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

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Answer 1 · 2018-06-01T19:02:08

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

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

We begin by isolating the $y$ $y$ 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.

How do you find the maximum or minimum of 2x^2-y=3x-2y2x2−y=3x−2y2x^2-y=3x-2y?