The #x#-coordinate of the vertex of the parabola #ax^2+bx+c# is:
#x=-b/(2a)#
In our case #a=3# and #b=-12#, which means the #x#-coordinate of the vertex is #-(-12)/(2(3))=2#. The #y#-coordinate can be found by evaluating #3x^2-12x+10# at #x=2#:
#y=3x^2-12x+10=3(2)^2-12(2)+10=-2#
Thus the vertex is #(2,-2)#.
Finding the intercepts will require use of the quadratic formula, which is:
#x=(-b+-sqrt(b^2-4ac))/(2a)#
We have #a=3#, #b=-12#, and #c=10#:
#x=(-(-12)+-sqrt((-12)^2-4(3)(10)))/(2(3))#
#color(white)(XX)=(12+-sqrt(144-120))/6#
#color(white)(XX)=2+-sqrt(24)/6#
#color(white)(XX)=2+-(2sqrt(6))/6#
#color(white)(XX)=2+-sqrt(6)/3#
The intercepts are #(2+sqrt(6)/3,0)# and #(2-sqrt(6)/3,0)#.
