zum Ursprung

Die Funktion


ist punktsymmstrisch, denn die Exponenten („3“ und „1“) sind ausschließlich ungerade Zahlen!

Die Funktion


ist NICHT punktsymmetrisch, da sie einen geraden Exponenten (2) und ein Absolutglied (Zahl die mit keiner Potenz von x multipliziert wird) hat!


Mathematischer Nachweis für Symmetrie zum Ursprung:

$f(-x) = -f(x)$

Natürlich gibt es einen mathematischen Nachweis für punktsymmetrische Funktionen.

Hierzu bildet man $f(-x)$ und vergleicht dieses Ergebnis mit $-f(x)$. Stimmen die beiden gebildeten Funktionen überein, dann ist die vorliegende Ausgangsfunktion symmetrisch zum Ursprung, andernfalls nicht.

Doch wie wird $f(-x)$ gebildet?

Hierzu ersetzt man jedes vorkommdende $x$ in der Ausgangsfunktion durch $-x$. Dieser Ausdruck wird dann vereinfacht. Beachte bitte, dass beim Vereinfachen die Potenzen Vorrang haben. Wenn man also $2(-x)^3$ vereinfachen soll, dann liegt das erste Augenmerk auf $(-x)^3$. $(-x)^3$ steht für $(-x)\cdot(-x)\cdot(-x)$ und dies ergibt $-x^3$. Das Ganze mit zwei multipliziert ist $-2x^3$, weil etwas Negatives mal etwas Positives insgesamt negativ ist.

Wie wird $-f(x)$ gebildet?

Hierzu schreibt man ein Minus-Zeichen vor die komplette Ausgangsfunktion, indem die Ausgangsfunktion in Klammern gesetzt wird!


\begin{align*}f(x)&=3x^3+5x\\\\f(-x)&=3\cdot(-x)^3+5\cdot(-x)=-3x^3-5x\\\\-f(x)&=-(3x^3+5x)=-3x^3-5x \end{align*}

Da hier f(-x) = -f(x) gilt, ist die vorliegende Funktion symmetrisch zum Ursprung!

\begin{align*}f(x)&=4x^4-6x^3+7x^2-3x-1\\\\f(-x)&=4\cdot(-x)^4-6\cdot(-x)^3+7\cdot(-x)^2-3\cdot(-x)-1\\&=4x^4+6x^3+7x^2+3x-1\\\\-f(x)&=-(4x^4-6x^3+7x^2-3x-1)\\&=-4x^4+6x^3-7x^2+3x+1 \end{align*}

Bei diesem Beispiel stimmen f(-x) und -f(x) nicht überein. Diese Funktion ist also nicht symmetrisch zum Ursprung!


[spoiler title=‘ 1. Entscheide anhand der Exponenten, welche der Funktionen punktsymmetrisch zum Ursprung sind    a)$f(x)=3x^2+6$     b) $g(x)=3x^3+4x$    $c) h(x)=2x^3+4x+1$     (Lösung anzeigen *klick*)‘ style=’blue‘ collapse_link=’true‘]

a) f(x) hat den Exponenten 2 und ein Absolutglied (=reine Zahl). Da punktsymmetrische Funktionen nur ungerade Exponenten haben dürfen und kein Absolutglied, ist f(x) nicht punktsymmetrisch zum Ursprung.

b) g(x) hat die Exponenten 3 und 1 und kein Absolutglied (=reine Zahl). Da punktsymmetrische Funktionen nur ungerade Exponenten haben dürfen und kein Absolutglied, ist g(x) punktsymmetrisch zum Ursprung.

c) h(x) hat die Exponenten 3 und 1 und ein Absolutglied (=reine Zahl). Da punktsymmetrische Funktionen nur ungerade Exponenten haben dürfen und kein Absolutglied, ist h(x) nicht punktsymmetrisch zum Ursprung. Das Absolutglied verhindert somit die Punktsymmetrie zum Ursprung.



[spoiler title=‘ 2. Zeige durch Rechnung, dass $g(x)=3x^3+4x$ punktsymmetrisch zum Urspung ist     (Lösung anzeigen *klick*)‘ style=’blue‘ collapse_link=’true‘]


1. Schritt: $g(-x)$ bilden:


2. Schritt: $-g(x)$ bilden:


3.Schritt: $g(-x)$ und $-g(x)$

$g(-x)$ und $-g(x)$ sind gleich, somit ist die Funktion $g(x)$ punktsymmetrisch zum Ursprung



  • Punktsymmetrie zum Ursprung = Drehung um 180° mit dem Punkt (0/0) als Drehzentrum
  • ganzrationale Funktionen, die zum Ursprung punktsymmetrisch sind, haben ausschließlich ungerade Exponenten und kein Absolutglied
  • rechnerischer Nachweis: $f(-x)=-f(x)$

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