Difficulty turning a quadratic equation to "vertex"-form I'm having difficulty reducing a quadratic equation to its "vertex-form" by following my textbook and nearly every tutorial I can find online.
The starting equation is:
$$f(x) = -2x^2 + 16x - 24$$
Next, I divided each term by $-2$ (to leave the first term as $x^2$):
$$ = x^2 - 8x + 12$$
Next, to complete the square, I divide $x$'s coefficient by $2$, then square it ($(b/2)^2$):
$$\left(\frac{8}{2}\right)^2 = 16$$
Which gives me $16$. I add $16$ to complete the square, but also subtract it so it doesn't affect the value:
$$ = x^2 - 8x + 16 + 12 - 16$$
Finally, I complete the square:
$$ = (x - 4)^2 + 12 - 16$$
Then simplify:
$$= (x - 4)^2 - 4$$
According to by book though,the answer is:
$$-2(x - 4)^2 + 8$$
Unfortunately, the text skips over steps and makes it very unclear how they got the answer that they did. Can anyone point out where I went wrong above? I compiled the above steps from several videosand tutorials, but I must have gone wrong somewhere.
 A: You made your error when you divided by $-2$.  When you divided $f(x) = -2x^2 + 16x - 24$ by $-2$, you obtained $\color{red}{f(x) = x^2 - 8x + 12}$.  The expression in red is not equal to the original function since the $y$-intercept of the original function is $-24$, while the $y$-intercept of the function in red is $12$.   
You want to transform the equation $f(x) = -2x^2 + 16x - 24$ into the form $f(x) = a(x - h)^2 + k$.  We first extract a factor of $-2$ from the quadratic and linear terms so that the expression in parentheses has the form $x^2 + 2kx$.  We do this since we want to complete the square on $x^2 + kx$ to form the perfect square $x^2 + 2kx + k^2 = (x + k)^2$.
\begin{align*}
f(x) & = -2x^2 + 16x - 24\\
     & = -2(x^2 - 8x) - 24
\end{align*}
The term in parentheses now has the form $x^2 + 2kx$, where $2k = -8$, so $k = -4$, and $k^2 = 16$.  If we add $16$ inside the parentheses, we will obtain the perfect square $x^2 - 8x + 16 = (x - 4)^2$.  However, adding $16$ inside the parentheses adds $-2 \cdot 16 = -32$ to the original expression, we must add $32$ to the expression in order to balance the equation.
\begin{align*}
f(x) & = -2(x^2 - 8x) - 24\\
     & = -2(x^2 - 8x + 16) - 24 + 32\\
     & = -2(x - 4)^2 + 8
\end{align*}
which tells us that the graph of $f(x)$ has vertex $(4, 8)$ and opens downwards.  Note that the $y$-intercept is 
$$f(0) = -2(0 - 4)^2 + 8 = -2(16) + 8 = -32 + 8 = -24$$
as we would expect.
Check: We verify that the vertex form is equal to the original function. 
\begin{align*}
f(x) & = -2(x - 4)^2 + 8\\
     & = -2(x^2 - 8x + 16) + 8\\
     & = -2x^2 + 16x - 32 + 8\\
     & = -2x^2 + 16x - 24
\end{align*}
Completing the square refers to the process of adding $k^2$ to an expression of the form $x^2 + 2kx$ to form the perfect square $x^2 + 2kx + k^2$.  One purpose of completing the square is to transform a quadratic function into vertex form.  What confused you is that it can also be used to solve for the $x$-intercepts.  When we solve for the $x$-intercepts, we set $f(x) = 0$ and solve for $x$.  It is here that we divide by $-2$ (or, more generally, the leading coefficient).  Continuing from above yields
\begin{align*}
f(x) & = 0\\
-2(x - 4)^2 + 8 & = 0\\
(x - 4)^2 - 4 & = 0\\
(x - 4)^2 & = 4\\
x - 4 & = \pm 2\\
x & = 4 \pm 2
\end{align*}
\begin{align*}
x & = 4 + 2 & x & = 4 - 2\\
x & = 6 & x & = 2
\end{align*}
You can verify that these solutions are correct by substituting them into the equation $f(x) = -2x^2 + 16x - 24$.  
