# What is the equation of the line normal to  f(x)=(x-1)(x+2)  at  x=0?

Jan 14, 2018

$y = x - 2$

#### Explanation:

$f \left(x\right) = \left(x - 1\right) \left(x + 2\right)$

Firstly, just expand the brackets so:

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

Differentiate:

$f ' \left(x\right) = 2 x + 1$

Find $f ' \left(0\right)$

$f ' \left(0\right) = 2 \left(0\right) + 1 = 1$

So we now the gradient of the line is is $1$.

We need a point on the line and we know it should pass through the point: $\left(0 , f \left(0\right)\right)$. So find this point:

$f \left(0\right) = {\left(0\right)}^{2} + \left(0\right) - 2 = - 2$

So $m = 1$ and $\left(a , b\right) = \left(0 , - 2\right)$.

Substitute into the equation of a line:

$y - b = m \left(x - a\right)$

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

$\to y = x - 2$