We are given a graph of a quadratic function, and we need to determine its equation. The graph shows a parabola with the following key points:

• The vertex is at $(-2, -6)$.
• Another point on the parabola is $(0, -2)$.

We can use the vertex form of a quadratic equation, which is:

$f(x) = a(x - h)^2 + k$

where $(h, k)$ is the vertex. In this case, the vertex is $(-2, -6)$, so we can substitute these values into the equation:

$f(x) = a(x + 2)^2 - 6$

Next, we use the other given point $(0, -2)$ to find the value of $a$. Substituting $x = 0$ and $f(x) = -2$ into the equation:

$-2 = a(0 + 2)^2 - 6$

Simplifying:

$-2 = a(4) - 6$ $-2 + 6 = 4a$ $4 = 4a$ $a = 1$

Thus, the quadratic function is:

$f(x) = (x + 2)^2 - 6$

Expanding this:

$f(x) = (x^2 + 4x + 4) - 6$ $f(x) = x^2 + 4x - 2$

So, the quadratic function that describes the graph is:

$f(x) = x^2 + 4x - 2$

Would you like more details on how I expanded or solved this equation? Here are some related questions for further practice:

1. How do you derive the vertex form of a quadratic equation?
2. What are the steps to expand $(x + 2)^2 - 6$ into standard form?
3. How can you determine the axis of symmetry for this quadratic function?
4. How does the value of $a$ affect the shape of the parabola?
5. Can you find the x-intercepts of this quadratic function?

Tip: When solving for the constant $a$, always use another point on the graph to substitute into the equation.