Wiki-Quellcode von Lösung Funktionsterm bestimmen
Zuletzt geändert von Simone Hochrein am 2025/10/01 13:30
Verstecke letzte Bearbeiter
| author | version | line-number | content |
|---|---|---|---|
| |
4.1 | 1 | (%class=abc%) |
| 2 | 1. (((Mit den Nullstellen {{formula}}x_1=-3{{/formula}} und {{formula}}x_2=2{{/formula}}: | ||
| |
1.1 | 3 | |
| 4 | {{formula}} | ||
| 5 | \begin{aligned} | ||
| |
5.1 | 6 | &f(x) &= a \cdot (x + 3) (x - 2) ~~~(1) |
| |
1.1 | 7 | \end{aligned} |
| 8 | {{/formula}} | ||
| 9 | |||
| 10 | Mit {{formula}}P(0|3){{/formula}}: | ||
| 11 | |||
| 12 | {{formula}} | ||
| 13 | \begin{aligned} | ||
| |
2.1 | 14 | &f(0) &= -3 ~~~(2) |
| |
1.1 | 15 | \end{aligned} |
| 16 | {{/formula}} | ||
| |
2.1 | 17 | |
| 18 | {{formula}}(2){{/formula}} in {{formula}}(1){{/formula}}: | ||
| 19 | |||
| 20 | {{formula}} | ||
| 21 | \begin{aligned} | ||
| |
5.1 | 22 | a \cdot (0 + 3) (0 - 2) &= -3\\ |
| |
2.1 | 23 | -6a &= -3 \\ |
| 24 | ~a &= 0,5 \\ | ||
| 25 | \end{aligned} | ||
| 26 | {{/formula}} | ||
| 27 | |||
| |
4.1 | 28 | {{formula}} |
| 29 | \Rightarrow f(x)=0,5\cdot (x+3)(x-2) | ||
| 30 | {{/formula}} | ||
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7.1 | 31 | |
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4.1 | 32 | ))) |
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6.1 | 33 | 1. (((Mit {{formula}}S(2|4){{/formula}} und {{formula}}a=-2{{/formula}}: |
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2.1 | 34 | |
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6.1 | 35 | {{formula}} |
| 36 | \Rightarrow f(x)=-2\cdot (x-2)^2+4 | ||
| 37 | {{/formula}} | ||
| 38 | |||
| 39 | ))) | ||
| |
7.1 | 40 | 1. (((Mit {{formula}}S(3|1){{/formula}}: |
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6.1 | 41 | |
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7.1 | 42 | {{formula}} |
| 43 | f(x) = a \cdot (x-3)^2+1 | ||
| 44 | {{/formula}} | ||
| 45 | |||
| 46 | Mit {{formula}}A(5|2){{/formula}}: | ||
| 47 | |||
| 48 | {{formula}} | ||
| 49 | \begin{aligned} | ||
| 50 | a \cdot (5-3)^2+1&=2 \\ | ||
| 51 | a \cdot 4+1&=2 \\ | ||
| 52 | 4a&=1 \\ | ||
| 53 | a&=0,25 | ||
| 54 | \end{aligned}{{/formula}} | ||
| 55 | ))) | ||
| 56 |