Find the Domain of (x-12½)²-4: Complete Function Analysis

To solve the problem of finding the positive and negative domains of the function $y = \left(x - 12\frac{1}{2}\right)^2 - 4$, follow these steps:

• Start by setting the quadratic equation to zero:
  $(x - 12\frac{1}{2})^2 - 4 = 0$.

• Add 4 to both sides:
  $(x - 12\frac{1}{2})^2 = 4$.

• Take the square root of both sides to find the x-values where the parabola intersects the x-axis:
  $x - 12\frac{1}{2} = \pm 2$.

Solve for $x$ in both cases:

• For $x - 12\frac{1}{2} = 2$:
  $x = 12\frac{1}{2} + 2 = 14\frac{1}{2}$.

• For $x - 12\frac{1}{2} = -2$:
  $x = 12\frac{1}{2} - 2 = 10\frac{1}{2}$.

Thus, the roots of the quadratic are $x = 10\frac{1}{2}$ and $x = 14\frac{1}{2}$. These points divide the x-axis into three intervals: x < 10\frac{1}{2} , 10\frac{1}{2} < x < 14\frac{1}{2} , and x > 14\frac{1}{2} .

Next, solve for where the function is positive or negative in these intervals:

• Interval x < 10\frac{1}{2} :
  Choose a test point $x = 0$.
  The function value is $(0 - 12\frac{1}{2})^2 - 4 = (12\frac{1}{2})^2 - 4 = 156.25 - 4 = 152.25$.
  Since 152.25 is positive, y > 0 for this interval.

• Interval 10\frac{1}{2} < x < 14\frac{1}{2} :
  Choose a test point $x = 12$.
  The function value is $(12 - 12\frac{1}{2})^2 - 4 = (0.5)^2 - 4 = 0.25 - 4 = -3.75$.
  Since $-3.75$ is negative, y < 0 in this interval.

• Interval x > 14\frac{1}{2} :
  Choose a test point $x = 15$.
  The function value is $(15 - 12\frac{1}{2})^2 - 4 = (2.5)^2 - 4 = 6.25 - 4 = 2.25$.
  Since 2.25 is positive, y > 0 for this interval.

Thus, the function is negative for 10\frac{1}{2} < x < 14\frac{1}{2} and positive for x < 10\frac{1}{2} and x > 14\frac{1}{2} .

Therefore, the positive and negative domains are:

Positive domain: x < 10\frac{1}{2} or x > 14\frac{1}{2}

Negative domain: 10\frac{1}{2} < x < 14\frac{1}{2}

The correct answer is choice 4.