1. The problem asks which values can be added to the set \{(-4,6), (5,-1), (2,0)\} so that it remains a function. 2. A function cannot have two different outputs for the same input. So, the x-values must be unique. 3. Check each option: - a. (-4,7): x = -4 already exists with y = 6, so adding (-4,7) would give two different y-values for x = -4. Not allowed. - b. (3,6): x = 3 is new, so this can be added. - c. (-2,7): x = -2 is new, so this can be added. - d. (5,-9): x = 5 already exists with y = -1, so adding (5,-9) would give two different y-values for x = 5. Not allowed. Answer: b and c can be added. 4. Sketch a graph that IS a function and one that IS NOT a function. - A function graph passes the vertical line test: no vertical line intersects the graph more than once. - A graph that is NOT a function fails this test. 5. Given \(f(x) = \frac{1}{3}x - 6\) and \(g(x) = 3(x - 4) + x\), find: - \(f(-12) = \frac{1}{3}(-12) - 6 = -4 - 6 = -10\) - \(g(6) = 3(6 - 4) + 6 = 3(2) + 6 = 6 + 6 = 12\) - \(f(0) = \frac{1}{3}(0) - 6 = 0 - 6 = -6\) 6. Complete the tables and find the equations. Table 1: - x: 2, 4, 6, 7 - y: 9, 12, 15, ? Find the pattern: Difference in y: 12 - 9 = 3, 15 - 12 = 3, so slope \(m = 3/2 = 1.5\) per 1 x unit? Check slope: Between x=2 and x=4, \(\Delta x = 2\), \(\Delta y = 3\), so slope \(m = \frac{3}{2} = 1.5\) Equation form: \(y = mx + b\) Use point (2,9): \(9 = 1.5(2) + b \Rightarrow 9 = 3 + b \Rightarrow b = 6\) So, \(y = 1.5x + 6\) Find y when x=7: \(y = 1.5(7) + 6 = 10.5 + 6 = 16.5\) Table 2: - x: 2, 5, 6, 8 - y: 14, 20, 22, ? Find slope between points: Between (2,14) and (5,20): \(m = \frac{20 - 14}{5 - 2} = \frac{6}{3} = 2\) Between (5,20) and (6,22): \(m = \frac{22 - 20}{6 - 5} = \frac{2}{1} = 2\) So slope \(m = 2\) Equation: \(y = 2x + b\) Use point (2,14): \(14 = 2(2) + b \Rightarrow 14 = 4 + b \Rightarrow b = 10\) Equation: \(y = 2x + 10\) Find y when x=8: \(y = 2(8) + 10 = 16 + 10 = 26\)