# What is the vertex form of y=x^2 - 19x +14 ?

The vertex form is ${\left(x - k\right)}^{2} = 4 p \left(y - k\right)$
${\left(x - \frac{19}{2}\right)}^{2} = y - - \frac{305}{4}$
with vertex at $\left(h , k\right) = \left(\frac{19}{2} , \frac{- 305}{4}\right)$

#### Explanation:

Start from the given equation $y = {x}^{2} - 19 x + 14$

Divide 19 by 2 then square the result to obtain $\frac{361}{4}$. Add and subtract $\frac{361}{4}$ to the right side of the equation right after $- 19 x$

$y = {x}^{2} - 19 x + 14$

$y = {x}^{2} - 19 x + \frac{361}{4} - \frac{361}{4} + 14$

the first three terms form a PERFECT SQUARE TRINOMIAL

$y = \left({x}^{2} - 19 x + \frac{361}{4}\right) - \frac{361}{4} + 14$

$y = {\left(x - \frac{19}{2}\right)}^{2} - \frac{361}{4} + 14$

$y = {\left(x - \frac{19}{2}\right)}^{2} - \frac{361}{4} + \frac{56}{4}$

$y = {\left(x - \frac{19}{2}\right)}^{2} - \frac{305}{4}$

$y - - \frac{305}{4} = {\left(x - \frac{19}{2}\right)}^{2}$

${\left(x - \frac{19}{2}\right)}^{2} = y - - \frac{305}{4}$

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