To determine the range of the function [tex]\( f(x) = 3x^2 + 6x - 8 \)[/tex], we will analyze the properties of this quadratic function:
1. Identify the parabola's direction:
- Since the coefficient of [tex]\( x^2 \)[/tex] (which is 3) is positive, the parabola opens upwards.
2. Find the vertex of the parabola:
- The vertex form of a parabola [tex]\( ax^2 + bx + c \)[/tex] can help us find the minimum or maximum point. For the quadratic equation [tex]\( f(x) = ax^2 + bx + c \)[/tex], the x-coordinate of the vertex is given by [tex]\( x = -\frac{b}{2a} \)[/tex].
- Here, [tex]\( a = 3 \)[/tex] and [tex]\( b = 6 \)[/tex], so:
[tex]\[
x = -\frac{6}{2 \cdot 3} = -\frac{6}{6} = -1
\][/tex]
3. Calculate the y-coordinate of the vertex by plugging [tex]\( x = -1 \)[/tex] back into the function:
[tex]\[
f(-1) = 3(-1)^2 + 6(-1) - 8 = 3(1) - 6 - 8 = 3 - 6 - 8 = -11
\][/tex]
Thus, the vertex of the parabola is [tex]\((-1, -11)\)[/tex].
4. Determine the range of the function:
- Since the parabola opens upwards, the vertex represents the minimum point on the graph.
- The y-coordinate of the vertex is [tex]\( -11 \)[/tex], which means the function can take any value greater than or equal to [tex]\( -11 \)[/tex].
Therefore, the range of the function [tex]\( f(x) = 3x^2 + 6x - 8 \)[/tex] is:
[tex]\[
\{y \mid y \geq -11\}
\][/tex]
So, the correct answer is:
[tex]\[
\boxed{\{y \mid y \geq -11\}}
\][/tex]
