Question

what is the range of the function f(x) = 3x² + 6x − 8?
○ {y|y ≥ −1}
○ {y|y ≤ −1}
○ {y|y ≥ −11}
○ {y|y ≤ −11}

Explanation:

Step1: Analyze the parabola's direction

The function \( f(x) = 3x^2 + 6x - 8 \) is a quadratic function. The coefficient of \( x^2 \) is \( 3 \), which is positive. So, the parabola opens upwards. This means the vertex is the minimum point of the parabola, and the range will be all real numbers greater than or equal to the \( y \)-coordinate of the vertex.

Step2: Find the vertex of the parabola

For a quadratic function in the form \( f(x) = ax^2 + bx + c \), the \( x \)-coordinate of the vertex is given by \( x = -\frac{b}{2a} \). Here, \( a = 3 \) and \( b = 6 \). So,

$$ x = -\frac{6}{2 \times 3} = -\frac{6}{6} = -1 $$

Now, substitute \( x = -1 \) into the function to find the \( y \)-coordinate of the vertex:

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

So, the vertex of the parabola is at \( (-1, -11) \).

Step3: Determine the range

Since the parabola opens upwards (because \( a = 3 > 0 \)) and the vertex is at \( y = -11 \), the function's minimum value is \( -11 \), and it increases without bound as \( x \) moves away from the vertex. Therefore, the range of the function is all real numbers \( y \) such that \( y \geq -11 \).

Answer:

\(\{y|y \geq -11\}\) (the third option)