Find the Domain of y=(x-4)²-4: Analyzing Positive and Negative Regions

The function given is $y = (x-4)^2 - 4$, which is a quadratic function in vertex form. This indicates a parabola that opens upwards.

The vertex of this function is $(4, -4)$, indicating the minimum point of the parabola. Since the parabola opens upwards, $y \geq -4$ for all $x$. The parabola crosses the x-axis where $y = 0$, so we solve for these points:

Set $y = 0$:

$(x-4)^2 - 4 = 0$

Add 4 to both sides:

$(x-4)^2 = 4$

Take the square root of both sides:

$x - 4 = \pm 2$

Thus, $x = 6$ or $x = 2$. These are the roots of the equation, indicating where the parabola crosses the x-axis.

For the positive domain (where $y > 0$), analyze the intervals:

• For $x < 2$, the parabola is above the x-axis (check any point like $x = 0$ to see $y = 12 > 0$).

• For $x > 6$, the parabola is above the x-axis (check any point like $x = 8$ to see $y = 12 > 0$).

Therefore, in terms of the positive domain, the function is positive for $x < 2$ (or $x > 6$).

For the negative domain (where $y < 0$), analyze the interval between roots:

• For $2 < x < 6$, the parabola is below the x-axis (check any point like $x = 4$ to see $y = -4 < 0$).

Therefore, in terms of the negative domain, the function is negative for $2 < x < 6$.

The solution to the problem is:

$x > 6$ or $x < 0 : x < 2$

$x < 0 : 2 < x < 6$