How do you write $y= -2x^2+4x+5$ into vertex form?

Question

18980 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-04-30T05:45:48

The vertex form of a quadratic function is given by
$y = a(x - h)^2 + k$ , where $(h, k)$ is the vertex of the parabola.

We can use the process of Completing the Square to get this into the Vertex Form.

$y=-2x^2+4x+5$

$-> y - 5 = -2x^2 + 4x$ (Transposed 5 to the Left Hand Side)

$-> y - 5 = -2(x^2 - 2x)$ (Made the coefficient of $x^2$ as 1)

Now we subtract $2$ from each side to complete the square

$-> y - 5 - 2 = -2(x^2 - 2x ) - 2$

$-> y - 5 - 2 = -2(x^2 - 2x + 1)$

$-> y - 5 - 2 = -2(x^2 - 2x + 1^2)$

$-> y - 7 = -2(x-1)^2$

$-> color(green)( y = -2(x-1)^2 + 7$ is the Vertex Form

The vertex of the Parabola is ${1 , 7}$

How do you write y= -2x^2+4x+5y= -2x^2+4x+5 into vertex form?