# How do you identify all the critical points for  x^2 – x + y^2 + y = 0?

Nov 2, 2016

Please see the explanation for a description of how one does it.

#### Explanation:

Use implicit differentiation, to find the first derivative, $\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)$:

$2 x - 1 + 2 y \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) + \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

Separate $\frac{\mathrm{dy}}{\mathrm{dx}}$ from everything else

$2 x - 1 + \left(2 y + 1\right) \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 0$

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 - 2 x}{2 y + 1}$

The critical points occur when the first derivative is equal to zero:

$0 = \frac{1 - 2 x}{2 y + 1}$

$0 = \left(1 - 2 x\right)$

$2 x = 1$

$x = \frac{1}{2}$

Find the two values of y where this occurs:

${y}^{2} + y - \frac{1}{2} + \frac{1}{4} = 0$

${y}^{2} + y - \frac{1}{4} = 0$

check the discriminant:

${b}^{2} - 4 \left(a\right) \left(c\right) = 1 - 4 \left(1\right) \left(- \frac{1}{4}\right) = 2$

$y = - \frac{1}{2} + \frac{\sqrt{2}}{2} \mathmr{and} y = - \frac{1}{2} - \frac{\sqrt{2}}{2}$

The critical points are $\left(\frac{1}{2} , - \frac{1}{2} + \frac{\sqrt{2}}{2}\right)$ and $\left(\frac{1}{2} , - \frac{1}{2} - \frac{\sqrt{2}}{2}\right)$