| 27 | ||
jest równa 3 zbiór wartości funkcji f jest przedział < − | , +∞) | |
| 4 |
| x1 + x2 | 3 | |||
xw = | = | |||
| 2 | 2 |
| 3 | 27 | |||
f(x) = a(x − | )2 − | |||
| 2 | 4 |
| 9 | 27 | 1 | ||||
f(0) = | a − | = −6 ⇒ a = − | ||||
| 4 | 4 | 3 |
| 1 | 3 | 27 | 1 | 3 | 9 | |||||||
f(x) = | (x − | )2 − | = | [(x − | )2 − ( | )2] = | ||||||
| 3 | 2 | 4 | 3 | 2 | 2 |
| 1 | ||
= | (x − 6)(x + 3) | |
| 3 |
| 27 | b | |||
f(x)=ax2+bx+c, c=−6 q=− | − | =3 ⇒ b=−3a | ||
| 4 | a |
| b | −3a | 3 | ||||
f(x)=ax2−3ax−6, p=− | =− | = | ||||
| 2a | 2a | 2 |
| 3 | 27 | |||
f(p)=q, f( | )=− | |||
| 2 | 4 |
| 9 | 9 | 27 | 1 | ||||
a− | a−6=− | ⇒ a= | |||||
| 4 | 2 | 4 | 3 |
| 1 | ||
f(x)= | x2−x−6 | |
| 3 |
| 1 | 3 | 27 | ||||
f(x)= | (x− | )2− | ||||
| 3 | 2 | 4 |
| 27 | ||
yw=− | ||
| 2 |
| x1+x2 | 3 | |||
xw= | = | |||
| 2 | 2 |
| 3 | 27 | |||
2) f(x)=a*(x− | )2− | |||
| 2 | 4 |
| 3 | 27 | 3 | 9 | |||||
−6=a*(0− | )2− | ⇔ | =a* | |||||
| 2 | 4 | 4 | 4 |
| 1 | ||
a= | ||
| 3 |
| 1 | 3 | 27 | ||||
f(x)= | (x− | )2− | − postać kanoniczna | |||
| 3 | 2 | 4 |
| 1 | 3 | 27 | |||
(x− | )2− | =0 | |||
| 3 | 2 | 4 |
| 1 | 3 | 27 | |||
(x− | )2= | ||||
| 3 | 2 | 4 |
| 3 | 81 | |||
(x− | )2= | |||
| 2 | 4 |
| 3 | 9 | 3 | 9 | |||||
x− | = | lub x− | =− | |||||
| 2 | 2 | 2 | 2 |
| 1 | ||
f(x)= | *(x−6)*(x+3)− postac iloczynowa | |
| 3 |
| 1 | 1 | |||
a = − | a potem jest napisane że a = | |||
| 3 | 3 |
| 1 | ||
co na kartce... @ICSP na pewno miał na myśli | , bo na dole przepisał już dobrze. | |
| 3 |
| −27 | 9 | ||
= −6 − | a /*4 | ||
| 4 | 4 |
| 1 | ||
y= | (x−6)(x+3) | |
| 3 |
, zerkałaś na zadanie @kasia0948, bo wygląda fajnie, ale nie wiem
czy wpadnę na rozwiązanie o tej godzinie hah