Answer:
Explanation:
Given:
[tex]f(x)=3x^2+6x-2[/tex]First, we do completing the square on the given function to express it into vertex form. So,
We write it in the form:
[tex]\begin{gathered} x^2+2ax+a^2 \\ \end{gathered}[/tex]And, factor out 3: So,
[tex]\begin{gathered} 3(x^2+2x-\frac{2}{3}) \\ \text{where:} \\ 2a=2\text{ or a=1} \\ \text{Hence} \\ 3(x^2-2x-\frac{2}{3}+1^2-1^2) \end{gathered}[/tex]Since:
[tex]\begin{gathered} x^2+2ax+a^2=(x+a)^2 \\ So, \\ x^2+2x+1^2=(x+1)^2 \end{gathered}[/tex]Then,
[tex]\begin{gathered} 3(x^2-2x-\frac{2}{3}+1^2-1^2) \\ =3((x+1)^2-\frac{2}{3}-1^2) \\ \text{Simplify} \\ f(x)=3(x+1)^2-2-3 \\ f(x)=3(x+1)^2-5 \end{gathered}[/tex]Therefore, the answer is:
[tex]f(x)=3(x+1)^2-5[/tex]