Let's analyze the given quadratic function:
$f(x) = 2(x + 4)^2 + 3$

Step 1: Identify the Vertex

The given equation is in vertex form:
$f(x) = a(x-h)^2 + k$
where $(h, k)$ is the vertex.

From $f(x) = 2(x + 4)^2 + 3$, we see that:

• $h = -4$ (since $x + 4 = x - (-4)$)
• $k = 3$

Thus, the vertex is $(-4,3)$.

Step 2: Axis of Symmetry

The axis of symmetry is the vertical line that passes through the vertex.
So, the axis of symmetry is $x = -4$.

Step 3: Domain

Quadratic functions have a domain of all real numbers:
$\text{Domain: } (-\infty, \infty)$

Step 4: Range

Since the coefficient of $(x + 4)^2$ is positive (2), the parabola opens upward.
The minimum value of $f(x)$ occurs at the vertex $(-4, 3)$, meaning the lowest y-value is 3.

Thus, the range is:
$\text{Range: } [3, \infty)$

Step 5: Y-Intercept

The y-intercept occurs at $x = 0$.
Substituting $x = 0$ into the equation:

$f(0) = 2(0 + 4)^2 + 3$

$= 2(16) + 3 = 32 + 3 = 35$

So, the y-intercept is $(0,35)$.

Final Answers:

• Axis of Symmetry: $x = -4$
• Vertex: $(-4,3)$
• Y-Intercept: $(0,35)$
• Domain: $(-\infty, \infty)$
• Range: $[3, \infty)$

Would you like any further clarifications?

Related Questions:

1. How do you determine if a quadratic function has a maximum or minimum value?
2. How can we find the x-intercepts of this function?
3. What effect does changing the coefficient of $(x+4)^2$ have on the graph?
4. What happens to the vertex if we change $+3$ to $-5$ in the equation?
5. How do you rewrite this function in standard form?

Tip:

Always check whether the coefficient of $x^2$ is positive or negative to determine if the parabola opens up (minimum) or down (maximum).