Let’s analyze the given problem, where you are asked to sketch the graphs of parabolas based on the provided characteristics for each part.

Part (a):

1. Range: $y \leq 5$, $y \in \mathbb{R}$.
   This tells us the maximum $y$-value of the parabola is 5, and the parabola opens downward.

2. Axis of symmetry: $x = -2$.
   The parabola is symmetric about the line $x = -2$.

3. y-intercept: $(0, 1)$.
   The parabola crosses the y-axis at the point $(0, 1)$.

   To sketch this:

   • Since the axis of symmetry is at $x = -2$, the vertex will be along this line.
   • The maximum point (vertex) should be at $y = 5$, on the line $x = -2$.
   • The parabola will pass through $(0, 1)$, which lies 2 units to the right of the axis of symmetry.
   • Reflect this across the axis of symmetry to get another point at $(-4, 1)$.

Part (b):

1. **Range