QuestionAugust 10, 2025

En la función y=4x^2-4x-3 las coordenadas de su vértice son: (2,-4) ((1)/(2),-4) (2,4) (-(1)/(2),-4) (-(1)/(2),4)

En la función y=4x^2-4x-3 las coordenadas de su vértice son: (2,-4) ((1)/(2),-4) (2,4) (-(1)/(2),-4) (-(1)/(2),4)
En la función y=4x^2-4x-3 las coordenadas de
su vértice son:
(2,-4)
((1)/(2),-4)
(2,4)
(-(1)/(2),-4)
(-(1)/(2),4)

Solution
4.0(286 votes)

Answer

(\frac{1}{2}, -4) Explanation 1. Identify the vertex formula for a quadratic function The vertex of a parabola y = ax^2 + bx + c is given by x = -\frac{b}{2a}. 2. Calculate the x-coordinate of the vertex For y = 4x^2 - 4x - 3, a = 4 and b = -4. So, x = -\frac{-4}{2 \cdot 4} = \frac{1}{2}. 3. Calculate the y-coordinate of the vertex Substitute x = \frac{1}{2} into the function: y = 4(\frac{1}{2})^2 - 4(\frac{1}{2}) - 3 = 1 - 2 - 3 = -4.

Explanation

1. Identify the vertex formula for a quadratic function<br /> The vertex of a parabola $y = ax^2 + bx + c$ is given by $x = -\frac{b}{2a}$.<br /><br />2. Calculate the x-coordinate of the vertex<br /> For $y = 4x^2 - 4x - 3$, $a = 4$ and $b = -4$. So, $x = -\frac{-4}{2 \cdot 4} = \frac{1}{2}$.<br /><br />3. Calculate the y-coordinate of the vertex<br /> Substitute $x = \frac{1}{2}$ into the function: $y = 4(\frac{1}{2})^2 - 4(\frac{1}{2}) - 3 = 1 - 2 - 3 = -4$.
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