To sketch the graph of the function $f(x) = \frac{1}{4}x^2$, follow these steps:

Key Features of the Graph:

1. Shape: The function is a quadratic equation and has a parabolic shape. Since the coefficient of $x^2$ is positive, the parabola opens upwards.
2. Vertex: The vertex of the parabola is at $(0, 0)$, since there is no linear term or constant term to shift it horizontally or vertically.
3. Axis of Symmetry: The axis of symmetry is the vertical line $x = 0$, which passes through the vertex.
4. Width: The coefficient $\frac{1}{4}$ makes the parabola wider compared to $x^2$. The larger the denominator, the wider the parabola.

Steps to Sketch the Graph:

• Plot the vertex at $(0, 0)$.
• Plot a few points by substituting values of $x$ into $f(x)$. For example:
  • When $x = -2$, $f(-2) = \frac{1}{4}(-2)^2 = 1$.
  • When $x = 2$, $f(2) = \frac{1}{4}(2)^2 = 1$.
  • When $x = -4$, $f(-4) = \frac{1}{4}(-4)^2 = 4$.
  • When $x = 4$, $f(4) = \frac{1}{4}(4)^2 = 4$.
• Draw a smooth curve through these points to form the parabola, ensuring it opens upwards and is symmetric around the y-axis.

Would you like me to generate this graph for you?

Questions for deeper understanding:

1. How does the coefficient $\frac{1}{4}$ affect the width of the parabola compared to the standard $x^2$?
2. What would happen to the graph if we changed the function to $-\frac{1}{4}x^2$?
3. How can you determine the vertex of a parabola for a more general quadratic function?
4. What is the significance of the axis of symmetry in a quadratic graph?
5. How can you calculate the points where the parabola intersects the y-axis?

Tip: For any quadratic function $ax^2 + bx + c$, the graph will always be a parabola. The sign of $a$ determines the direction it opens, and the value of $a$ influences how "wide" or "narrow" the parabola is.