The given problem asks you to graph the function $f(x) = 3x^2 + 6x$. Let's break it down step by step:

1. Identify the Vertex of the Parabola

The given function is in standard quadratic form:
$f(x) = ax^2 + bx + c$
Here, $a = 3$, $b = 6$, and $c = 0$.

The formula for the x-coordinate of the vertex is:
$x_{\text{vertex}} = -\frac{b}{2a}$
Substitute $a = 3$ and $b = 6$:
$x_{\text{vertex}} = -\frac{6}{2(3)} = -1$

To find the y-coordinate, substitute $x = -1$ into the function:
$f(-1) = 3(-1)^2 + 6(-1) = 3(1) - 6 = -3$

So, the vertex is at $(-1, -3)$.

2. Plot Another Point on the Parabola

Choose another x-value near the vertex, for example, $x = 0$.
Substitute $x = 0$ into the function:
$f(0) = 3(0)^2 + 6(0) = 0$
So, another point on the parabola is $(0, 0)$, which is also the y-intercept.

3. Draw the Parabola

1. Plot the vertex $(-1, -3)$.
2. Plot the second point $(0, 0)$.
3. Since parabolas are symmetric, another point on the opposite side of the vertex can be reflected over $x = -1$. For example, since $(0, 0)$ is 1 unit to the right of the vertex, the point $(-2, 0)$ will be 1 unit to the left of the vertex.

Final Points for Plotting

• Vertex: $(-1, -3)$
• Point: $(0, 0)$
• Symmetric Point: $(-2, 0)$

Would you like me to provide additional details or clarifications?

Related Questions:

1. How do you convert a quadratic function to vertex form?
2. What is the significance of the axis of symmetry in a parabola?
3. How do the values of $a$, $b$, and $c$ in a quadratic function affect its graph?
4. Can you determine the direction of a parabola from its equation? How?
5. How would adding a constant to $f(x)$ affect the graph?

Tip:

The vertex formula $x = -\frac{b}{2a}$ is always useful when dealing with quadratic functions in standard form!