Write the quadratic function in the form $g(x)= a(x-h)^2 + k$ Write the quadratic function in the form $g(x)= a(x-h)^2 + k$ 
Then, give the vertex of its graph.
$g(x)= 2x^2-16x+35$
~
I tried, but still lost:
$(2x^2-16x)+35$
$2(x^2-4^2-16)+35$
$2(x^2-4^2-32+35$
 A: $2x^2-16x+35$
$=2(x^2-8x)+35$
$=2(x^2-8x+16)-2\cdot 16+35$
$=2(x-4)^2+3$
There you go. Completing the square.
The vertex is easy from this. It's just x=4 since squares are always nonnegative so to make the square part zero you let x=4. (to be more specific the vertex is (4, 3))
A: I'm not too sure what you did, one of your brackets is missing, but I believe that your error lies in $2(x^2-4^2-16)+35$.
When you complete the square for $x^2+bx$, you take half of $b$ and then square that. However, starting with $x^2+16x$, you subtracted $16$, and then "turned" the $8x$ into a $4^2$. Not possible!
Instead, we're supposed to have$$\begin{align*}2x^2-16x+35 & =2(x^2-8x)+35\\ & =2\left\{x^2-8x+\left(\dfrac {8}2\right)^2-\left(\dfrac 82\right)^2\right\}+35\\ & =2\left(x^2-8x+16\right)+35-32\\ & =2(x-4)^2+3\end{align*}$$
Thus, the vertex of $g(x)$ is $(4,3)$.
A: You have few mistakes.
It should be 
$2(x^2 + 4^2 - 8x) + 35 - 2 \cdot 4^2$
= $2(x - 4)^2 + 35 - 32$
= $2(x - 4)^2 + 3$
A: It may be prudent to write out all of the steps without any shortcuts.
$$\begin{align}
g(x)&=2x^2 - 16x + 35\\
g(x)&=2(x^2 - 8x) + 35\\
g(x)&=2(x^2 - 8x) + 35\\
g(x)&=2(x^2 - 4x -4x) + 35\\
g(x)&=2(x^2 - 4x -4x + 0) + 35\\
g(x)&=2(x^2 - 4x -4x + 16 - 16) + 35\\
g(x)&=2(x^2 - 4x -4x + 16) -32 + 35\\
g(x)&=2(x(x - 4) -4(x - 4)) +3\\
g(x)&=2(x - 4)(x - 4) +3\\
g(x)&=2(x - 4)^2 +3\\
\end{align}$$
You know that the vertex will occur when $x-h = 0$. Maybe draw a table of values, or draw a graph, or both to determine the coordinates of the vertex.
