As you are using # f(x) # I am assuming you are at a higher level of Mathematics.
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(brown)("Using Calculus with shortcuts")#
#color(blue)("To find "x_("vertex")#
Given: #f(x)=-3x^2-6x-7#..............(1)
#f'(x)=-6x-6#
Equating to zero gives #color(blue)(x_("vertex")=-1)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("To find "y_("vertex")#
Substitute #x=-1# into equation (1) giving
#y_("vertex")= -3(-1)^2-6(-1)-7#
#color(blue)(y_("vertex")= -4)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Vertex#->(x,y)->(-1,-4)#
'~~~~~~~~~~~~~~~~~~~~~~~~
This is a maximum as #-3x^2# being negative is indicative of an inverted U shape. Also the 2nd differential is negative which is also indicative of a maximum
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(brown)("NOT using Calculus")#
#color(blue)("I am going to show you a really cool trick!")#
Write as:# y= (-3x^2-6x)-7#
Factor the -3 out
#y= -3(x^2+2x)-7#............................(1)
Now consider the #+2" from "2x# inside the bracket
Multiply this by #(-1/2)#
#(-1/2)xx(+2)=-1#
This is the value you are after for #x# so:
#color(blue)(x_("vertex")=-1)#
Substitute this x value into the original equation to find the value of #y_("vertex")#