Final answer:
The minimum value of the function y = 2x^2 - 12x + 1 on the interval [0, 7] is -11.
Step-by-step explanation:
The function y = 2x^2 - 12x + 1 is a quadratic function, which means it forms a parabola when graphed. To find the minimum and maximum values for this function on the interval [0, 7], we need to determine the vertex of the parabola.
The vertex of a quadratic function with the equation y = ax^2 + bx + c is given by the coordinates (-b/2a, f(-b/2a)), where f(x) is the value of the function at x.
In the given function, a = 2, b = -12, and c = 1. So, the x-coordinate of the vertex is -(-12)/(2*2) = 3. Substituting this value into the function, we get y = 2(3)^2 - 12(3) + 1 = -11.
Therefore, the vertex of the parabola and the minimum value of the function on the interval [0, 7] is (3, -11).