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