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The equation given is:

\[
f(x) = 2(x + 3)^2
\]

This is a quadratic function, so let's break it down step by step.

### a) Graph the function \( f(x) = 2(x + 3)^2 \)

This equation is in vertex form:

\[
f(x) = a(x-h)^2 + k
\]

where \( (h, k) \) is the vertex of the parabola, and \( a \) determines the direction and the width of the parabola. In this case:

- \( a = 2 \), so the parabola opens upwards and is vertically stretched by a factor of 2.
- \( h = -3 \) and \( k = 0 \), so the vertex is at \( (-3, 0) \).

The general shape of this graph is a parabola that opens upwards, with its vertex shifted to the left by 3 units. The scaling factor of 2 makes it narrower than a standard parabola.

### b) Identify the vertex

From the equation, we can directly identify the vertex:

\[
\text{Vertex} = (-3, 0)
\]

### c) Identify the axis of symmetry

The axis of symmetry for any parabola is the vertical line that passes through the vertex. So for this function:

\[
x = -3
\]

This is the equation for the axis of symmetry.

Would you like me to create a plot for the graph? Also, let me know if you have any questions on these parts! Here are 5 related questions:

1. What is the impact of the coefficient \( a \) in a parabola's equation?
2. How would the graph change if the equation were \( f(x) = -2(x + 3)^2 \)?
3. How do you find the x-intercepts for a quadratic equation?
4. What is the general form of a quadratic equation, and how is it different from the vertex form?
5. Can you explain how to calculate the y-intercept for the function \( f(x) = 2(x + 3)^2 \)?

**Tip**: The vertex form of a quadratic equation is extremely useful for quickly identifying the vertex and the direction of the parabola.

Graph the function f(x) = 2(x + 3)^2, identify the vertex, and use the graph to find the axis of symmetry.

Learn how to graph the quadratic function f(x) = 2(x + 3)^2, identify its vertex, and find the axis of symmetry. This problem covers key algebraic concepts related to quadratic functions, suitable for Grades 9-11.

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