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: Rewrite the function in vertex form

The function is \( f(x) = 3x^2 + 6x - 8 \). First, factor out the coefficient of \( x^2 \) from the first two terms: \( f(x) = 3(x^2 + 2x) - 8 \). Then, complete the square inside the parentheses. To complete the square for \( x^2 + 2x \), we take half of the coefficient of \( x \) (which is \( 2/2 = 1 \)), square it (which is \( 1^2 = 1 \)), and add and subtract it inside the parentheses: \( f(x) = 3(x^2 + 2x + 1 - 1) - 8 \). This can be rewritten as \( f(x) = 3((x + 1)^2 - 1) - 8 \).

Step2: Simplify the vertex form

Distribute the 3: \( f(x) = 3(x + 1)^2 - 3 - 8 \). Then combine like terms: \( f(x) = 3(x + 1)^2 - 11 \).

Step3: Determine the range

Since the coefficient of \( (x + 1)^2 \) is 3 (which is positive), the parabola opens upwards. The vertex of the parabola is at \( (-1, -11) \), and the minimum value of the function is \( -11 \). Therefore, the range of the function is all real numbers greater than or equal to \( -11 \), which is \( \{y | y \geq -11\} \).

Answer:

\(\{y | y \geq -11\}\) (corresponding to the option "O \(\{y | y \geq -11\}\)")