Let's break down the problem step by step:

We are given the quadratic function:

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

(a) Find the vertex

The vertex of a quadratic function $f(x) = ax^2 + bx + c$ can be found using the vertex formula. The x-coordinate of the vertex is given by:

$x = \frac{-b}{2a}$

Here, $a = 2$ and $b = 6$.

Substituting these values into the formula:

$x = \frac{-6}{2(2)} = \frac{-6}{4} = -\frac{3}{2}$

Now, to find the y-coordinate of the vertex, substitute $x = -\frac{3}{2}$ back into the original function:

$f\left(-\frac{3}{2}\right) = 2\left(-\frac{3}{2}\right)^2 + 6\left(-\frac{3}{2}\right) + 6$

First, calculate each term:

$2\left(-\frac{3}{2}\right)^2 = 2 \times \frac{9}{4} = \frac{18}{4} = 4.5$

$6 \times \left(-\frac{3}{2}\right) = -9$

Now, sum these:

$f\left(-\frac{3}{2}\right) = 4.5 - 9 + 6 = 1.5$

Thus, the vertex is:

$\left(-\frac{3}{2}, 1.5\right)$

(b) Find the axis of symmetry

The axis of symmetry is the vertical line that passes through the x-coordinate of the vertex. Therefore, the equation for the axis of symmetry is:

$x = -\frac{3}{2}$

(c) Maximum or minimum value

Since the coefficient of $x^2$ (which is 2) is positive, the parabola opens upwards. This means the function has a minimum value at the vertex. The minimum value of the function is the y-coordinate of the vertex, which is:

$\text{Minimum value} = 1.5$

(d) Graph the function

To graph the function, plot the vertex $\left(-\frac{3}{2}, 1.5\right)$, then plot additional points on either side of the vertex by choosing nearby x-values and calculating the corresponding y-values. The parabola will open upwards because the coefficient of $x^2$ is positive.

Would you like more details on any part of the solution?

Here are 5 related questions to consider:

1. How would the vertex change if the coefficient of $x^2$ was negative?
2. What is the significance of the axis of symmetry in the graph of a parabola?
3. How can you find the x-intercepts (if any) of this quadratic function?
4. How does the value of the leading coefficient (2 in this case) affect the width of the parabola?
5. What would happen if you shifted the graph of this function vertically or horizontally?

Tip: When graphing quadratic functions, plotting the vertex and a few points on each side helps create an accurate sketch.