By completing the square, find the greatest value of−7 + 12x − 3x2.

To find the maximum value of the quadratic function -3x^2 + 12x - 7, we first need to rewrite it in vertex form, which is f(x) = a(x-h)^2 + k, where (h, k) is the vertex of the parabola.

Step 1: Rewrite the function in the form -3(x^2 - 4x) - 7.

Step 2: Complete the square inside the parentheses. To do this, take half of the coefficient of x, square it, and add it inside the parentheses. Half of -4 is -2, and (-2)^2 = 4. So we add and subtract 4 inside the parentheses to get -3(x^2 - 4x + 4 - 4) - 7.

Step 3: Simplify to get -3[(x - 2)^2 - 4] - 7.

Step 4: Distribute the -3 to get -3(x - 2)^2 + 12 - 7.

Step 5: Simplify to get -3(x - 2)^2 + 5.

Now the function is in vertex form, and we can see that the vertex is (2, 5). Since the coefficient of (x - 2)^2 is negative, the parabola opens downwards, and the maximum value is the y-coordinate of the vertex, which is 5.

To find the maximum value of the quadratic function -3x^2 + 12x - 7, we first need to rewrite it in vertex form, which is f(x) = a(x-h)^2 + k, where (h, k) is the vertex of the parabola.

Step 1: Rewrite the function in the form -3(x^2 - 4x) - 7.

Step 2: To complete the square, we need to add and subtract (4/2)^2 = 4 inside the parentheses: -3[(x - 2)^2 - 4] - 7.

Step 3: Simplify to get the function in vertex form: -3(x - 2)^2 + 5.

So, the vertex of the parabola is (2, 5), and since the coefficient of x^2 is negative, the parabola opens downwards, meaning that the maximum value of the function is the y-coordinate of the vertex, which is 5.