Question

10. O vértice da parábola f(x) = 1 ponto
-3x2 + 2x + 8 é: *
O (1/3, -25/3)
O (-1/3, 25/3)
O (1/3,25/3)
O (-1/3, -25/3)​

Answer 1

$\underbrace{Veja:}$

$\sf f(x)=-3x^2+2x+8$

coeficientes:

• a = -3
• b = 2
• c = 8

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Vamos calcular primeiramente o discriminante pois precisaremos mais pra frente

$\sf \Delta=b^2-4ac$

$\sf \Delta=(2)^2-4.(-3).(8)$

$\sf \Delta=4+96$

$\sf\Delta=100$

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x do vértice

$\Rightarrow~~\sf x_v=\dfrac{-b}{2a}$

$\Rightarrow~~\sf x_v=\dfrac{-(2)}{2.(-3)}$

$\Rightarrow~~\sf x_v=\dfrac{-2}{-6}$

• podemos simplificar por 2

$\Rightarrow~~\sf x_v=\dfrac{-2^{~:~2}}{-6^{~:~2}}$

$\Rightarrow~~\sf x_v=\dfrac{-1}{-3}$

• lembre-se, sinais iguais: positivo

$\therefore~~\boxed{\sf x_v=\dfrac{1}{3}}$

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y do vértice

$\Rightarrow~~\sf y_v=\dfrac{-\Delta}{4a}$

$\Rightarrow~~\sf y_v=\dfrac{-(100)}{4.(-3)}$

$\Rightarrow~~\sf y_v=\dfrac{-100}{-12}$

• podemos simplificar por 4

$\Rightarrow~~\sf y_v=\dfrac{-100^{~:~4}}{-12^{~:~4}}$

$\Rightarrow~~\sf y_v=\dfrac{-25}{-3}$

• lembre-se, sinais iguais: positivo

$\therefore~~\boxed{\sf y_v=\dfrac{25}{3}}$

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Resposta: Portanto, o vértice da parábola é:

$\sf V(x_v~~,~~yv)~~\to~~\boxed{\sf V\bigg(\dfrac{1}{3}~~,~~\dfrac{25}{3}\bigg)}$