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

Jun 1, 2018

The vertex is $\left(\frac{3}{4} , \frac{3}{4}\right)$ a maximum point.

#### Explanation:

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

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

$2 {x}^{2} - y + 2 y = 3 x \cancel{- 2 y} \cancel{+} 2 y$

$2 {x}^{2} + y = 3 x$

$\cancel{2 {x}^{2}} + y \cancel{- 2 {x}^{2}} = 3 x - 2 {x}^{2}$

Now we have the equation

$y = - 2 {x}^{2} + 3 x$

in the proper for of $y = a {x}^{2} + b x + c$

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

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 $- \frac{b}{2 a} = x$ of the vertex

$x = - \frac{3}{2 \left(- 2\right)}$
$x = \frac{3}{4}$

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

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

$y = - \frac{6}{4} + \frac{9}{4}$

$y = \frac{3}{4}$

The vertex is $\left(\frac{3}{4} , \frac{3}{4}\right)$ a maximum point.