| 1 | 1 | |||
funkcji jest równy 100, a wierzchołkiem paraboli jest punkt W (−1 | , 12 | ) | ||
| 2 | 2 |
| 1 | 1 | |||
W = ( − 1 | ; 12 | ) | ||
| 2 | 2 |
| b | 3 | |||
p = − | = − | ⇒ b = 3 a | ||
| 2a | 2 |
| 3 | 3 | 9 | 3 | 25 | ||||||
q = f( p) = a*( − | )2 + b*( − | ) + c = | a − | b + c = | ||||||
| 2 | 2 | 4 | 2 | 2 |
| 9 | 3 | 15 | |||
a − | * 3 a + c = | ||||
| 4 | 2 | 2 |
| 9 | 9 | 15 | |||
a − | a + c = | ||||
| 4 | 2 | 2 |
| 15 | 9 | |||
c = | + | a | ||
| 2 | 4 |
| 15 | 9 | |||
Δ = ( 3a)2 − 4 a*[ | + | a] = 100 | ||
| 2 | 4 |
| 100 | 10 | |||
a = − | = − | |||
| 30 | 3 |
| 15 | 9 | 10 | 15 | 15 | ||||||
c = | + | *( − | ) = | − | = 0 | |||||
| 2 | 4 | 3 | 2 | 2 |
| 10 | ||
y = − | x2 − 10 x | |
| 3 |
| 15 | 25 | |||
Pomyłkowo wpisałem | zamiast | |||
| 2 | 2 |
| 9 | 3 | 25 | |||
a − | *3a + c = | ||||
| 4 | 2 | 2 |
| 25 | 9 | |||
c = | + | a | ||
| 2 | 4 |
| 25 | 9 | |||
Δ = (3a)2 − 4a*[ | + | ] = 100 | ||
| 2 | 4 |
| 25 | 9 | 25 | 9 | 16 | ||||||
c = | + | *(−2) = | − | = | = 8 | |||||
| 2 | 4 | 2 | 2 | 2 |