1. **State the problem:** We need to complete the table for the function $$f(x) = -2x^2 + 5x + 3$$ for the domain $$-2 \leq x \leq 5$$. 2. **Recall the function:** $$f(x) = -2x^2 + 5x + 3$$ means for each value of $$x$$, calculate $$-2x^2$$, $$5x$$, and add 3, then sum all to get $$f(x)$$. 3. **Calculate each term for each $$x$$:** - For $$x = -2$$: - $$-2x^2 = -2(-2)^2 = -2(4) = -8$$ - $$5x = 5(-2) = -10$$ - Constant term = 3 - $$f(-2) = -8 + (-10) + 3 = -15$$ - For $$x = -1$$: - $$-2x^2 = -2(-1)^2 = -2(1) = -2$$ - $$5x = 5(-1) = -5$$ - Constant term = 3 - $$f(-1) = -2 + (-5) + 3 = -4$$ - For $$x = 0$$: - $$-2x^2 = -2(0)^2 = 0$$ - $$5x = 5(0) = 0$$ - Constant term = 3 - $$f(0) = 0 + 0 + 3 = 3$$ - For $$x = 1$$: - $$-2x^2 = -2(1)^2 = -2(1) = -2$$ - $$5x = 5(1) = 5$$ - Constant term = 3 - $$f(1) = -2 + 5 + 3 = 6$$ - For $$x = 2$$: - $$-2x^2 = -2(2)^2 = -2(4) = -8$$ - $$5x = 5(2) = 10$$ - Constant term = 3 - $$f(2) = -8 + 10 + 3 = 5$$ - For $$x = 3$$: - $$-2x^2 = -2(3)^2 = -2(9) = -18$$ - $$5x = 5(3) = 15$$ - Constant term = 3 - $$f(3) = -18 + 15 + 3 = 0$$ - For $$x = 4$$: - $$-2x^2 = -2(4)^2 = -2(16) = -32$$ - $$5x = 5(4) = 20$$ - Constant term = 3 - $$f(4) = -32 + 20 + 3 = -9$$ - For $$x = 5$$: - $$-2x^2 = -2(5)^2 = -2(25) = -50$$ - $$5x = 5(5) = 25$$ - Constant term = 3 - $$f(5) = -50 + 25 + 3 = -22$$ 4. **Final completed table:** | x | -2x^2 | 5x | 3 | f(x) | |----|-------|-----|---|------| | -2 | -8 | -10 | 3 | -15 | | -1 | -2 | -5 | 3 | -4 | | 0 | 0 | 0 | 3 | 3 | | 1 | -2 | 5 | 3 | 6 | | 2 | -8 | 10 | 3 | 5 | | 3 | -18 | 15 | 3 | 0 | | 4 | -32 | 20 | 3 | -9 | | 5 | -50 | 25 | 3 | -22 |