Question

dada a funçao f(x) = -x2 + 4x - 2 Calcule as coordenadas do vértice

Answer 1

Explicação passo-a-passo:

As coordenadas do vértice são dadas por;

• $\sf x_V=\dfrac{-b}{2a}$

• $\sf y_V=\dfrac{-\Delta}{4a}$

Temos:

• $\sf x_V=\dfrac{-b}{2a}$

$\sf x_V=\dfrac{-4}{2\cdot(-1)}$

$\sf x_V=\dfrac{-4}{-2}$

$\sf \red{x_V=2}$

• $\sf y_V=\dfrac{-\Delta}{4a}$

$\sf \Delta=4^2-4\cdot(-1)\cdot(-2)$

$\sf \Delta=16-8$

$\sf \Delta=8$

$\sf y_V=\dfrac{-\Delta}{4a}$

$\sf y_V=\dfrac{-8}{4\cdot(-1)}$

$\sf y_V=\dfrac{-8}{-4}$

$\sf \red{y_V=2}$

O vértice é V(2, 2)

Answer 2

• As coordenadas do vértice são dadas pelo Y do vértice e o X do vértice.

Yv = - delta/4a

Yv = - (4² - 4 . (- 1) . (- 2))/(- 4)

Yv = (16 - 8)/4

Yv = 8/4

Yv = 2

Xv = - b/a

Xv = - 4/(- 2)

Xv = 4/2

Xv = 2

Resposta: (2,2)