Question

transform the quadratic function y=-2x^2-12x-10 into vertex form. find the vertex, x and y intercepts.

Answer 1

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

Answer 2

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


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


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