Wykres f(x) = −(x + 1)²(x − 2) - odbicie od osi i ujemny współczynnik

Szkoła średnia łatwe

Naszkicuj wykres funkcji

$f(x) = -(x + 1)^2(x - 2)$

i odczytaj z niego zbiory argumentów, dla których $f(x) > 0$ oraz $f(x) \ge 0$ .

Rozwiązanie

Pokaż rozwiązanie krok po krokuUkryj rozwiązanie

Krok 1. Krańce wykresu. Wyraz najwyższego stopnia: $-x^2 \cdot x = -x^3$ . Stopień $3$ , współczynnik $a = -1 < 0$ , więc prawy koniec wykresu idzie w dół, a lewy w górę.

Krok 2. Miejsca zerowe. $x = -1$ jest pierwiastkiem podwójnym (czynnik $(x + 1)^2$ ) - wykres dotyka osi i się od niej odbija. $x = 2$ jest pojedynczy - wykres przecina oś.

Krok 3. Punkt na osi $OY$ . $f(0) = -(1)^2 \cdot (0 - 2) = 2$ .

Krok 4. Punkty pomocnicze.

$f(1) = -(2)^2 \cdot (-1) = 4, \qquad f(-2) = -(-1)^2 \cdot (-4) = 4.$

Krok 5. Szkic. Od prawej: wykres przychodzi z dołu, przecina oś w $2$ , wznosi się nad oś (przez $(1,\ 4)$ i $(0,\ 2)$ ), schodzi do $-1$ , dotyka osi i wraca nad nią, a dalej w lewo idzie w górę.

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cy="123" r="3.4"/><circle cx="258" cy="123" r="3.4"/><circle cx="162" cy="93" r="3.4"/><circle cx="210" cy="63" r="3.4"/><circle cx="66" cy="63" r="3.4"/></g><g fill="currentColor" font-size="10" stroke="var(--bg-page)" stroke-width="2" paint-order="stroke"><text x="18" y="137" text-anchor="middle">−3</text><text x="66" y="137" text-anchor="middle">−2</text><text x="114" y="137" text-anchor="middle">−1</text><text x="210" y="137" text-anchor="middle">1</text><text x="258" y="137" text-anchor="middle">2</text><text x="306" y="137" text-anchor="middle">3</text><text x="156" y="216.5" text-anchor="end">−6</text><text x="156" y="186.5" text-anchor="end">−4</text><text x="156" y="156.5" text-anchor="end">−2</text><text x="156" y="96.5" text-anchor="end">2</text><text x="156" y="66.5" text-anchor="end">4</text><text x="156" y="36.5" text-anchor="end">6</text></g><g fill="currentColor" font-size="12" opacity="0.85" stroke="var(--bg-page)" stroke-width="2" paint-order="stroke"><text x="317.6" y="116" text-anchor="end">x</text><text x="169" y="22">y</text></g><g fill="currentColor" stroke="var(--bg-page)" stroke-width="2" paint-order="stroke"><text x="169" y="106" font-size="10.5">(0, 2)</text><text x="210" y="54" font-size="10.5" text-anchor="middle">(1, 4)</text><text x="73" y="61" font-size="10.5">(−2, 4)</text><text x="22.8" y="198" font-size="11" fill="var(--accent)">f(x) = −(x + 1)²(x − 2)</text></g>

Wykres f(x) = −(x + 1)²(x − 2): lewy koniec w górze, w podwójnym miejscu zerowym −1 wykres dotyka osi i się odbija, w pojedynczym 2 przecina oś, prawy koniec w dole.

Krok 6. Odczyt. Wykres leży nad osią na lewo od $2$ z wyjątkiem punktu $-1$ , w którym dotyka osi. Stąd:

$f(x) > 0$ dla $x \in (-\infty;\ -1) \cup (-1;\ 2)$ - punkt $-1$ trzeba wyciąć, bo $f(-1) = 0$ ;
$f(x) \ge 0$ dla $x \in (-\infty;\ 2\rangle$ - tu $-1$ i $2$ należą do zbioru, więc przedziały się sklejają.

Uwaga: w $-1$ wykres nie przechodzi pod oś. Kto narysuje tam przecięcie, dostanie błędny znak na całej półprostej $(-\infty;\ -1)$ .

Odpowiedź. $f(x) > 0 \iff x \in (-\infty;\ -1) \cup (-1;\ 2)$ , a $f(x) \ge 0 \iff x \in (-\infty;\ 2\rangle$ .

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