To determine the range of the quadratic function [tex]\( f(x) = 3x^2 + 6x - 8 \)[/tex], we need to analyze its properties and find its minimum or maximum value, as it will guide us in understanding the possible values that [tex]\( f(x) \)[/tex] can take.
### Step-by-Step Solution:
1. Identify the general form of the quadratic function:
The given function, [tex]\( f(x) = 3x^2 + 6x - 8 \)[/tex], is a quadratic function of the form [tex]\( ax^2 + bx + c \)[/tex].
2. Determine the orientation of the parabola:
For a quadratic function in the form [tex]\( ax^2 + bx + c \)[/tex]:
- If [tex]\( a > 0 \)[/tex], the parabola opens upwards.
- If [tex]\( a < 0 \)[/tex], the parabola opens downwards.
Here, [tex]\( a = 3 \)[/tex], which is positive, so the parabola opens upwards. This means the function has a minimum value, which occurs at its vertex.
3. Find the x-coordinate of the vertex:
The x-coordinate of the vertex of a parabola given by [tex]\( ax^2 + bx + c \)[/tex] is found using the formula:
[tex]\[
x = -\frac{b}{2a}
\][/tex]
Substituting [tex]\( a = 3 \)[/tex] and [tex]\( b = 6 \)[/tex] into the formula:
[tex]\[
x = -\frac{6}{2 \times 3} = -\frac{6}{6} = -1
\][/tex]
4. Find the y-coordinate of the vertex:
Substitute [tex]\( x = -1 \)[/tex] back into the function to find [tex]\( f(-1) \)[/tex]:
[tex]\[
f(-1) = 3(-1)^2 + 6(-1) - 8
\][/tex]
Simplifying this:
[tex]\[
f(-1) = 3(1) + 6(-1) - 8 = 3 - 6 - 8 = 3 - 14 = -11
\][/tex]
So, the vertex of the parabola is at [tex]\( (-1, -11) \)[/tex].
5. Determine the range:
Since the parabola opens upwards and the minimum value occurs at [tex]\( y = -11 \)[/tex], the range of the function includes all values greater than or equal to this minimum value.
Therefore, the range of the function [tex]\( f(x) \)[/tex] is:
[tex]\[
\{y \mid y \geq -11\}
\][/tex]
### Conclusion:
The correct answer is:
[tex]\[
\{y \mid y \geq -11\}
\][/tex]
