1. Let's start by stating the problem: You are trying to complete the square and simplify the quadratic expression $$y = 2x^2 - 8x + \_ + 6$$ to the form $$y = (x - 4)^2 - 26$$. 2. The formula for completing the square for a quadratic expression $$ax^2 + bx + c$$ is to rewrite it as $$a(x - h)^2 + k$$ where $$h = -\frac{b}{2a}$$ and $$k$$ is the constant term adjusted accordingly. 3. First, factor out the coefficient of $$x^2$$ from the terms involving $$x$$: $$y = 2(x^2 - 4x + \_) + 6$$ 4. To complete the square inside the parentheses, take half of the coefficient of $$x$$, which is $$-4$$, divide by 2 to get $$-2$$, then square it: $$(-2)^2 = 4$$ 5. Add and subtract 4 inside the parentheses to keep the expression equivalent: $$y = 2(x^2 - 4x + 4 - 4) + 6$$ 6. Group the perfect square trinomial and simplify: $$y = 2((x - 2)^2 - 4) + 6 = 2(x - 2)^2 - 8 + 6 = 2(x - 2)^2 - 2$$ 7. Your mistake was in the value added inside the parentheses for completing the square. You used 16 instead of 4, which changed the vertex and the shape of the parabola. 8. The correct vertex form is: $$y = 2(x - 2)^2 - 2$$ 9. This parabola opens upwards (since 2 > 0), is narrower than $$y = x^2$$, and has its vertex at $$(2, -2)$$. 10. The equation $$y = (x - 4)^2 - 26$$ corresponds to a different parabola with vertex at $$(4, -26)$$ and leading coefficient 1, so it does not match the original quadratic. Final answer: The correct completed square form is $$y = 2(x - 2)^2 - 2$$, not $$y = (x - 4)^2 - 26$$.