Question

If the vertex of the curve $y=-2x^2-4ax-k$ is $(-2,7),$ then the value of $k$ is

Answer 1

Comparing
$y=-2x^2-4ax-k$ with $y=p{x}^2+qx+r$
We get
$p=-2,q=-4a,r=-k$
So, the coordinates of the vertex is
$\therefore (-\frac{q}{2p},-\frac{D}{4p})\equiv (-2,7)\Rightarrow -\frac{(-4a)}{-4}=-2,-\frac{(16a^2-8k)}{-8}=7\Rightarrow a=2\Rightarrow -\frac{(16a^2-8k)}{-8}=7\Rightarrow 16\times 4-8k=56\therefore k=1$