# What is the equation of the line that is normal to f(x)= ln(x^2-x+1)/x at  x= 1 ?

Oct 18, 2016

$y = 1 - x$

#### Explanation:

Let $y = f \left(x\right) = \ln \frac{{x}^{2} - x + 1}{x}$

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

Find the y coordinate at x = 1:

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

The slope, m, of the tangent line is $\frac{\mathrm{dy}}{\mathrm{dx}}$ evaluated at $x = 1$:

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

$m = 1$

The slope, n, of the normal line is:

$n = - \frac{1}{m}$

$n = - \frac{1}{1}$

$n = - 1$

The normal line has a slope of -1 and passes through the point (1, 0). Using the point-slope form of the equation of a line, we obtain the following equation:

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

$y = 1 - x$