Determine x for the Quadratic Function y = x² + 16 When f(x) > 0

Given the function:

$y=x^2+16$

Determine for which values of x is $f(x) > 0$ true

To solve the problem of determining for which values of $x$ the function $y = x^2 + 16$ is positive, we proceed as follows:

Step 1: Analyze the function $y = x^2 + 16$.
The expression $x^2$ is non-negative (i.e., $x^2 \geq 0$) for all real numbers $x$. Therefore, the smallest value $x^2$ can take is 0.

Step 2: Evaluate the function at its minimum value.
Substituting the minimum value of $x^2$ into the function gives us:

$y = x^2 + 16 = 0 + 16 = 16$.

Step 3: Determine for which $x$ the function is positive.
Since the minimum value of the function is 16, which is greater than 0, the function $y = x^2 + 16$ is greater than 0 for all real numbers $x$.

Thus, the solution to the problem is that the function is positive for all $x$.