\( The\ vertex\ form\ of\ y=ax^2+bx+c: y=a(x-h)^2+k\\\\where\ h=\frac{-b}{2a}\ and\ k=f(h)=\frac{-(b^2-4ac)}{4a}\\-\\\\y=-2x^2-12x-10\\\\a=-2;\ b=-12;\ c=-10\\\\h=\frac{-(-12)}{2(-2)}=\frac{12}{-4}=-3\\\\k=f(-3)=-2(-3)^2-12(-3)-10=-2\cdot9+36-10=-18+26=8\\\\therefore: y=-2(x-(-3))^2+8\to\boxed{y=-2(x+3)^2+8}\leftarrow The\ vertex\ form \)

\( y-intercept\to f(0)=-2(0^2)-12(0)-10=\boxed{-10}\leftarrow y-intercept \)

\( x-intercept\ if\ y=0\\\\-2x^2-12x-10=0\ \ \ \ \ |divide\ both\ sides\ by\ (-2)\\x^2+6x+5=0\\x^2+x+5x+5=0\\x(x+1)+5(x+1)=0\\(x+1)(x+5)=0\iff x+1=0\ or\ x+5=0\\\\\boxed{x=-1\ or\ x=-5}\leftarrow x-intercept \)

Y=-2x^2-12x-10

to find the x value of the vertex find the axis of symmetry which is

-b/2a

=12/2(-2)

=-3

to find the y value of the vertex sub the axis of symmetry back into the original equation.

y=-2(-3)^2-12(-3)-10

y=8**vertex=(-3,8)**

to put it into vertex form use y=a(x-h)^2 +k where h and k are the x and y values of the vertex respectively.

y=-2(x-3)^2 +8**y=-2(x+3)^2+8**

to find the x intercept let y=0

-2x^2-12x-10=0

x^2+6x+5=0

use quad formula and you will get **x=-5 and -1**

to find the y intercept let x=0

y= -2(0)^2-12(0)-10**y=-10**