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The vertex of the parabola [tex]y = -2x^{2} + 8x - 6[/tex] is at:

A. [tex]\((2, 2)\)[/tex]

B. [tex]\((2, -2)\)[/tex]

C. [tex]\((-2, -2)\)[/tex]

D. [tex]\((-2, 2)\)[/tex]


Sagot :

To determine the vertex of the parabola given by the equation [tex]\(y = -2x^2 + 8x - 6\)[/tex], we need to follow a few steps.

1. Understanding the Form: The given equation is in the standard form of a quadratic equation, [tex]\(y = ax^2 + bx + c\)[/tex].

2. Identify Coefficients:
- [tex]\(a = -2\)[/tex]
- [tex]\(b = 8\)[/tex]
- [tex]\(c = -6\)[/tex]

3. Formula for the Vertex:
The vertex form of a parabola, [tex]\( y = ax^2 + bx + c \)[/tex], has its vertex at the point [tex]\( (h, k) \)[/tex]. The [tex]\(x\)[/tex]-coordinate of the vertex can be found using the formula:
[tex]\[ x = -\frac{b}{2a} \][/tex]

4. Calculate the [tex]\(x\)[/tex]-coordinate of the Vertex:
Substitute [tex]\(a\)[/tex] and [tex]\(b\)[/tex] into the formula:
[tex]\[ x = -\frac{8}{2(-2)} = -\frac{8}{-4} = 2 \][/tex]

5. Calculate the [tex]\(y\)[/tex]-coordinate of the Vertex:
To find the [tex]\(y\)[/tex]-coordinate, substitute [tex]\(x = 2\)[/tex] back into the original equation:
[tex]\[ y = -2(2)^2 + 8(2) - 6 \][/tex]
Simplify this step-by-step:
[tex]\[ y = -2(4) + 16 - 6 \][/tex]
[tex]\[ y = -8 + 16 - 6 \][/tex]
[tex]\[ y = 2 \][/tex]

So, the coordinates of the vertex are [tex]\((2, 2)\)[/tex].

Therefore, the vertex of the parabola [tex]\(y = -2x^2 + 8x - 6\)[/tex] is at:
[tex]\[ \boxed{(2, 2)} \][/tex]