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You're on a roll — Here are the solutions for questions 8 and 9. Question 8: a) Draw the graph for the relation y = x^2 - 3x - 5 using the interval -3 x 6. From the graph, find the; (i) coordinates at the turning point. (ii) the minimum value (iii) the truthset of y = x^2 - 3x - 5 = 0 b) Find the equation whose roots are -6 and 13. Step 1: Create a table of values for y = x^2 - 3x - 5 for x from -3 to 6. |c|c|c|c|c|c|c|c|c|c|c| x & -3 & -2 & -1 & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ y & 13 & 5 & -1 & -5 & -7 & -7 & -5 & -1 & 5 & 13 \\ Note: The graph would be drawn by plotting these points and drawing a smooth curve through them. Step 2: Find the coordinates of the turning point. The turning point (vertex) of a parabola y = ax^2 + bx + c is at x = (-b)/(2a). For y = x^2 - 3x - 5, we have a=1, b=-3, c=-5. x = (-(-3))/(2(1)) = (3)/(2) = 1.5 Substitute x=1.5 into the equation to find y: y = (1.5)^2 - 3(1.5) - 5 = 2.25 - 4.5 - 5 = -7.25 (i) The coordinates at the turning point (read from the graph) are approximately (1.5, -7.25). Step 3: Find the minimum value. (ii) The minimum value (read from the graph) is the y-coordinate of the turning point: -7.25. Step 4: Find the truthset of y = x^2 - 3x - 5 = 0. The truthset consists of the x-intercepts where y=0. From the table, y changes from negative to positive between x=-1 and x=0, and between x=4 and x=5. Using the quadratic formula x = -b ± sqrt(b^2 - 4ac)2a: x = -(-3) ± sqrt((-3)^2 - 4(1)(-5))2(1) x = 3 ± sqrt(9 + 20)2 x = 3 ± sqrt(29)2 x_1 = 3 - sqrt(29)2 ≈ (3 - 5.385)/(2) ≈ -1.19 x_2 = 3 + sqrt(29)2 ≈ (3 + 5.385)/(2) ≈ 4.19 (iii) The truthset of y = x^2 - 3x - 5 = 0 (read from the graph) is approximately \-1.19, 4.19\. Part 8b: Step 5: Find the equation whose roots are -6 and 13. If the roots are and , the quadratic equation is x^2 - ( + )x + = 0. Given roots = -6 and = 13. Sum of roots: + = -6 + 13 = 7. Product of roots: = (-6)(13) = -78. Substitute these values into the general equation: x^2 - (7)x + (-78) = 0 The equation is x^2 - 7x - 78 = 0. Question 9: A total of 50 items consisting of pens and pencils were bought for GH¢190. A pen cost GH¢ and a pencil costs GH¢3. a. form a linear equation to represent the situation b. find the number of pen and pencils bought c. how much will a customer pay for 8 pens and 4 pencils. Assumption: The cost of a pen is missing from the question. We will assume the cost of a pen is GH¢4. Step 1: Define variables and form linear equations. Let p be the number of pens and c be the number of pencils. The total number of items is 50: p + c = 50 (1) The cost of a pen is GH¢4 and a pencil is GH¢3. The total cost is GH¢190: 4p + 3c = 190 (2) a) The linear equations representing the situation are: p + c = 50 4p + 3c = 190 Step 2: Solve the system of equations to find the number of pens and pencils. From equation (1), express c in terms of p: c = 50 - p Substitute this into equation (2): 4p + 3(50 - p) = 190 4p + 150 - 3p = 190 p + 150 = 190 p = 190 - 150 p = 40 Now find c: c = 50 - 40 = 10 b) The number of pens bought is 40 and the number of pencils bought is 10. Step 3: Calculate the cost for 8 pens and 4 pencils. Cost of 8 pens = 8 × GH¢4 = GH¢32. Cost of 4 pencils = 4 × GH¢3 = GH¢12. Total cost = Cost of 8 pens +