# How do you find the axis of symmetry, and the maximum or minimum value of the function  y=-2x^2 +4x +2 ?

Mar 20, 2017

The axis of symmetry is $x = 1$
The maximum value is $V = \left(1 , 4\right)$

#### Explanation:

We need

${a}^{2} - 2 a b + {b}^{2} = {\left(a - b\right)}^{2}$

We complete the square and factorise

$y = - 2 {x}^{2} + 4 x + 2$

$y = - 2 \left({x}^{2} - 2 x\right) + 2$

$y = - 2 \left({x}^{2} - 2 x + 1\right) + 2 + 2$

$y = - 2 {\left(x - 1\right)}^{2} + 4$

The axis of symmetry is $x = 1$

The vertex is $V = \left(1 , 4\right)$

The y-intercept is when $x = 0$

$y = - 2 \cdot 1 + 4 = 2$

Therefore,

The maximum value is $V = \left(1 , 4\right)$

graph{(y+2x^2-4x-2)(y-1000(x-1))=0 [-11.25, 11.25, -5.625, 5.625]}