The vertex is pretty easy. We just need to complete the square .
First, the leading coefficient of the polynomial must be #1#, so we need to factor the #-1#. That leaves us with #y=-1(x^2-8x-2)#. Now, the purpose of completing the square is to find a constant that makes #x^2-8x# a perfect square. To that, we use this formula: #c=(1/2*b)^2# or #(1/2*8)^2#, which is #16#.
Now we know that we have to add #16# to make it a pefect square, but because we cannot just add something on one side of the equation, we need to "get rid of it" too. We could add #16# on both sides, or we can just add #16# and then subtract it immediately, which is the same thing. Either way works :)
#y=-1(x^2-8xcolor(green)(+16-16)-2)#
#y=-((x^2-8x+16)-16--2)#
#x^2-8x+16# is a perfect square, so let's symplify it
#y=-((x-4)^2-16-2)#
#y=-((x-4)^2-18)#
Now we just distribute the negative:
#y=-(x-4)^2+18#
The equation is now in vertex form.
It's easy to find the vertex from this point:
#y=-(x-color(red)(4))^2+color(purple)(18)#
#(color(red)(4), color(purple)(18))#.
Finding the #x#-interecpt means setting #y=0# and solving for #x#:
#0=-(x-4)^2+18#
#-18=-(x-4)^2#
#18=(x-4)^2#
#+-sqrt(18)=x-4#
#4+-sqrt(18)=x#
or #x=4+-3sqrt2#
Those are the exact values. If you want the estimated values, they're #x~~8.243# and #x~~-.0.243#
To find the #y#-intercept we just set #x=0# and solve for #y#:
#y=-(0-4)^2+18#
#y=-(-4)^2+18#
#y=-16+18#
#y=2#
