# What is the vertex form of #y=3x^2 - 4x +17?

The Vertex Form

$\textcolor{red}{{\left(x - \frac{2}{3}\right)}^{2} = \frac{1}{3} \left(y - \frac{47}{3}\right)}$

#### Explanation:

Start from the given

$y = 3 {x}^{2} - 4 x + 17$

from the numerical coefficients
$a = 3$ and $b = - 4$ and $c = 17$

The vertex can be computed

$h = \frac{- b}{2 a}$
$h = \frac{- \left(- 4\right)}{2 \cdot 3}$

$h = \frac{2}{3}$

for the $k$
$k = c - {b}^{2} / \left(4 a\right)$

$k = 17 - {\left(- 4\right)}^{2} / \left(4 \cdot 3\right)$

$k = \frac{47}{3}$

$p = \frac{1}{4 a} = \frac{1}{4 \cdot 3} = \frac{1}{12}$

The vertex form

${\left(x - h\right)}^{2} = + 4 p \left(y - k\right)$

${\left(x - \frac{3}{2}\right)}^{2} = 4 \left(\frac{1}{12}\right) \left(y - \frac{47}{3}\right)$

${\left(x - \frac{3}{2}\right)}^{2} = \frac{1}{3} \left(y - \frac{47}{3}\right)$

By completing the square method

$y = 3 {x}^{2} - 4 x + 17$
$y = 3 \left({x}^{2} - \frac{4}{3} x\right) + 17$

$y = 3 \left({x}^{2} - \frac{4}{3} x + \frac{4}{9} - \frac{4}{9}\right) + 17$

$y = 3 \left(\left({x}^{2} - \frac{4}{3} x + \frac{4}{9}\right) - \frac{4}{9}\right) + 17$

$y = 3 \left({\left(x - \frac{2}{3}\right)}^{2} - \frac{4}{9}\right) + 17$

$y = 3 {\left(x - \frac{2}{3}\right)}^{2} - \frac{4}{3} + 17$

$y = 3 {\left(x - \frac{2}{3}\right)}^{2} + \frac{47}{3}$

$y - \frac{47}{3} = 3 {\left(x - \frac{2}{3}\right)}^{2}$

$\frac{1}{3} \left(y - \frac{47}{3}\right) = {\left(x - \frac{2}{3}\right)}^{2}$

The Vertex Form

$\textcolor{red}{{\left(x - \frac{2}{3}\right)}^{2} = \frac{1}{3} \left(y - \frac{47}{3}\right)}$

God bless....I hope the explanation is useful.