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.