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

Question

Given the function:

y=x2+16 y=x^2+16

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

Step-by-Step Solution

To solve the problem of determining for which values of x x the function y=x2+16 y = x^2 + 16 is positive, we proceed as follows:

Step 1: Analyze the function y=x2+16 y = x^2 + 16 .
The expression x2 x^2 is non-negative (i.e., x20 x^2 \geq 0 ) for all real numbers x x . Therefore, the smallest value x2 x^2 can take is 0.

Step 2: Evaluate the function at its minimum value.
Substituting the minimum value of x2 x^2 into the function gives us:

y=x2+16=0+16=16 y = x^2 + 16 = 0 + 16 = 16 .

Step 3: Determine for which x x the function is positive.
Since the minimum value of the function is 16, which is greater than 0, the function y=x2+16 y = x^2 + 16 is greater than 0 for all real numbers x x .

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

Answer

All x