Find the Domain of y=-(x-2)²+1/6: Complete Function Analysis

To solve this problem, we will follow these steps:

• Step 1: Identify the vertex and the direction in which the parabola opens.

• Step 2: Set the function equal to zero to find critical x-values.

• Step 3: Solve for these x-values to find the specific points where the function changes signs.

• Step 4: Determine which intervals on the x-axis correspond to the function being positive and which are negative.

Now, let's work through each step:
Step 1: The given function is $y = -\left(x-2\right)^2+\frac{1}{6}$. The vertex is at point $(2, \frac{1}{6})$, and since the leading coefficient is negative, the parabola opens downwards.
Step 2: We set the equation equal to zero: $-\left(x-2\right)^2+\frac{1}{6} = 0$.
Step 3: Solving for when the function is zero, we have: $\begin{aligned} -\left(x-2\right)^2 & = -\frac{1}{6}\\ (x-2)^2 & = \frac{1}{6}\\ x-2 & = \pm\sqrt{\frac{1}{6}} \end{aligned}$ This gives us two solutions for x: $x = 2+\sqrt{\frac{1}{6}}$ and $x = 2-\sqrt{\frac{1}{6}}$.
Step 4: We will test intervals determined by these points to see where the function is positive or negative.

• For x < 2-\sqrt{\frac{1}{6}} , plug an x-value less than $2-\sqrt{\frac{1}{6}}$ into the function; the function will be negative because the entire parabola opens downward from the vertex.

• For 2-\sqrt{\frac{1}{6}} < x < 2+\sqrt{\frac{1}{6}} , the function is positive.

• For x > 2+\sqrt{\frac{1}{6}} , again, the function returns to being negative.

Therefore, the intervals are:
- Negative domain: x < 0 : x < 2-\sqrt{\frac{1}{6}} and x > 2+\sqrt{\frac{1}{6}}
- Positive domain: x > 0 : 2 -\sqrt{\frac{1}{6}} < x < 2+\sqrt{\frac{1}{6}} .

The correct answer is:

x > 2+\sqrt{\frac{1}{6}} or x < 0 : x < 2-\sqrt{\frac{1}{6}}

x > 0 : 2 -\sqrt{\frac{1}{6}} < x < 2+\sqrt{\frac{1}{6}}