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

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Feb 21, 2018

Hence, the vertex is
I have approached by the method of calculus (maxima and minima)
$V \equiv \left(x , y\right) = V \equiv \left(- \frac{1}{4} , - \frac{34}{16}\right)$

#### Explanation:

I have approached by the method of calculus (maxima and minima)
The curve is symmetrical about an axis parallel to y axis.
The vertex is the point where $\frac{\mathrm{dy}}{\mathrm{dx}} = 0$

Given:

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

Differentiating wrt x

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 2 x - 2 - 3 \times 2 \left(\frac{x}{3} - \frac{2}{3}\right) \times \frac{1}{3}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$- 2 x - 2 - 3 \times 2 \left(\frac{x}{3} - \frac{2}{3}\right) \times \frac{1}{3} = 0$

$- 2 x - 2 - \frac{2}{3} x + \frac{4}{3} = 0$

$- 2 x - \frac{2}{3} x = 2 - \frac{4}{3}$
$- \frac{6}{3} x - \frac{2}{3} x = \frac{6}{3} - \frac{4}{3}$

$- 6 x - 2 x = 6 - 4$

$- 8 x = 2$

$\frac{8}{8} x = - \frac{2}{8}$

$x = - \frac{1}{4}$

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

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

$= - \frac{1}{16} + \frac{1}{2} - 3 {\left(- \frac{1}{12} - \frac{2}{3}\right)}^{2}$

$= - \frac{1}{16} + \frac{8}{16} - 3 {\left(- \frac{1}{12} - \frac{8}{12}\right)}^{2}$

$= \frac{- 1 + 8}{16} - 3 {\left(\frac{- 1 - 8}{12}\right)}^{2}$

$= \frac{7}{16} - 3 {\left(- \frac{9}{12}\right)}^{2}$

$= \frac{7}{16} - 3 {\left(- \frac{3}{4}\right)}^{2}$

$= \frac{7}{16} - 3 \times \frac{9}{16}$

$\frac{7}{16} - \frac{27}{16}$

$y = - \frac{34}{16}$

Hence, the vertex is

$V \equiv \left(x , y\right) = V \equiv \left(- \frac{1}{4} , - \frac{34}{16}\right)$

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