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

Jan 17, 2016

$\therefore x = - 1$

#### Explanation:

The slope ${m}_{n} = - \frac{1}{m} _ \left({x}_{0}\right)$

${m}_{{x}_{0}} = f ' \left({x}_{0}\right)$

and the generic line equation is:

$\left(y - {y}_{0}\right) = {m}_{n} \left(x - {x}_{0}\right)$

with ${y}_{0} = f \left({x}_{0}\right)$

• We have to find $f ' \left(x\right)$ using the Chain Rule

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

$f ' \left({x}_{0}\right) = f ' \left(- 1\right) = 2 \left(- 1 + 1\right) = 0$

$\therefore {x}_{0} = - 1$ it's a critical point of $f \left(x\right)$

${m}_{n} = - \frac{1}{m} _ \left({x}_{0}\right) = - \frac{1}{0} \implies \cancel{\exists}$

then in ${x}_{0}$ the normal line has no slope and it's

$x = {x}_{0}$

$\therefore x = - 1$

Alternatively:

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

It's a parabola with two cincident real roots.

The roots are $x = - 1 = {x}_{0}$

Then the vertex of the parabola it's $V \left(- 1 , 0\right)$. The vertex it's a local min (because if we rewrite $f \left(x\right)$ as $y = a {x}^{2} + b x + c$ we have $a > 0$ ), where $f ' \left(x\right) = 0$. The tangent line it's $y = 0$ therefore the normal it's $x = - 1$