1. The problem is to graph the function $g(x) = \frac{1}{4}x^2$ and understand its shape. 2. The formula for a quadratic function is $g(x) = ax^2 + bx + c$. Here, $a = \frac{1}{4}$, $b = 0$, and $c = 0$. 3. Since $a > 0$, the parabola opens upwards. 4. The vertex of the parabola is at the origin $(0,0)$ because $b=0$ and $c=0$. 5. To plot points, substitute values of $x$: - For $x=0$, $g(0) = \frac{1}{4} \times 0^2 = 0$ - For $x=2$, $g(2) = \frac{1}{4} \times 2^2 = \frac{1}{4} \times 4 = 1$ - For $x=-2$, $g(-2) = \frac{1}{4} \times (-2)^2 = 1$ 6. The parabola is symmetric about the y-axis. 7. The graph starts at the origin and gradually increases as $x$ moves away from zero in both directions. Final answer: The graph of $g(x) = \frac{1}{4}x^2$ is a parabola opening upwards with vertex at $(0,0)$ and points like $(2,1)$ and $(-2,1)$ on the curve.