1. **State the problem:** We are given two quadratic functions $a(x) = x^2 + 3x + 1$ and $b(x) = (x + 3)(x + 1)$ and asked to evaluate each at chosen $x$-values: $-2, -1, 0, 1, 2$. 2. **Formula and strategy:** - For $a(x)$, use direct substitution and simplify. - For $b(x)$, either expand first or substitute directly and multiply. 3. **Evaluate $a(x)$:** - $a(-2) = (-2)^2 + 3(-2) + 1 = 4 - 6 + 1 = -1$ - $a(-1) = (-1)^2 + 3(-1) + 1 = 1 - 3 + 1 = -1$ - $a(0) = 0^2 + 3(0) + 1 = 0 + 0 + 1 = 1$ - $a(1) = 1^2 + 3(1) + 1 = 1 + 3 + 1 = 5$ - $a(2) = 2^2 + 3(2) + 1 = 4 + 6 + 1 = 11$ 4. **Evaluate $b(x)$:** - $b(-2) = (-2 + 3)(-2 + 1) = (1)(-1) = -1$ - $b(-1) = (-1 + 3)(-1 + 1) = (2)(0) = 0$ - $b(0) = (0 + 3)(0 + 1) = (3)(1) = 3$ - $b(1) = (1 + 3)(1 + 1) = (4)(2) = 8$ - $b(2) = (2 + 3)(2 + 1) = (5)(3) = 15$ 5. **Strategy discussion:** - For $a(x)$, substitution into the expanded form is straightforward. - For $b(x)$, using the factored form allows quick multiplication after substituting. - Both methods are valid; choice depends on convenience. **Final tables:** | $x$ | $a(x)$ | $b(x)$ | |-----|--------|--------| | -2 | -1 | -1 | | -1 | -1 | 0 | | 0 | 1 | 3 | | 1 | 5 | 8 | | 2 | 11 | 15 |