1. Problem a): Identify domain, range, and whether the relation is a function for the set $\{(-3,0), (-1,1), (0,1), (4,5), (0,6)\}$.\n\nStep 1: Domain is the set of all first coordinates: $\{-3, -1, 0, 4\}$. Note $0$ appears twice with different second coordinates.\nStep 2: Range is the set of all second coordinates: $\{0, 1, 5, 6\}$.\nStep 3: Since the input $0$ maps to both $1$ and $6$, this relation is \textbf{not} a function (functions cannot assign multiple outputs to the same input).\n\n2. Problem b): Given the equation $y = 4 - x$, find domain, range, and whether it is a function.\n\nStep 1: This is a linear equation, defined for all real numbers, so domain is $\mathbb{R}$.\nStep 2: The range is also all real numbers $\mathbb{R}$, since as $x$ varies, $y$ can take on any real value.\nStep 3: This is a function because for every input $x$ there is exactly one output $y$.\n\n3. Problem c): Given the table:\n$x$: $-4, -2, 0, 2, 4$\n$y$: $4, 2, 0, -2, -4$\n\nStep 1: Domain is $\{-4, -2, 0, 2, 4\}$.\nStep 2: Range is $\{4, 2, 0, -2, -4\}$.\nStep 3: Each input corresponds to exactly one output, so this relation is a function.\n\n4. Problem d): Equation $x^2 + y^2 = 16$ (circle with radius 4)\n\nStep 1: Domain consists of all $x$ where $-4 \le x \le 4$ because otherwise $y^2$ would be negative.\nStep 2: For each $x$ in domain, $y = \pm \sqrt{16 - x^2}$. This means two possible $y$ values for most $x$, except at $x = \pm 4$ where $y=0$.\nStep 3: Range is $[-4,4]$, the $y$ values on the circle.\nStep 4: This relation is not a function since some $x$ values correspond to two $y$ values, violating the definition of function.\n\nFinal summary:\n- a) Domain: $\{-3,-1,0,4\}$, Range: $\{0,1,5,6\}$, Not a function\n- b) Domain: $\mathbb{R}$, Range: $\mathbb{R}$, Is a function\n- c) Domain: $\{-4,-2,0,2,4\}$, Range: $\{4,2,0,-2,-4\}$, Is a function\n- d) Domain: $[-4,4]$, Range: $[-4,4]$, Not a function