To solve the given problem step-by-step, let's analyze the function $f(x) = |(x + 2)^3| - 1$ and answer each of the questions provided.

1. Finding the zeros of the function $f(x)$:

   The function $f(x) = |(x + 2)^3| - 1$ will be zero when: $|(x + 2)^3| - 1 = 0$ This simplifies to: $|(x + 2)^3| = 1$ Solving for $x$, we consider two cases:

   • Case 1: $(x + 2)^3 = 1$
     • Taking the cube root on both sides, $x + 2 = 1 \Rightarrow x = -1$.
   • Case 2: $(x + 2)^3 = -1$
     • Taking the cube root, $x + 2 = -1 \Rightarrow x = -3$.

   Therefore, the zeros of $f(x)$ are at $x = -1$ and $x = -3$.

2. Finding the y-intercept of the function:

   The y-intercept occurs when $x = 0$: $f(0) = |(0 + 2)^3| - 1 = |8| - 1 = 8 - 1 = 7$ So, the y-intercept is at $(0, 7)$.

3. Recording the extrema (maximums) of the function in the interval $[-4, 1]$:

   To find the extrema, we should examine the behavior of $f(x)$ within $[-4, 1]$ and determine if there are any critical points within this interval.

   • At $x = -4$: $f(-4) = |(-4 + 2)^3| - 1 = |(-2)^3| - 1 = | -8 | - 1 = 8 - 1 = 7$
   • At $x = -3$ (a zero, so this is a critical point): $f(-3) = 0$
   • At $x = -1$ (another zero): $f(-1) = 0$
   • At $x = 1$: $f(1) = |(1 + 2)^3| - 1 = |3^3| - 1 = |27| - 1 = 26$

   Therefore, the maximum value in $[-4, 1]$ occurs at $x = 1$, where $f(1) = 26$.

4. Recording the interval where the function is decreasing and positive in the interval $[-4, 1]$:

   To determine where the function is decreasing, we examine the derivative of $f(x)$ and the behavior of $|(x + 2)^3|$. The function $f(x)$ decreases in the interval $[-4, -3]$, as $(x+2)^3$ is negative and becomes less negative as $x$ approaches -3.

5. Identifying where the function is positive only for $x \leq 0$:

   From the zeros calculated in part 1, the function $f(x)$ is positive in the interval $[-4, -3) \cup (-3, -1) \cup (-1, 1]$ where it is strictly above zero.

6. Identifying where the function is positive only for $x > 0$:

   In the interval $x > 0$, the function is positive for $x > 0$ as $f(x) \to \infty$ as $x$ grows.