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f(x)=x2+3x+2
f(x)=2x2+5x+3
f(x)=12x2+2x−2
f(x)=3x2+2x+1
f(x)=(x+3)2+1
f(x)=−3(x−1)2−2
f(x)=12(x+4)2−1
f(x)=14(x+1)2−5
f(x)=2(x−1)(x+2)
f(x)=−3(x−5)2
f(x)=2(x+0)(x+5)
f(x)=−3x(x−8)
| 1 | 1 | |||
1) Δ=1 x1=−2 x2=−1 p=−1 | q=− | i teraz | ||
| 2 | 4 |
| 1 | 1 | |||
postac kan y=1(x+1 | )2− | |||
| 2 | 4 |