How do you find the maximum or minimum of #2x^2-y=3x-2y#?

1 Answer
Jun 1, 2018

The vertex is #(3/4,3/4)# a maximum point.

Explanation:

To find the maximum or minimum value for the equation
#2x^2-y=3x-2y#

We begin by isolating the #y# value using the additive inverse.

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

#2x^2+y=3x#

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

Now we have the equation

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

in the proper for of #y=ax^2+bx+c#

#a=-2#
#b=3#
#c=0#

Since the #a# value is negative we know that this parabola will open downward and the vertex point will be a maximum.

To solve for the vertex we use the formula #-b/(2a) = x# of the vertex

#x = -3/(2(-2))#
#x = 3/4#

Now plug the #x# value into the equation to solve for #y#

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

#y = -6/4 + 9/4#

#y = 3/4#

The vertex is #(3/4,3/4)# a maximum point.