We start with:
#y = 2x^2 -12x#.
In order to find the #y#-intercept, we must substitute #x# by 0, as the #y#-axis is #x = 0#. So,
#y = 2(0)^2 - 12(0)#,
#y = 0 - 0#,
#y = 0#.
In order to find the #x#-intercept, we must substitute #y# by 0, as the #x#-axis is #y = 0#. So,
#0 = 2x^2 - 12x#,
#2(x^2 - 6x) = 0#,
#x^2 - 6x = 0#, according to #x = (-b +- sqrt(b^2 -4ac))/(2a)#,
#x = (- (-6) +- sqrt((-6)^2 -4(1)(0))) / (2(1))#,
#x = (6 +- sqrt(36))/2#,
#x = (6+6)/2#,
#x = 12/2#,
#x = 6#,
#x = (6-6)/2#,
#x = 0/2#,
#x = 0#.
Hope it Helps! :D .
