Wiki-Quellcode von Lösung Funktionsterm bestimmen
Version 5.1 von Simone Hochrein am 2025/10/01 12:20
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| author | version | line-number | content |
|---|---|---|---|
| 1 | (%class=abc%) | ||
| 2 | 1. (((Mit den Nullstellen {{formula}}x_1=-3{{/formula}} und {{formula}}x_2=2{{/formula}}: | ||
| 3 | |||
| 4 | {{formula}} | ||
| 5 | \begin{aligned} | ||
| 6 | &f(x) &= a \cdot (x + 3) (x - 2) ~~~(1) | ||
| 7 | \end{aligned} | ||
| 8 | {{/formula}} | ||
| 9 | |||
| 10 | Mit {{formula}}P(0|3){{/formula}}: | ||
| 11 | |||
| 12 | {{formula}} | ||
| 13 | \begin{aligned} | ||
| 14 | &f(0) &= -3 ~~~(2) | ||
| 15 | \end{aligned} | ||
| 16 | {{/formula}} | ||
| 17 | |||
| 18 | {{formula}}(2){{/formula}} in {{formula}}(1){{/formula}}: | ||
| 19 | |||
| 20 | {{formula}} | ||
| 21 | \begin{aligned} | ||
| 22 | a \cdot (0 + 3) (0 - 2) &= -3\\ | ||
| 23 | -6a &= -3 \\ | ||
| 24 | ~a &= 0,5 \\ | ||
| 25 | \end{aligned} | ||
| 26 | {{/formula}} | ||
| 27 | |||
| 28 | {{formula}} | ||
| 29 | \Rightarrow f(x)=0,5\cdot (x+3)(x-2) | ||
| 30 | {{/formula}} | ||
| 31 | ))) |