The line \( x=\frac{2}{3} \) through the turning point is called the axis of the curve and the curve is symmetry about this line. Example 8.6 .2 Express \( 5-x-2 x^{2} \) in the form \( a-b(x+c)^{2} \) and hence or otherwise find its maximum value and the value of x where this occurs and sketch the graph.

We start with the function   f(x) = 5 – x – 2x². Step 1. Rewrite the Function in a More Standard Form It is often easier to complete the square when the quadratic is written in the form   f(x) = ax² + bx + c. So write:   f(x) = –2x² – x + 5. Step 2. Factor Out the Coefficient of x² Factor –2 from the x² and x terms:   f(x) = –2 (x² + (1/2)x) + 5. Step 3. Complete the Square To complete the square for the expression in parentheses, look at   x² + (1/2)x. Recall that to complete the square, we add and subtract (half of the coefficient of x)². Here the coefficient of x is 1/2, so half of that is 1/4, and (1/4)² = 1/16. Thus:   x² + (1/2)x    = (x + 1/4)² – 1/16. Substitute this back into f(x):   f(x) = –2 [(x + 1/4)² – 1/16] + 5. Step 4. Distribute and Simplify Distribute the –2:   f(x) = –2 (x + 1/4)² + 2/16 + 5. Since 2/16 simplifies to 1/8, we have:   f(x) = –2 (x + 1/4)² + 1/8 + 5. Now combine the constant terms (writing 5 as 40/8):   1/8 + 40/8 = 41/8. Thus, the function becomes   f(x) = –2 (x + 1/4)² + 41/8. This is now in the form   a – b(x + c)²   with a = 41/8, b = 2, and c = 1/4. Step 5. Find the Maximum Value and the x‐Value at Which It Occurs In the vertex form   f(x) = –2 (x + 1/4)² + 41/8, the vertex (or turning point) occurs when the squared term is zero, which happens at:   x + 1/4 = 0 ⟹ x = –1/4. At x = –1/4, the value of f(x) is   f(–1/4) = 41/8. Since the coefficient of the squared term (–2) is negative, the parabola opens downward and this represents the maximum value of the function. Step 6. Sketching the Graph Key points for the graph: 1. Vertex (Turning Point): (–1/4, 41/8). 2. Axis of Symmetry: The graph is symmetric about the vertical line x = –1/4. 3. y-Intercept: Find by letting x = 0:   f(0) = 5 – 0 – 0 = 5.   So, the point is (0, 5). 4. x-Intercepts: Solve f(x) = 0.   We have: –2x² – x + 5 = 0.   Multiply both sides by –1: 2x² + x – 5 = 0.   Using the quadratic formula, where a = 2, b = 1, c = –5,    x = [–1 ± √(1² – 4·2·(–5))] / (2·2)    x = [–1 ± √(1 + 40)]/4    x = [–1 ± √41] / 4.   So the x-intercepts are at x = (–1 + √41)/4 and x = (–1 – √41)/4. 5. End Behavior: Since the coefficient of x² is –2, as x → ±∞, f(x) → –∞. A rough sketch would show a downward-opening parabola with a vertex at (–0.25, 5.125), crossing the y-axis at (0, 5), and intercepting the x-axis at the two values computed. Final Answer: - The function in the form a – b(x + c)² is   f(x) = –2 (x + 1/4)² + 41/8. - Its maximum value is 41/8, which occurs at x = –1/4.