From this form of the parabola equation, i.e. #ax^2+bx+c#, the #x#-coordinate of the vertex is:
#x=(-b)/(2a)#
Here, #a=1# and #b=-12#, and #c# is #-2# therefore:
#x=(-(-12))/(2(1))=12/2=6#
Now we plug that into the equation to find #y#:
#y=6^2-12(6)-2=36-72-2=36-74=-38#
Vertex: #(6,-38)#
To find the #x#-intercepts, you have to set #y# (the equation) equal to #0# and solve for #x#
#x^2-12x-2=0#
We cannot factor this. Let's use the quadratic formula:
#x=(-b+-sqrt(b^2-4ac))/(2a)#
#x=(-(-12)+-sqrt((-12)^2-4(1)(-2)))/(2(1)#
#x=(12+-sqrt(144+8))/2=(12+-sqrt152)/2=(12+-sqrt(2^2*38))/2=(12+-2sqrt38)/2=(2(6+-sqrt38))/2=(cancel2(6+-sqrt38))/cancel2#
#x=6+-sqrt38#
Therefore, the #x#-intercepts are:
#(6+sqrt38, 0)# and #(6-sqrt38, 0)#
To find the #y#-intercept, you have to plug in #0# for #x#:
#y=0^2-12(0)-2=-2#
Therefore, the #y#-intercept is:
#(0, -2)#
