How do I use the vertex formula to determine the vertex of the graph of the function and write it in standard form for #2x^2+6x-1#?

1 Answer
Jun 2, 2015

#y = 2x^2+6x-1# is already in standard form

#color(white)("XXXXX")#General standard form for a parabola is #y=ax^2+bx+c#

Vertex form for a parabola is #y =m(x-a)+b#
#color(white)("XXXXX")#Note: the #a and b# are not the same as in the standard form but they are constants.

#y=2x^2+6x+1#
#color(white)("XXXXX")#extract #m#
#y= 2(x^2+3x) + 1#
#color(white)("XXXXX")#complete the square
#y=2(x^2+3x+(3/2)^2) +1 - 9/2#

#y = 2(x+3/2)^2 -7/2#

#y = 2(x-(-3))^2+(-7/2)#
#color(white)("XXXXX")#which is in vertex form with a vertex at #(-3,-7/2)#