Here’s the solution to the problem based on the graph provided:

(a) Sign of the leading coefficient

The parabola opens downward, which means the leading coefficient is negative.

(b) Vertex

The vertex is the highest point on the parabola, which is located at:
$(-3, 1)$.

(c) Axis of symmetry

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

(d) Intervals where $f$ is increasing and decreasing

1. The function is increasing to the left of the vertex (as $x$ approaches $-3$)
   → Interval: $(-\infty, -3)$.
2. The function is decreasing to the right of the vertex (as $x$ moves past $-3$)
   → Interval: $(-3, \infty)$.

(e) Domain and Range

1. The domain of a parabola is all real numbers because it extends infinitely in both directions:
   $(-\infty, \infty)$.
2. The range of the parabola is restricted by its maximum value (at $y = 1$), and it decreases infinitely downward:
   $(-\infty, 1]$.

Final Answers Summary:

1. (a) Negative
2. (b) $(-3, 1)$
3. (c) $x = -3$
4. (d) Increasing: $(-\infty, -3)$, Decreasing: $(-3, \infty)$
5. (e) Domain: $(-\infty, \infty)$, Range: $(-\infty, 1]$

Would you like a deeper explanation for any part? Let me know! 😊

Related Questions:

1. What happens to the graph if the leading coefficient is positive?
2. How would you write the equation of this parabola in vertex form?
3. What is the effect of moving the vertex to a new point, say $(0, 0)$?
4. How do you determine the axis of symmetry from the standard form of a quadratic equation?
5. If the parabola opens downward, what does that tell us about the vertex’s role in the graph?

Tip:

The vertex of a parabola in standard form $y = ax^2 + bx + c$ can be found using $x = -\frac{b}{2a}$. For vertex form $y = a(x-h)^2 + k$, the vertex is directly $(h, k)$.