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

Apr 10, 2017

Thus the vertex$\to \left(x , y\right) = \left(\frac{5}{4} , \frac{15}{8}\right)$

#### Explanation:

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We need to include the $x$ that is outside the brackets

Expanding the brackets we have:
$y = 2 {\left(x - 1\right)}^{2} \text{ } \textcolor{w h i t e}{.} + 3 + x$
$y = 2 {x}^{2} - 4 x + 2 + 3 - x$

$y = 2 {x}^{2} - 5 x + 5$

As the question presents a part vertex form equation it is reasonable to assume that the questioner's intention is for you to continue using vertex form format.

$y = 2 \left({x}^{2} - \frac{5}{2} x\right) + 5 + k$

Where $k$ is a correction constant

$y = 2 {\left(x - \frac{5}{4}\right)}^{2} + 5 + k$

Set $\text{ "2(-5/4)^2+k=0" "=>" } k = - \frac{25}{8}$ giving:

$y = 2 {\left(x - \frac{5}{4}\right)}^{2} + 5 - \frac{25}{8}$

$y = 2 {\left(x - \frac{5}{4}\right)}^{2} + \frac{15}{8}$

Thus the vertex$\to \left(x , y\right) = \left(\frac{5}{4} , \frac{15}{8}\right)$