# What is the vertex form of y= (2x+7)(3x-1) ?

Apr 17, 2017

$y = 6 {\left(x - - \frac{19}{12}\right)}^{2} - \frac{529}{24}$

#### Explanation:

Given: $y = \left(2 x + 7\right) \left(3 x - 1\right) \text{ [1]}$

The vertex form of a parabola of this type is:

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

We know that the "a" in the vertex form is the same as the coefficient $a {x}^{2}$ in standard form. Please observe the product of the first terms of the binomials:

$2 x \cdot 3 x = 6 {x}^{2}$

Therefore, $a = 6$. Substitute 6 for "a" into equation [2]:

$y = 6 {\left(x - h\right)}^{2} + k \text{ [3]}$

Evaluate equation [1] at $x = 0$:

$y = \left(2 \left(0\right) + 7\right) \left(3 \left(0\right) - 1\right)$

$y = 7 \left(- 1\right)$

$y = - 7$

Evaluate equation [3] at $x = 0 \mathmr{and} y = - 7$:

$- 7 = 6 {\left(0 - h\right)}^{2} + k$

$- 7 = 6 {h}^{2} + k \text{ [4]}$

Evaluate equation [1] at $x = 1$:

$y = \left(2 \left(1\right) + 7\right) \left(3 \left(1\right) - 1\right)$

$y = \left(9\right) \left(2\right)$

$y = 18$

Evaluate equation [3] at $x = 1$ and $y = 18$:

$18 = 6 {\left(1 - h\right)}^{2} + k$

$18 = 6 \left(1 - 2 h + {h}^{2}\right) + k$

$18 = 6 - 12 h + 6 {h}^{2} + k \text{ [5]}$

Subtract equation [4] from equation [5]:

$25 = 6 - 12 h$

$19 = - 12 h$

$h = - \frac{19}{12}$

Use equation [4] to find the value of k:

$- 7 = 6 {h}^{2} + k$

$k = - 6 {h}^{2} - 7$

$k = - 6 {\left(- \frac{19}{12}\right)}^{2} - 7$

$k = - \frac{529}{24}$

Substitute these values into equation [3]:

$y = 6 {\left(x - - \frac{19}{12}\right)}^{2} - \frac{529}{24}$