Question

what is the range of the function \\(f(x) = 3x^2 + 6x - 8\\)?

\\(\bigcirc\\) \\(\\{y | y \ge -1\\}\\)
\\(\bigcirc\\) \\(\\{y | y \le -1\\}\\)
\\(\bigcirc\\) \\(\\{y | y \ge -11\\}\\)
\\(\bigcirc\\) \\(\\{y | y \le -11\\}\\)

Explanation:

Identify the function and direction of opening

The given quadratic function is:

$$f(x) = 3x^2 + 6x - 8$$

Since the leading coefficient \(a = 3\) is positive (\(3 > 0\)), the parabola opens upward. This means the function has a minimum value at its vertex, and its range will be all real numbers greater than or equal to the \(y\)-coordinate of the vertex.

Find the x-coordinate of the vertex

Using the Quadratic Vertex knowledge point

$$ x = -\frac{b}{2a} = -\frac{6}{2(3)} = -1 $$

Find the y-coordinate of the vertex

Using the Quadratic Vertex knowledge point

$$ y = f(-1) = 3(-1)^2 + 6(-1) - 8 = 3 - 6 - 8 = -11 $$

Determine the range of the function

Since the parabola opens upward and has its minimum value at the vertex \((-1, -11)\), the range of the function is all values of \(y\) that are greater than or equal to \(-11\).

$$\text{Range} = \{y \mid y \ge -11\}$$

Answer:

• \(\{y \mid y \ge -1\}\)
• \(\{y \mid y \le -1\}\)
• \(\{y \mid y \ge -11\}\) (Correct answer)
• \(\{y \mid y \le -11\}\)