The function is \( y = -2x^2 - 4x + 2 \).

This is a quadratic function with a negative leading coefficient (\( a = -2 < 0 \)), so the parabola opens downwards and has a maximum value (not a minimum). The minimum value would be \( -\infty \) as \( x \to \pm \infty \).

Method 1: Vertex formula
For \( y = ax^2 + bx + c \), the vertex occurs at \( x = -\frac{b}{2a} \).

Here, \( a = -2 \), \( b = -4 \), \( c = 2 \).

\( x = -\frac{-4}{2(-2)} = \frac{4}{-4} = -1 \)

Substitute \( x = -1 \):
\( y = -2(-1)^2 - 4(-1) + 2 = -2(1) + 4 + 2 = -2 + 4 + 2 = 4 \)

Method 2: Completing the square
\( y = -2x^2 - 4x + 2 = -2(x^2 + 2x) + 2 \)

Complete the square inside: \( x^2 + 2x = (x + 1)^2 - 1 \)

\( y = -2[(x + 1)^2 - 1] + 2 = -2(x + 1)^2 + 2 + 2 = -2(x + 1)^2 + 4 \)

The maximum value is 4 (when \( x = -1 \)), and \( y \leq 4 \).

Method 3: Calculus (derivative)
\( \frac{dy}{dx} = -4x - 4 \)

Set to zero: \( -4x - 4 = 0 \) → \( x = -1 \)

Second derivative \( \frac{d^2y}{dx^2} = -4 < 0 \), confirming a maximum.

\( y(-1) = 4 \)

Conclusion:
The function has "no minimum value" (it decreases without bound).
The "maximum value" is 4.