Respuesta :
well, let's start off by doing some grouping, what we'll be doing is so-called "completing the square" as in a perfect square trinomial, since that's what the vertex form of a quadratic uses.
[tex]\bf f(x) = (x^2+6x)+14\implies f(x) = (x^2+6x+\boxed{?}^2)+14[/tex]
well, darn, we have a missing number for our perfect trinomial, however let's recall that in a perfect square trinomial the middle term is really the product of 2 times the term on the left and the term on the right without the exponent, so then we know that
[tex]\bf 2(x)\boxed{?} = 6x\implies \boxed{?}=\cfrac{6x}{2x}\implies \boxed{?}=3[/tex]
well then, that's our mystery guy, now, let's recall all we're doing is borrowing from our very good friend Mr Zero, 0, so if we add 3², we also have to subtract 3².
[tex]\bf ~~~~~~\textit{vertical parabola vertex form} \\\\ y=a(x- h)^2+ k\qquad \begin{cases} \stackrel{vertex}{(h,k)}\\\\ \stackrel{"a"~is~negative}{op ens~\cap}\qquad \stackrel{"a"~is~positive}{op ens~\cup} \end{cases} \\\\[-0.35em] ~\dotfill\\\\ f(x) = (x^2+6x)+14\implies f(x) = (x^2+6x+3^2-3^2)+14 \\\\\\ f(x) = (x^2+6x+3^2)+14-3^2\implies f(x) = (x+3)^2+5~\hfill \stackrel{vertex}{(-3,5)}[/tex]
