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Dobierz wzór do wykresu wielomianu stopnia czwartego

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Na rysunku przedstawiono wykres wielomianu WW stopnia czwartego.

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Wykres wielomianu W: przecina oś OX w punktach −2 i 3, dotyka jej w punkcie 1, oś OY przecina w punkcie (0, −3). Oba końce wykresu idą w górę.

Który wzór opisuje ten wielomian?

A. W(x)=12(x−2)(x+1)2(x+3)W(x) = \tfrac{1}{2}(x - 2)(x + 1)^2(x + 3)  B. W(x)=12(x+2)(x−1)2(x−3)W(x) = \tfrac{1}{2}(x + 2)(x - 1)^2(x - 3)

C. W(x)=−12(x+2)(x−1)2(x−3)W(x) = -\tfrac{1}{2}(x + 2)(x - 1)^2(x - 3)  D. W(x)=(x+2)(x−1)2(x−3)W(x) = (x + 2)(x - 1)^2(x - 3)

Rozwiązanie

Pokaż rozwiązanie krok po krokuUkryj rozwiązanie

Krok 1. Miejsca zerowe. Wykres przecina oś OXOX w −2-2 i 33, a w 11 jej dotyka. Potrzebne są więc czynniki (x+2)(x + 2), (x−3)(x - 3) i (x−1)(x - 1) w potędze parzystej - stąd (x−1)2(x - 1)^2. Wzór A ma miejsca zerowe 22, −1-1, −3-3 (znaki odwrócone) - odpada.

Krok 2. Krańce wykresu. Oba końce idą w górę: stopień parzysty i współczynnik wiodący dodatni. Wzór C ma współczynnik −12<0-\tfrac{1}{2} < 0, czyli oba końce w dół - odpada.

Krok 3. Punkt na osi OYOY. Zostały B i D, różniące się tylko współczynnikiem. Rozstrzyga punkt (0, −3)(0,\ -3):

B: 12⋅2⋅(−1)2⋅(−3)=−3,D: 2⋅(−1)2⋅(−3)=−6.\text{B: } \tfrac{1}{2} \cdot 2 \cdot (-1)^2 \cdot (-3) = -3, \qquad \text{D: } 2 \cdot (-1)^2 \cdot (-3) = -6.

Tylko B daje W(0)=−3W(0) = -3.

Uwaga: same miejsca zerowe i kształt nie wystarczają - wzory B i D mają identyczne miejsca zerowe i krańce. Współczynnik aa wyznacza dopiero dodatkowy punkt.

Odpowiedź. B: W(x)=12(x+2)(x−1)2(x−3)W(x) = \tfrac{1}{2}(x + 2)(x - 1)^2(x - 3).