Discover the Symmetry in the Quadratic f(x) = 1/2x^2

To determine the symmetry (vertex) point of the quadratic function $f(x) = \frac{1}{2}x^2$, we will use the formula for the x-coordinate of the vertex (or axis of symmetry) for a general quadratic function $f(x) = ax^2 + bx + c$, which is given by:

• $x = -\frac{b}{2a}$

In this problem, the coefficients are $a = \frac{1}{2}$, $b = 0$, and $c = 0$. By substituting these values into the vertex formula:

$x = -\frac{0}{2 \times \frac{1}{2}} = 0$

This tells us that the x-coordinate of the vertex is $0$. To find the y-coordinate of the vertex, we substitute $x = 0$ back into the function $f(x) = \frac{1}{2}(x^2)$:

$f(0) = \frac{1}{2}(0)^2 = 0$

Thus, the vertex of the function, also its symmetry point, is at the coordinate $(0,0)$.

Therefore, the symmetry point of the function $f(x) = \frac{1}{2}x^2$ is $(0, 0)$.