| 4x | ||
a) f(x)= | ||
| 2x + 1 |
| −2x | ||
a) f(x)= | ||
| x + 1 |
| 2(2x+1)−2 | ||
a) f(x)= | ||
| 2x+1 |
| 2 | ||
f(x)=2− | ||
| 2x+1 |
| 2(2x+1)−2 | 2 | ||
=2− | |||
| 2x+1 | 2x+1 |
| 4x | 4x+2−2 | 2(2x+1)−2 | ||||
a) f(x)= | = | = | = | |||
| 2x+1 | 2x+1 | 2x+1 |
| 2(2x+1) | −2 | |||
= | + | = | ||
| 2x+1 | 2(x+12) |
| −1 | ||
= 2+ | − szukana postać kanoniczna . | |
| x+12 |
| −1 | ||
wykres funkcji y= | i przesuwasz go o wektor [−12,2]. | |
| x |
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