Wielomian, którego wykres dotyka osi OX w −1 i 3 - wzór i zbiór wartości

Szkoła średnia średnie

Wykres wielomianu $W$ stopnia czwartego dotyka osi $OX$ w punktach $-1$ i $3$ oraz przechodzi przez punkt $(1,\ -8)$ .

a) Wyznacz wzór wielomianu $W$ .

b) Naszkicuj jego wykres i podaj zbiór wartości.

c) Rozwiąż nierówność $W(x) < 0$ .

Rozwiązanie

Pokaż rozwiązanie krok po krokuUkryj rozwiązanie

Krok 1. Postać wzoru. „Dotyka osi” oznacza pierwiastek o krotności parzystej, czyli co najmniej $2$ . Dwa takie pierwiastki dają razem stopień co najmniej $4$ - a stopień to dokładnie $4$ , więc innych czynników nie ma:

$W(x) = a(x + 1)^2(x - 3)^2.$

Krok 2. Współczynnik $a$ . Podstawiamy punkt $(1,\ -8)$ :

$W(1) = a \cdot 2^2 \cdot (-2)^2 = 16a = -8 \quad\Longrightarrow\quad a = -\frac{1}{2}.$

Krok 3. Szkic. Stopień parzysty, $a < 0$ - oba końce w dół. W $-1$ i $3$ wykres dotyka osi od dołu. Na osi $OY$ : $W(0) = -\tfrac{1}{2} \cdot 1 \cdot 9 = -\tfrac{9}{2}$ .

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y="62" text-anchor="middle">4</text><text x="122" y="201.5" text-anchor="end">−10</text><text x="122" y="171.5" text-anchor="end">−8</text><text x="122" y="141.5" text-anchor="end">−6</text><text x="122" y="111.5" text-anchor="end">−4</text><text x="122" y="81.5" text-anchor="end">−2</text></g><g fill="currentColor" font-size="12" opacity="0.85" stroke="var(--bg-page)" stroke-width="2" paint-order="stroke"><text x="328" y="41" text-anchor="end">x</text><text x="135" y="22">y</text></g><g fill="currentColor" stroke="var(--bg-page)" stroke-width="2" paint-order="stroke"><text x="172" y="185" font-size="10.5" text-anchor="middle">(1, −8)</text><text x="321.6" y="30" font-size="11" text-anchor="end" fill="var(--accent)">f(x) = −½(x + 1)²(x − 3)²</text></g>

Wykres f(x) = −½(x + 1)²(x − 3)²: dotyka osi OX w punktach −1 i 3 (pierwiastki podwójne), poza nimi leży pod osią. Największa wartość to 0.

Krok 4. Zbiór wartości. Iloczyn $(x + 1)^2(x - 3)^2$ jest nieujemny dla każdego $x$ , więc po pomnożeniu przez $-\tfrac{1}{2}$ mamy $W(x) \le 0$ . Wartość $0$ jest osiągana (w $-1$ i $3$ ), a oba końce wykresu schodzą w dół bez ograniczenia. Zbiór wartości to $(-\infty;\ 0\rangle$ .

Krok 5. Nierówność $W(x) < 0$ . $W(x) \le 0$ zawsze, a równość zachodzi tylko w $-1$ i $3$ . Wystarczy więc wyrzucić te dwa punkty.

Odpowiedź. a) $W(x) = -\tfrac{1}{2}(x + 1)^2(x - 3)^2$ ; b) zbiór wartości $(-\infty;\ 0\rangle$ ; c) $x \in \mathbb{R} \setminus \{-1,\ 3\}$ .

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