To find the maximum and minimum values of the quadratic expression 3x^2 + 2x + 11, we can use the concept of vertex.
The quadratic expression is in the form ax^2 + bx + c, where a = 3, b = 2, and c = 11. The vertex of a quadratic function is given by the formula x = -b/2a.
Using this formula, we can calculate the x-coordinate of the vertex:
x = -2 / (2 * 3),
x = -2 / 6,
x = -1/3.
Now, to find the corresponding y-coordinate of the vertex, we substitute the x-coordinate back into the quadratic expression:
y = 3(-1/3)^2 + 2(-1/3) + 11,
y = 3(1/9) - 2/3 + 11,
y = 1/3 - 2/3 + 11,
y = 10 + 11,
y = 21.
Therefore, the vertex of the quadratic expression is (-1/3, 21).
Since the coefficient of x^2 (a) is positive, the parabola opens upward. This means that the vertex represents the minimum value of the quadratic expression.
Hence, the minimum value of the quadratic expression 3x^2 + 2x + 11 is y = 21 at x = -1/3.
Since the parabola opens upward and the coefficient of x^2 (a) is positive, there is no maximum value for the quadratic expression.