# How do you identify the critical points for (x-1)^2 = (y+1)^2?

Feb 27, 2017

There are critical points at $\left(1 , 0\right)$; $\left(3 , 1\right)$ and $\left(- 1 , 1\right)$.

#### Explanation:

Start by expanding.

${x}^{2} - 2 x + 1 = {y}^{2} + 2 y + 1$

${x}^{2} - {y}^{2} - 2 x - 2 y + 1 - 1 = 0$

${x}^{2} - {y}^{2} - 2 x - 2 y = 0$

Now differentiate.

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

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

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

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

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

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

Critical points will occur whenever the derivative equals $0$ or is undefined. It will be undefined when $y = 1$. When the derivative equals $0$, $x = 1$.

Let's solve for the corresponding x-and y values.

For $y = 1$

${\left(x - 1\right)}^{2} = {\left(1 + 1\right)}^{2}$

${\left(x - 1\right)}^{2} = 4$

$x - 1 = \pm 2$

$x = 3 \mathmr{and} - 1$

For $x = 1$

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

$0 = {y}^{2}$

$y = 0$

The critical points are therefore $\left(1 , 0\right)$, $\left(3 , 1\right)$ and $\left(- 1 , 1\right)$.

Hopefully this helps!