The equation you are tasked with graphing is:

$y = x^2 - 9$

This is a quadratic equation, which will produce a parabola when graphed. The general form of a quadratic equation is $y = ax^2 + bx + c$, where in your case, $a = 1$, $b = 0$, and $c = -9$.

Steps to graph the equation:

1. Vertex: The vertex of the parabola occurs at the point $(0, -9)$. This is because the quadratic is in the form $y = x^2 + c$, where the vertex lies at $x = 0$, and $y = -9$.

2. Plot Points:

   • When $x = -3$ or $x = 3$, $y = 0$ (these are the x-intercepts).
   • When $x = 2$ or $x = -2$, $y = -5$.
   • When $x = 1$ or $x = -1$, $y = -8$.

3. Symmetry: Since this is a parabola, it is symmetric about the y-axis (i.e., for every point $(x, y)$, there is a corresponding point $(-x, y)$).

4. Graph Shape: The parabola opens upwards because the coefficient of $x^2$ is positive.

Key points to plot:

• $(0, -9)$ (the vertex)
• $(3, 0)$ and $(-3, 0)$ (the x-intercepts)
• $(2, -5)$, $(-2, -5)$
• $(1, -8)$, $(-1, -8)$

Once you have these points, you can draw a smooth curve through them to form the parabola.

Would you like a detailed explanation on how to compute each point, or do you have any other questions?

Here are 5 questions you could consider for further exploration:

1. How do we determine the axis of symmetry for any quadratic equation?
2. What happens to the graph if we change the constant in the equation (e.g., $y = x^2 - 5$)?
3. How can we find the vertex of a quadratic equation algebraically?
4. What is the role of the leading coefficient in determining the direction of a parabola?
5. How do we determine the x-intercepts algebraically?

Tip: When graphing quadratic functions, always start by finding the vertex, as it provides the center of symmetry for the parabola.