#color(brown)("Note that the question does not state the method to be used.")#
#color(blue)("A quick trick to find the axis of symmetry.")#
#y=1/2(x^2-2x)-1#
Using the -2 from #-2x# we have:
#color(blue)(x_("vertex")=(-1/2)xx(-2) = +1) color(red)(larr"axis of symmetry " x=1)#
The above process is part of completing the square.
#color(blue)(y_("vertex") = 1/2(1)^2-(1)-1 = - 1 1/2->-3/2)#
#color(blue)(=>"Vertex "->(x,y)=(1,-3/2))#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
To find the x-intercepts
Using the formula
#y=ax^2+bx+c" "=>" "x=(-b+-sqrt(b^2-4ac))/(2a)#
#a=1/2; b=-1; c=-1#
#=>x=(-1)+-sqrt((-1)^2-4(1/2)(-1))/(2(1/2))#
#x=-1+-sqrt(3)#
#x=0.73205..." and "x=-2.73205...#
