The given function is:

$y = \frac{1}{5}(x + 3)^2 - 2$

Let's analyze this function step by step:

1. Vertex Form

This is a quadratic function in vertex form:

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

where $(h, k)$ is the vertex, and $a$ determines the shape and direction of the parabola. For the given equation:

• $h = -3$ (the sign is flipped inside the parentheses),
• $k = -2$ (the vertical shift),
• $a = \frac{1}{5}$, which means the parabola opens upwards and is relatively wide due to the small value of $a$.

So, the vertex of the parabola is at $(-3, -2)$.

2. Parabola Characteristics

• Direction: Since $a > 0$, the parabola opens upward.
• Width: The factor $\frac{1}{5}$ means the parabola is wider than the standard parabola, which opens with a narrower curve.
• Vertex: The vertex is at $(-3, -2)$, so the parabola is shifted to the left by 3 units and down by 2 units.

3. Graph Matching

Now, let's compare this with the provided graphs:

• We need to look for a graph with its vertex at $(-3, -2)$ and opening upwards with a wide curve.

From the given graphs, the bottom left graph has the vertex at $(-3, -2)$ and opens upwards. This matches the equation provided.

Answer: The graph that corresponds to the given equation is the bottom left graph.

Would you like more details or have any questions about this?

Here are some related questions to explore:

1. How do you find the axis of symmetry for a parabola in vertex form?
2. What role does the coefficient $a$ play in determining the width of a parabola?
3. How does shifting the graph horizontally and vertically affect its equation?
4. What is the general form of a quadratic equation, and how does it differ from vertex form?
5. How can you convert a quadratic equation from standard form to vertex form?

Tip: When graphing parabolas, always check the vertex and the direction of opening first to make matching with the graph easier.