How do you write the quadratic in vertex form given $y=2x^2+3x-8$ ?

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Answer 1 · 2015-04-29T07:33:20

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.

$y=2x^2+3x-8$

$-> y + 8 = 2x^2 +3x$ (Transposed -8 to the Left Hand Side)

$-> y + 8 = 2(x^2 + (3/2)x)$ (Made the coefficient of $x^2$ as 1#

Now we add $2*(3/4)^2$ to each side to complete the square

$-> y + 8 + 2*(3/4)^2 = 2(x^2 + (3/2)x + (3/ 4)^2)$

$-> y + 8 + 9/8 = 2(x+3/4)^2$

$-> y + 73/8 = 2{x - (-3/4)}^2$

$-> color(green)(y = 2{x - (-3/4)}^2 + (-73/8)$ is the Vertex Form

The vertex of the Parabola is ${-3/4 , -73/8}$

How do you write the quadratic in vertex form given y=2x^2+3x-8y=2x^2+3x-8?