Find Domains of y=-x²+4x-4: Positive and Negative Regions

To solve the problem, we'll follow these steps:

• Step 1: Identify the roots of the quadratic function $y = -x^2 + 4x - 4$ using the quadratic formula.
• Step 2: Use the calculated roots to form intervals on the x-axis.
• Step 3: Evaluate the sign of the function in each interval.

Let's begin:

Step 1: Find the roots of the quadratic function.
To find the roots, use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Here, $a = -1$, $b = 4$, and $c = -4$.
Calculate the discriminant $b^2 - 4ac = 4^2 - 4(-1)(-4) = 16 - 16 = 0$.

Since the discriminant is zero, there is one (repeated) root:
$x = \frac{-4 \pm \sqrt{0}}{-2} = \frac{-4}{-2} = 2.$ Thus, the root is $x = 2$.

Step 2: Define the intervals.
The root at $x = 2$ divides the x-axis into two intervals: $x < 2$ and $x > 2$.

Step 3: Evaluate the sign of the function within each interval.

For $x < 2$, choose a test point, such as $x = 0$.
Substitute into the function: $y = -(0)^2 + 4(0) - 4 = -4$.
Since $-4$ is negative, the function is negative for $x < 2$.

For $x > 2$, choose a test point, such as $x = 3$.
Substitute into the function: $y = -(3)^2 + 4(3) - 4 = -9 + 12 - 4 = -1$.
Since $-1$ is also negative, the function is negative for $x > 2$.

The quadratic function does not have positive values in any interval on the real number line.

Thus, the positive and negative domains are:
$x < 0 : x\ne2$ (negative domain)
$x > 0 :$ none (as all values are negative)

Therefore, the solution to the problem is $x < 0 : x\ne2$ and $x > 0 :$ none.