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

Explanation:

Step1: Rewrite the function in vertex - form

The general form of a quadratic function is $y = ax^{2}+bx + c$. For $f(x)=3x^{2}+6x - 8$, where $a = 3$, $b = 6$, and $c=-8$. We use the completing - the - square method. First, factor out the coefficient of $x^{2}$ from the first two terms: $y=3(x^{2}+2x)-8$. Then complete the square inside the parentheses. $x^{2}+2x=(x + 1)^{2}-1$. So $y=3((x + 1)^{2}-1)-8$. $y=3(x + 1)^{2}-3-8=3(x + 1)^{2}-11$.

Step2: Determine the range

Since $a = 3>0$, the parabola opens upward. The vertex of the parabola $y = a(x - h)^{2}+k$ is $(h,k)$. Here, the vertex is $(-1,-11)$. The minimum value of the function occurs at the vertex. So the range of the function is all $y$ values such that $y\geq - 11$.

Answer:

${y|y\geq - 11}$