If the standard form of an equation is #y=1/2x^2-2x+5#, what is the vertex form?

1 Answer
Nov 17, 2016

#y=1/2(x-2)^2+3#

Explanation:

To convert from the standard form to the vertex form, you must use completing the square, and in order to use it, the value of #a# must be "#1#". In this case

#a=1/2,# #b=-2,# #c=5#

The value of #a# here is #1/2#, so we must factor out #1/2# from the expression as follows

#y=1/2x^2-2x+5 -> y=1/2(x^2-4x+10)#

The values of #a,b,c# in the expression #x^2-4x+10# are

#a=1,# #b=-4,# #c=10#

The expression #x^2-4x+10# is not a perfect square trinomial. We can make it a perfect square by finding the value of #c# which is equal to #(b/2)^2#

#color(red)(c)=(b/2)^2= ((-4)/2)^2=(-2)^2=4#

Next, we add #color(red)(c)# to the expression #x^2-4x+10#, but to keep the equation balanced, we must subtract #color(red)(c)# also, but without cancelling

#x^2-4x+10=x^2-4xcolor(red)(+4)+10color(blue)(-4)=x^2-4x+4+6#

So

#y=1/2(x^2-4x+4+6)#

Bring #6# outside from the brackets, and in order to do that you must multiply it by #1/2#. When you do that you will get

#y=1/2(x^2-4x+4)+3#

Finally, now you can factor #x^2-4x+4# into #(x-2)^2# and you get

#y=1/2(x-2)^2+3# Which is the vertex form.