Question #2cf05

2 Answers
Oct 12, 2017

#y=2(x-(3/2))^2-(3/2)#
#vertex (3/2,-(3/2))#

Explanation:

#y=2x^2-6x+3#
#y-3=2x^2-3x#
#y-3=2(x^2-(3/2)x)#
#y-3=2(x^2-(3/2)x++9/4)-(9/2)#
#y=2(x-(3/2))^2-(9/2)+3#
#y=2(x-(3/2))^2-(3/2)#

#h=3/2, k=-(3/2)#

Oct 12, 2017

The vertex of #y=2x^2-6x+3# is #y=2(x-3/2)^2+3/2#

Explanation:

To solve this problem, we need to convert this quadratic equation from standard form to vertex form.

Let's start with the original equation:

#y=2x^2-6x+3#

Now let's turn the coefficient of the #x^2# term into #1#. This way, we're setting the stage for the process of completing the square.

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

Now we can start completing the square:

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

Then we simplify the left side and write #x^2-3x+(-3/2)^2# as the square of a binomial (a two-term polynomial):

#1/2y+9/4=(x-3/2)^2+3/2#

We then subtract #9/4# from both sides and simplify:

#1/2y=(x-3/2)^2+3/2-9/4=(x-3/2)^2+6/4-9/4=(x-3/2)^2-3/4#

All we need to do now is multiply by #2# to isolate #y# and obtain our vertex form:

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

And so our vertex form is:

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