# How do you find the equation of the tangent and normal line to the curve y=x^2-x at x=1?

Nov 1, 2016

Tangent $y = x - 1$
Normal $y = - x + 1$

#### Explanation:

We have $y = {x}^{2} - x$

Differentiating wrt $x$ we get:
$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 x - 1$

When $x = 1 \implies y = 1 - 1 = 0$
And, $x = 1 \implies \frac{\mathrm{dy}}{\mathrm{dx}} = 2 - 1 = 1$

So the tangent passes through the point (1,0) and has gradient ${m}_{T} = 1$, Using $y - {y}_{1} = m \left(x - {x}_{1}\right)$, so the equation of the tangent is:

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

The normal is perpendicular to the tangent,so the product of their gradients is $- 1$, so the gradient of the normal is ${m}_{N} = - \frac{1}{1} = - 1$

So the normal passes through the point (1,0) and has gradient ${m}_{N} = - 1$, so the equation of the normal is:

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