Let's analyze the given quadratic function:

$f(x) = 2(x + 3)^2 - 4$

This function is in vertex form, $f(x) = a(x - h)^2 + k$, where:

• $(h, k)$ is the vertex of the parabola.
• $a$ determines the direction and width of the parabola.

Step 1: Identify the Vertex

From the equation, we can see:

• $h = -3$ and $k = -4$.
• Therefore, the vertex is at $(-3, -4)$.

Step 2: Determine the Axis of Symmetry

The axis of symmetry is a vertical line that passes through the vertex. The equation for the axis of symmetry is: $x = -3$

Step 3: Find the Intercepts

1. Y-Intercept: Set $x = 0$ and solve for $f(0)$: $f(0) = 2(0 + 3)^2 - 4 = 2(3)^2 - 4 = 2(9) - 4 = 18 - 4 = 14$ So, the y-intercept is $(0, 14)$.

2. X-Intercept(s): Set $f(x) = 0$ and solve for $x$: $0 = 2(x + 3)^2 - 4$ $4 = 2(x + 3)^2$ $2 = (x + 3)^2$ $x + 3 = \pm \sqrt{2}$ $x = -3 \pm \sqrt{2}$ Therefore, the x-intercepts are $x = -3 + \sqrt{2}$ and $x = -3 - \sqrt{2}$.

Step 4: Domain and Range

1. Domain: The domain of a quadratic function is always all real numbers: $\text{Domain: } (-\infty, \infty)$

2. Range: Since the parabola opens upwards (because $a = 2 > 0$), the vertex represents the minimum point. The y-coordinate of the vertex is -4, so: $\text{Range: } [-4, \infty)$

Summary:

• Vertex: $(-3, -4)$
• Axis of Symmetry: $x = -3$
• Y-Intercept: $(0, 14)$
• X-Intercepts: $(-3 + \sqrt{2}, 0)$ and $(-3 - \sqrt{2}, 0)$
• Domain: $(-\infty, \infty)$
• Range: $[-4, \infty)$

Would you like more details, or have any questions?

Related Questions:

1. How can we convert the quadratic function from vertex form to standard form?
2. What effect does the value of $a$ have on the shape of the parabola?
3. How do we determine the direction in which the parabola opens?
4. What are the key features of a parabola in standard form?
5. Can a quadratic function have more than two x-intercepts?

Tip:

When a quadratic function is in vertex form, it is easy to identify the vertex directly. This form is useful for graphing and understanding the basic properties of the parabola.