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 in vertex - form

For a quadratic function $y = ax^{2}+bx + c$, the vertex - form is $y=a(x - h)^{2}+k$. Given $f(x)=3x^{2}+6x - 8$, where $a = 3$, $b = 6$, $c=-8$. First, factor out the coefficient of $x^{2}$ from the first two terms: $f(x)=3(x^{2}+2x)-8$. Complete the square inside the parentheses. Since for $x^{2}+2x$, half of the coefficient of $x$ is 1 and $(x + 1)^{2}=x^{2}+2x + 1$, so $x^{2}+2x=(x + 1)^{2}-1$. Then $f(x)=3((x + 1)^{2}-1)-8$.

Step2: Expand and simplify

Expand $3((x + 1)^{2}-1)-8$: $f(x)=3(x + 1)^{2}-3-8=3(x + 1)^{2}-11$.

Step3: Determine the range

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

Answer:

{y|y≥ - 11}