Let's analyze the graph of the function to answer the questions.

1. The function has a maximum/minimum at $x =$:

   • The graph shows a parabola opening downward, indicating it has a maximum point at the vertex.
   • The vertex occurs at $x = -6$ with a value of $y = 4$.
   • So, the function has a maximum at $x = -6$.

2. The function is increasing on the interval(s):

   • A function increases when moving from left to right if the slope is positive.
   • The graph increases on the left side of the vertex, i.e., from $x = -\infty$ to $x = -6$.
   • The interval where the function is increasing is $(-\infty, -6]$.

3. The function is decreasing on the interval(s):

   • A function decreases when moving from left to right if the slope is negative.
   • The graph decreases on the right side of the vertex, i.e., from $x = -6$ to $x = \infty$.
   • The interval where the function is decreasing is $[-6, \infty)$.

4. The domain of the function is:

   • The domain of a quadratic function is all real numbers since there are no restrictions on the $x$-values.
   • The domain is $(-\infty, \infty)$.

5. The range of the function is:

   • The range is the set of all possible $y$-values.
   • Since the maximum value is $y = 4$ and the parabola opens downward, the range is $(-\infty, 4]$.

Summary of Answers:

• Maximum at $x = -6$
• Increasing on $(-\infty, -6]$
• Decreasing on $[-6, \infty)$
• Domain: $(-\infty, \infty)$
• Range: $(-\infty, 4]$

Would you like further clarification or additional details on any of these? Here are five related questions you could explore:

1. How can you find the vertex of a quadratic function algebraically?
2. Why does the parabola open downward in this case?
3. How would the domain change for a different type of function (e.g., a square root function)?
4. How can you determine the intervals of increase or decrease using calculus (derivatives)?
5. What real-world scenarios could this quadratic model represent?

Tip: Remember that the vertex form of a quadratic function, $f(x) = a(x-h)^2 + k$, directly gives the vertex as $(h, k)$.