To find the range of the function f(x) = 3x² + 6x + 8, we first need to analyze its shape and vertex, as it is a quadratic function.
The general form of a quadratic function is f(x) = ax² + bx + c. For our function, a = 3, b = 6, and c = 8. Since a is positive (3), the parabola opens upwards.
Next, we can find the vertex of the parabola, which will give us the minimum value of the function. The x-coordinate of the vertex can be calculated using the formula:
x = -b / (2a)
Plugging in the values of a and b:
x = -6 / (2 * 3) = -6 / 6 = -1
Now that we have the x-coordinate of the vertex, we substitute this value back into the function to find the corresponding y-coordinate:
f(-1) = 3(-1)² + 6(-1) + 8
Calculating this:
f(-1) = 3(1) – 6 + 8 = 3 – 6 + 8 = 5
The vertex of the parabola is at (-1, 5), which means the minimum value of the function is 5.
Since the parabola opens upwards, the function will reach its minimum at this point and can increase indefinitely. Therefore, the range of the function f(x) = 3x² + 6x + 8 is:
Range: [5, ∞)