Locate the Vertex of the Quadratic Function f(x) = 3x^2 + 6x

To find the symmetry point of the quadratic function $f(x) = 3x^2 + 6x$, follow these steps:

• Step 1: Identify the coefficients from the quadratic function $ax^2 + bx + c$. Here, $a = 3$ and $b = 6$.
• Step 2: Use the vertex formula for the symmetry point, which is given by $x = -\frac{b}{2a}$.
• Step 3: Substitute the values of $a$ and $b$ into the formula:

$x = -\frac{6}{2 \times 3} = -\frac{6}{6} = -1$

• Step 4: Substitute $x = -1$ back into the original function to find the corresponding $y$-coordinate:

$f(-1) = 3(-1)^2 + 6(-1) = 3 \times 1 - 6 = 3 - 6 = -3$

• Step 5: Therefore, the symmetry point of the function is $(-1, -3)$.

Thus, the symmetry point of the given quadratic function $f(x) = 3x^2 + 6x$ is $\boxed{(-1, -3)}$.