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

Apr 29, 2015

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

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

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

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

$\to y + 8 = 2 \left({x}^{2} + \left(\frac{3}{2}\right) x\right)$ (Made the coefficient of ${x}^{2}$ as 1

Now we add $2 \cdot {\left(\frac{3}{4}\right)}^{2}$ to each side to complete the square

$\to y + 8 + 2 \cdot {\left(\frac{3}{4}\right)}^{2} = 2 \left({x}^{2} + \left(\frac{3}{2}\right) x + {\left(\frac{3}{4}\right)}^{2}\right)$

$\to y + 8 + \frac{9}{8} = 2 {\left(x + \frac{3}{4}\right)}^{2}$

$\to y + \frac{73}{8} = 2 {\left\{x - \left(- \frac{3}{4}\right)\right\}}^{2}$

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

The vertex of the Parabola is$\left\{- \frac{3}{4} , - \frac{73}{8}\right\}$