Analyze the Quadratic y = 2x² - 4x + 2: Determining Positive and Negative Domains

To solve for the positive and negative domains of the function $y = 2x^2 - 4x + 2$, we follow these steps:

• Step 1: Find the roots of the equation using the quadratic formula. For $ax^2 + bx + c = 0$, the roots are found by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
• Step 2: For our specific function, identify: $a = 2$, $b = -4$, $c = 2$.

Let's calculate the discriminant: $b^2 - 4ac = (-4)^2 - 4 \times 2 \times 2 = 16 - 16 = 0$.
Since the discriminant is zero, we have a repeated root, which means the graph touches the x-axis at one point.

The root is found as follows:
$x = \frac{-(-4) \pm \sqrt{0}}{2 \times 2} = \frac{4}{4} = 1$.

This root $x = 1$ represents a vertex-touching parabola, with no intervals of $x$ such that $y < 0$.

To find the positive domain ($y > 0$), we note the parabola $y = 2x^2 - 4x + 2$ opens upwards (since $a = 2 > 0$) and only touches the x-axis at one point $(1, 0)$. Thus, the positive domain is $x \ne 1$.

The function does not take any negative values, since it opens upwards and only touches the x-axis.

Therefore, the positive domain is $x > 0 :x \ne 1$ and the negative domain is $x < 0 :$ none.

In conclusion, the solution to the problem is:

$x > 0 :x \ne 1$

$x < 0 :$ none