What is the range of the function f(x) = 3x² + 6x + 8?

The function f(x) = 3x² + 6x + 8 is a quadratic function, which means it has the general form of f(x) = ax² + bx + c, where a, b, and c are constants. In this case, a = 3, b = 6, and c = 8.

To find the range of this quadratic function, we first note that the coefficient of the x² term (which is 3) is positive. This indicates that the parabola opens upwards. Therefore, the minimum value of the function will occur at the vertex of the parabola. The formula to find the x-coordinate of the vertex for a quadratic function is given by:

x = -b / (2a)

Substituting in our values:

x = -6 / (2 * 3) = -1

Now we need to find the corresponding y-coordinate by plugging this x value back into the function:

f(-1) = 3(-1)² + 6(-1) + 8

       = 3(1) - 6 + 8

       = 3 - 6 + 8

       = 5

Thus, the vertex of the parabola is at the point (-1, 5), which is the minimum point of the function.

Since the parabola opens upwards, the range of the function is all values of y greater than or equal to the minimum value found at the vertex. Therefore, we conclude that:

Range of f(x): [5, ∞)

In summary, the range of the function f(x) = 3x² + 6x + 8 is all real numbers starting from 5 and going to positive infinity.