What is the vertex form of y=(2x-9)(3x-1)?

Sep 9, 2017

$y = 6 {\left(x - \frac{29}{24}\right)}^{2} - \frac{625}{24}$

Explanation:

Given -

$y = \left(2 x - 9\right) \left(3 x - 1\right)$

Vertex form of the equation is -

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

Find the vertex first -

$y = 6 {x}^{2} - 27 x - 2 x + 9$
$y = 6 {x}^{2} - 29 x + 9$

x coordinate of the vertex
$x = \frac{- b}{2 a}$

$x = \frac{- \left(- 29\right)}{2 \times 6} = \frac{29}{12}$

y-coordinate of the vertex

$y = 6 {\left(\frac{29}{12}\right)}^{2} - 29 \left(\frac{29}{12}\right) + 9$
$y = 6 \left(\frac{841}{144}\right) - \frac{841}{12} + 9$
$y = \frac{5046}{144} - \frac{841}{12} + 9$
$y = \frac{5046 - 10092 + 1296}{144} = - \frac{3750}{144} = - \frac{625}{24}$
Vertex $\left(\frac{29}{12} , - \frac{625}{24}\right)$

$a = 6$ [coefficient of ${x}^{2}$]
$h = \frac{29}{12}$ [x coordinae of the vertex]
$k = - \frac{625}{24}$

The vertex form of the parabola equation is -

$y = 6 {\left(x - \frac{29}{24}\right)}^{2} + \left(\frac{- 625}{24}\right)$
$y = 6 {\left(x - \frac{29}{24}\right)}^{2} - \frac{625}{24}$