Answer :
To determine the range of the function \(f(x) = 3x^2 + 6x - 8\), we need to analyze the behavior of the quadratic function. We will do this by finding the vertex of the parabola represented by this function.
### Step 1: Identify the coefficients
The given quadratic function is in the standard form \(ax^2 + bx + c\), where in this case:
- \(a = 3\)
- \(b = 6\)
- \(c = -8\)
### Step 2: Find the x-coordinate of the vertex
For a quadratic function \(ax^2 + bx + c\), the x-coordinate of the vertex \(h\) can be found using the formula:
[tex]\[ h = -\frac{b}{2a} \][/tex]
Plugging in the values from our function:
[tex]\[ h = -\frac{6}{2 \cdot 3} = -\frac{6}{6} = -1 \][/tex]
So, the x-coordinate of the vertex is \(x = -1\).
### Step 3: Find the y-coordinate of the vertex
To find the y-coordinate of the vertex, we substitute \(x = -1\) back into the original function \(f(x)\):
[tex]\[ f(-1) = 3(-1)^2 + 6(-1) - 8 \][/tex]
[tex]\[ f(-1) = 3(1) + 6(-1) - 8 \][/tex]
[tex]\[ f(-1) = 3 - 6 - 8 \][/tex]
[tex]\[ f(-1) = -11 \][/tex]
So, the y-coordinate of the vertex is \(y = -11\).
### Step 4: Determine the range
The parabola opens upwards because the coefficient of \(x^2\) (which is 3) is positive. This means the vertex represents the minimum value of the function.
Therefore, the minimum value of \(f(x)\) is \(-11\). Since the parabola opens upwards, the function values can be as small as \(-11\) and increase without bound. Thus, the range of \(f(x)\) is all \(y\) values such that \(y \geq -11\).
### Conclusion
The range of the function \(f(x) = 3x^2 + 6x - 8\) is:
[tex]\[ \{ y \mid y \geq -11 \} \][/tex]
So, the correct option is:
[tex]\[ \boxed{\{ y \mid y \geq -11 \}} \][/tex]
### Step 1: Identify the coefficients
The given quadratic function is in the standard form \(ax^2 + bx + c\), where in this case:
- \(a = 3\)
- \(b = 6\)
- \(c = -8\)
### Step 2: Find the x-coordinate of the vertex
For a quadratic function \(ax^2 + bx + c\), the x-coordinate of the vertex \(h\) can be found using the formula:
[tex]\[ h = -\frac{b}{2a} \][/tex]
Plugging in the values from our function:
[tex]\[ h = -\frac{6}{2 \cdot 3} = -\frac{6}{6} = -1 \][/tex]
So, the x-coordinate of the vertex is \(x = -1\).
### Step 3: Find the y-coordinate of the vertex
To find the y-coordinate of the vertex, we substitute \(x = -1\) back into the original function \(f(x)\):
[tex]\[ f(-1) = 3(-1)^2 + 6(-1) - 8 \][/tex]
[tex]\[ f(-1) = 3(1) + 6(-1) - 8 \][/tex]
[tex]\[ f(-1) = 3 - 6 - 8 \][/tex]
[tex]\[ f(-1) = -11 \][/tex]
So, the y-coordinate of the vertex is \(y = -11\).
### Step 4: Determine the range
The parabola opens upwards because the coefficient of \(x^2\) (which is 3) is positive. This means the vertex represents the minimum value of the function.
Therefore, the minimum value of \(f(x)\) is \(-11\). Since the parabola opens upwards, the function values can be as small as \(-11\) and increase without bound. Thus, the range of \(f(x)\) is all \(y\) values such that \(y \geq -11\).
### Conclusion
The range of the function \(f(x) = 3x^2 + 6x - 8\) is:
[tex]\[ \{ y \mid y \geq -11 \} \][/tex]
So, the correct option is:
[tex]\[ \boxed{\{ y \mid y \geq -11 \}} \][/tex]
