What is the the vertex of y = 2(x - 3)^2 ­- x+3?

Sep 20, 2016

Convert to standard form, which is $y = a {x}^{2} + b x + c , a \ne 0$.

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

$y = 2 \left({x}^{2} - 6 x + 9\right) - x + 3$

$y = 2 {x}^{2} - 12 x + 18 - x + 3$

$y = 2 {x}^{2} - 13 x + 21$

Now, to determine the vertex, convert to vertex form, which is $y = a {\left(x - p\right)}^{2} + q , a \ne 0$

$y = 2 {\left({x}^{2} - \frac{13}{2} x + m - m\right)}^{2} + 21$

The goal here is to convert to a perfect square. $m$ is given by ${\left(\frac{b}{2}\right)}^{2}$, where #b = (ax^2 + bx + ...) inside the parentheses.

$m = {\left(\frac{- \frac{13}{2}}{2}\right)}^{2} = \frac{169}{16}$

$y = 2 \left({x}^{2} - \frac{13}{2} x + \frac{169}{16} - \frac{169}{16}\right) + 21$

$y = 2 \left({x}^{2} - \frac{13}{2} x + \frac{169}{16}\right) - \frac{169}{8} + 21$

$y = 2 {\left(x - \frac{13}{4}\right)}^{2} - \frac{1}{8}$

In vertex form, $y = a {\left(x - p\right)}^{2} + q , a \ne 0$, the vertex is located at $\left(p , q\right)$. Hence, the vertex is at the coordinates $\left(\frac{13}{4} , - \frac{1}{8}\right)$.

Hopefully this helps!