For this problem we have 3 options.
1) We can use the completing the square the method to put the equations in Standard Vertex Form, #(x-h)^2-k=0 -> vertex (h,-k)#
2) We can use the expression #(-b/(2a),f(-b/(2a)))# where #a# and #b# are the coefficients found when the equation is in standard form, #ax^2+bx+c=0#
3) We could just graph it.
Let do options 1 & 2.
Completing the Square:
#x^2-2x-1 =0#
#x^2-2x=1#
#(-2/2)^2=(-1)^2=1#, We will add #1# to both sides of the equation.
#x^2-2x+1=1+1#
#x^2-2x+1=2#
#(x-1)^2=2#
#(x-1)^2-2=0 -> vertex -> (1,-2)# Solution
#(x-h)^2-k=0 -> vertex -> (h,-k)#
Using the expression #(-b/(2a),f(-b/(2a)))#:
#a=1, b=-2, c=-1#
#x=-b/(2a)=-(-2)/(2(1))=2/(2(1))=2/2=1#
#y=f(-b/(2a))=f(1)=(1)^2-2(1)-1=1-2-1=-2#
#vertex -> (1,-2)# Solution
Notice that both methods produced the same results.
