| −4 | 3 | |||
Proszę o pomoc.Jeśli wykres funkcji f(x)= | + | przesuniemy o wektor | ||
| x−p | 5 |
| 1 | 1 | 2 | 2 | |||||
v[2 | b+4 | , | b+ | ], to otrzymamy hiperbolę, której środkiem symetrii jest | ||||
| 2 | 2 | 3 | 5 |
| −4 | 3 | 2 | 2 | |||||
g(x)= | +( | + | b+ | ) | ||||
| x−(p+52b+92) | 5 | 3 | 5 |
| 3 | 2 | 2 | ||||
( | + | b+ | )=0 | |||
| 5 | 3 | 5 |
| 3 | ||
b=− | ||
| 2 |
| 5 | 3 | 9 | ||||
p+ | *(− | + | =0 | |||
| 2 | 2 | 2 |
| 15 | 18 | |||
p− | + | =0 | ||
| 4 | 4 |
| 3 | ||
p=− | ||
| 4 |
| −4 | 2 | |||
f(x)= | + | |||
| x+34 | 5 |
| −4 | 3 | |||
f(x)= | + | |||
| x+34 | 5 |
