Show that for a quadratic function y = ax2 + bx + c with a < 0, the function has a maximum value at x = -b/(2a) using the method of completing the square.

To show that if a < 0, the quadratic function y = ax^2 + bx + c has a maximum value when x = -(b)/(2a), we can use the method of completing the square. Step 1: Start with the given quadratic function. y = ax^2 + bx + c Step 2: Factor out a from the terms involving x. y = a(x^2 + (b)/(a)x) + c Step 3: Complete the square inside the parenthesis. To do this, add and subtract ((b)/(2a))^2. y = a(x^2 + (b)/(a)x + ((b)/(2a))^2 - ((b)/(2a))^2) + c y = a(x^2 + (b)/(a)x + (b^2)/(4a^2) - (b^2)/(4a^2)) + c Step 4: Group the perfect square trinomial and separate the constant term. y = a((x + (b)/(2a))^2 - (b^2)/(4a^2)) + c Step 5: Distribute a back into the terms inside the parenthesis. y = a(x + (b)/(2a))^2 - a((b^2)/(4a^2)) + c y = a(x + (b)/(2a))^2 - (b^2)/(4a) + c Step 6: Combine the constant terms to get the vertex form of the quadratic equation. y = a(x + (b)/(2a))^2 + (4ac - b^2)/(4a) This is the vertex form y = a(x-h)^2 + k, where the vertex is at (h, k). In this case, h = -(b)/(2a) and k = (4ac - b^2)/(4a). Step 7: Analyze the expression to determine if it's a maximum or minimum. The term (x + (b)/(2a))^2 is always greater than or equal to zero, i.e., (x + (b)/(2a))^2 0. The minimum value of this term is 0, which occurs when x + (b)/(2a) = 0, or x = -(b)/(2a). We are given that a < 0. Since a is negative and (x + (b)/(2a))^2 is non-negative, the product a(x + (b)/(2a))^2 will always be less than or equal to zero: a(x + (b)/(2a))^2 0 To maximize y, we need to make the term a(x + (b)/(2a))^2 as large as possible. Since it is always non-positive, its maximum value is 0. This occurs when (x + (b)/(2a))^2 = 0, which means x = -(b)/(2a). At x = -(b)/(2a), the function y reaches its maximum value: y_max = a(0) + (4ac - b^2)/(4a) = (4ac - b^2)/(4a) Therefore, if a < 0, the quadratic function y = ax^2 + bx + c has a maximum value when x = -(b)/(2a).