To graph the function $f(x) = -7x^2$, we need to follow these steps:

Step 1: Identify the Vertex

The given function is a quadratic in the form of $f(x) = ax^2 + bx + c$. In this case:

• $a = -7$
• $b = 0$
• $c = 0$

The vertex of a parabola in this form is at the point $(h, k)$, where:

• $h = \frac{-b}{2a} = 0$
• $k = f(h) = f(0) = 0$

So, the vertex is at $(0, 0)$.

Step 2: Plot the Vertex

Plot the vertex at the point $(0, 0)$ on the graph.

Step 3: Plot Another Point

Choose another value for $x$ to find a second point on the parabola. For example, let's use $x = 1$:

• $f(1) = -7(1)^2 = -7$ So, the point $(1, -7)$ is another point on the parabola.

Step 4: Plot the Point and Draw the Parabola

Now plot the point $(1, -7)$ on the graph. Since the parabola is symmetric, you can also plot the point $(-1, -7)$ on the other side of the vertex.

Final Step: Draw the Parabola

Draw a smooth curve through the points $(0, 0)$, $(1, -7)$, and $(-1, -7)$. The parabola opens downward because the coefficient of $x^2$ is negative.

Would you like further details or have any questions?

Here are some additional questions related to this topic:

1. What is the axis of symmetry of the parabola?
2. How would the graph change if the coefficient of $x^2$ were positive?
3. What are the x-intercepts of the function?
4. How would the graph change if a constant term $c$ were added to the function?
5. What effect does changing the value of $a$ (e.g., from -7 to -2) have on the shape of the parabola?

Tip: A parabola always opens upward if $a > 0$ and downward if $a < 0$.