Solve y = 3x² + 21: When is the Function Positive?

To solve the problem, we must determine when the function $f(x) = 3x^2 + 21$ is positive:

• Step 1: Analyze the quadratic function $f(x) = ax^2 + bx + c$. Since $a = 3$ and $b = 0$, the parabola opens upwards with vertex at $x = 0$.

• Step 2: Compute the function's value at the vertex. For $x = 0$, $f(0) = 3(0)^2 + 21 = 21$, which is positive.

• Step 3: Understand the behavior for any $x$. Since $3x^2$ is non-negative and $+21$ added ensures $f(x) > 0$, the entire graph of $f(x)$ lies above the x-axis.

Therefore, the solution is that for all values of $x$, the function $f(x) = 3x^2 + 21$ is always greater than 0.

The correct answer is All x.