first off, let's notice that the squared variable is the "x", that means is a vertically opening parabola.
notice the leading term's coefficient, is -2, is negative, that means the parabola is opening downwards, is facing down, so it goes up up up, reaches the U-turn and then down down down, is a "hump" or a maximum point.
[tex]\bf \textit{vertex of a vertical parabola, using coefficients}
\\\\
\begin{array}{lcccl}
y = & -2x^2& -8x& +10\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}
\qquad
\left(-\cfrac{ b}{2 a}\quad ,\quad c-\cfrac{ b^2}{4 a}\right)
\\\\\\
\left( -\cfrac{-8}{2(-2)}~~,~~10-\cfrac{(-8)^2}{4(-2)} \right)\implies \left( \cfrac{8}{-4}~~,~~10-\cfrac{64}{-8} \right)
\\\\\\
\left( -2~~,~~10+8 \right)\implies (-2~~,~~18)[/tex]
and the line of symmetry, well is a vertical parabola, mirroring itself at the line x = -2, which is the x-coordinate of the vertex.
