Question

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

Answer 1

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.