Summary

Zusammenfassung Mathematik J1/3

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Course
Mathematik

Institution
Gymnasium

In dieser Zusammenfassung, aus dem zweiten Halbjahr aus J1, habe ich anhand von Beispielaufgaben und bildlicher Veranschaulichung verschiedene Thematiken aus dem Mathematik Unterricht zusammengefasst und veranschaulicht.Durch diese sind die genannten Thematiken leicht zu verstehen und einfach zu er...

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Uploaded on January 24, 2023
Number of pages 8
Written in 2021/2022
Type Summary

optimierungsaufgaben
stammfunktionen
bestimmte integral
aufstellen von kurvengleichung
mathematische modellierung

Institution Secondary school
Education Gymnasium
Course
Mathematik
School year 2

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Zusammenfassung der kompletten J1

$ 54.38 $ 52.64 4 items

1. Summary - Zusammenfassung mathematik j1/1
2. Summary - Zusammenfassung mathematik j1/2
3. Summary - Zusammenfassung mathematik j1/3
4. Summary - Zusammenfassung integralrechnung/stochastik mathematik j1/4
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:

, Beispiel:

Das Schaubild einer ganzrationaler Funktion 3. Grades schneidet die y-Achse im Punkt Sy(0/1), verläuft durch den Punkt P(1/ ) und hat den
Hochpunkt H(-2/5).
Bestimmen Sie den Funktionsterm.

Ansatz: ƒ(x)= ax + bx 2 + cx + d 3

ƒ(x)= 3ax + 2bx + c 2

Sy(0/1): f(0)=1: 0a+0b+0c+d=1 (1)

s
LGS
P(1/ ): f(1)= : a + b + c + d= (2)
4Unbekannte
2

H(-2/5): f(-2)=5: -8a+4b--2c+ d=5 (3)
4Gleichungen
zwei
f´(-2)=0: > 12a—4b+c =0 (4)
Möglichkeiten

Lösen das LGS aus (1): d=1 einsetzen in (2) (3) (4)

a + b + c + 1 =7
-8a + 4b — 2c + 1 = 5 In vernünftige Form (Alle Buchstaben Links / Zahlen Rechts)
12a — 4b + c = 0

a + b + c = — 2
08 • ( -12 )

-8a + 4b — 2c = 4 +

12a — 4b + c = 0 +

a + b + c = —
2

12b + 6c = 0 • ¥ ± §
-16 — 11c= 6 +

a + b + c = —
2-
12b + 6c = 0
Dreiecks
— 3c = 6
Form

Letzte Zeile: -3c = 6 I:(-3) Zeile 2 : 12b — 12 = 0 Zeile 1: a + 1 — 2 = — 2-
c = —2 12b = 12 a=£
b=1

=> Ergebnis: ƒ(x)= x + x — 2x + 1
3 2

g.

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