Locate the Symmetry Point of the Quadratic Function f(x)=3+3x^2

To solve this problem, we need to find the symmetry point of the quadratic function given by $f(x) = 3 + 3x^2$.

• Step 1: Identify the type of quadratic equation. Here, it's $ax^2 + bx + c$ where $a = 3$, $b = 0$, and $c = 3$.
• Step 2: Use the formula for the symmetry point $x = -\frac{b}{2a}$. Since $b = 0$, the formula simplifies to $x = 0$.
• Step 3: Calculate the y-coordinate by substituting $x = 0$ into the original function: $f(0) = 3(0)^2 + 3 = 3$.

Thus, the symmetry point (also the vertex of the parabola) is $(0, 3)$.

Therefore, the solution to the problem is $(0, 3)$.