Question

What are the axis of symmetry and vertex of the quadratic function y=-2x^2-8x+5​

Answer 1

$\bf \textit{vertex of a vertical parabola, using coefficients} \\\\ y=\stackrel{\stackrel{a}{\downarrow }}{-2}x^2\stackrel{\stackrel{b}{\downarrow }}{-8}x\stackrel{\stackrel{c}{\downarrow }}{+5} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left( -\cfrac{(-8)}{2(-2)}~~,~~5-\cfrac{(-8)^2}{4(-2)} \right)\implies \left( \cfrac{8}{-4}~~,~~5-\cfrac{64}{-8} \right)\implies (-2~~,~~5+8) \\\\\\ (-2~~,~~13)~\hfill \stackrel{\textit{axis of symmetry}}{x=-2}$

recall that the axis of symmetry for a vertical parabola is simply the equation for the vertical line of the vertex's x-coordinate.