Domain Analysis of y = 1/4x² + 0.5: Finding Positive and Negative Regions

To solve this problem, we need to determine where the quadratic function $y = \frac{1}{4}x^2 + 0.5$ is positive, zero, or negative across its domain.

The given function is $y = \frac{1}{4}x^2 + 0.5$. This is a standard quadratic function where $a = \frac{1}{4}$ and $c = 0.5$. Because there is no $bx$ term, the parabola's vertex is directly on the $y$-axis at $x = 0$.

Since $a$ is positive ($\frac{1}{4} > 0$), the parabola opens upwards. The minimum point of the graph is at the vertex, $x = 0$. Evaluating the function at this point, we have:

$y = \frac{1}{4}(0)^2 + 0.5 = 0.5$

Here, the value of $y$ at $x = 0$ is $0.5$, which is positive.

For all other values of $x$, because the parabola opens upwards, $y$ will be greater than $0.5$. Therefore, $y$ is always positive for all real numbers $x$.

This means:

• For $x < 0$: The function is always positive because $y > 0.5$.
• For $x > 0$: The function is always positive because $y > 0.5$.

There are no values of $x$ where $y$ is negative.

Therefore, the positive and negative domains of the function are:

$x < 0 : \text{none}$

$x > 0 : \text{all } x$

Thus, the correct choice is:

$x < 0 :$ none

$x > 0 :$ all x