Looking at the graph you provided, we can observe that the quadratic function has a vertex and a clear parabolic shape.

Key observations:

• The vertex of the parabola is at $(2, -1)$, which suggests that the equation is in vertex form.

• The vertex form of a quadratic equation is: $y = a(x - h)^2 + k$ where $(h, k)$ is the vertex. From the graph, $(h, k) = (2, -1)$.

• We also notice that the parabola opens upward, which indicates that the coefficient $a > 0$.

Step 1: Substituting the vertex coordinates:

$y = a(x - 2)^2 - 1$

Step 2: Determine the value of $a$:

To find $a$, we can use another point from the graph. From the graph, we can see that the parabola passes through the point $(1, 0)$.

• Substituting $x = 1$ and $y = 0$ into the equation to solve for $a$: $0 = a(1 - 2)^2 - 1$ $0 = a(1)^2 - 1$ $0 = a - 1$ $a = 1$

Step 3: Final equation:

Now that we have $a = 1$, the equation of the parabola becomes: $y = (x - 2)^2 - 1$

This is the equation for the quadratic graph.

Would you like more details on any part of this process or have additional questions?

Here are 5 questions related to this:

1. What is the general form of a quadratic equation, and how can we convert from vertex form?
2. How can we determine the axis of symmetry for a quadratic function?
3. How do we find the x-intercepts (roots) of a quadratic equation?
4. What are the implications of the value of $a$ on the shape of a parabola?
5. How can we graph a quadratic equation using only its vertex and one additional point?

Tip: In quadratic equations, the sign of the coefficient $a$ determines whether the parabola opens upward ($a > 0$) or downward ($a < 0$).