#color(blue)("Determine the vertex")#
This is the Vertex Form of a quadratic equation so you can virtually directly read off the coordinates of the vertex.
#y=-3(xcolor(red)(-1))^2color(blue)(-1)#
#x_("vertex")=(-1)xx color(red)((-1)) = +1#
#y_("vertex")=color(blue)(-1)#
Vertex#->(x,y)=(1,-1)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine the y intercept")#
Set #x=0# giving
#y_("intercept")=-3(0-1)^2-1" "=" "-4#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine the x intercept")#
Set #y=0# giving
#0=-3(x-1)^2-1" " larr # add 1 to both sides
#=>1=-3(x-1)^2" "larr #multiply both sides by (-1)
#=>-1=+3(x-1)^2" "larr #divide both sides by 3
#=>-1/3=(x-1)^2" "larr # square root both sides
#=>+-sqrt(-1/3)=x-1" "larr # add 1 to both sides
#x=1+-sqrt(-1/3)" " rarr x in CC#
As the determinant is negative the curve does not cross the x-axis nor is the axis a tangent to the curve.
Thus the roots are in the number range of 'Complex Numbers'
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)(" The question lists intercepts as 'singular' thus the x-axis roots are not required")#
