# What is the equation of the normal line of f(x)= (x^2 + 2x - 1)/(4 - 3x)  at x=-1?

Apr 5, 2017

$y - \frac{49}{6} x - \frac{331}{42} = 0$ or

$42 y - 343 x - 331 = 0$

#### Explanation:

Before beginning, you should know that derivative of a function of

the form $g \frac{x}{f} \left(x\right)$ is given by $\frac{f \left(x\right) \cdot g ' \left(x\right) - g \left(x\right) \cdot f ' \left(x\right)}{f \left(x\right)} ^ 2$

$i . e .$ if $a \left(x\right) = g \frac{x}{f} \left(x\right)$ then,

$\textcolor{red}{\frac{d}{\mathrm{dx}} a \left(x\right) = \frac{f \left(x\right) \cdot \frac{d}{\mathrm{dx}} g \left(x\right) - g \left(x\right) \cdot \frac{d}{\mathrm{dx}} f \left(x\right)}{f \left(x\right)} ^ 2}$

Coming back to the question,

At $x = - 1$; $f \left(x\right) = \frac{1 - 2 - 1}{4 + 3} = - \frac{2}{7}$

$\therefore$ we have to find equation of normal at $\left(- 1 , - \frac{2}{7}\right)$

slope ${m}_{1}$ of $f \left(x\right)$ at any general point $\left(x , y\right)$ is given by $f ' \left(x\right)$

$i . e . {m}_{1} = f ' \left(x\right)$

$= \frac{\left(4 - 3 x\right) \cdot \frac{d}{\mathrm{dx}} \left({x}^{2} + 2 x - 1\right) - \left({x}^{2} + 2 x - 1\right) \cdot \frac{d}{\mathrm{dx}} \left(4 - 3 x\right)}{4 - 3 x} ^ 2$

$= \frac{\left(4 - 3 x\right) \left(2 x + 2\right) - \left({x}^{2} + 2 x - 1\right) \left(- 3\right)}{4 - 3 x} ^ 2$

$\therefore$ slope of $f \left(x\right)$ at $x = - 1$ is given by m_1]_(x=-1)

m_1]_(x=-1)=((4+3)(-cancel(2)+cancel(2))-(1-2-1)(-3))/(4+3)^2

implies m_1]_(x=-1)= (3*(-2))/49 = -6/49

Let the slope of normal at the given point be ${m}_{2.}$

When two curves are normal/perpendicular to each other at some point then the product of their slopes at that point equals $- 1$.

$\therefore$ at $\left(- 1 , - \frac{2}{7}\right)$; ${m}_{1} \cdot {m}_{2} = - 1 \implies {m}_{2} = - \frac{1}{m} _ 1 = \frac{49}{6}$

Hence equation of normal at given point can be written in slope point form as

$y - \left(- \frac{2}{7}\right) = {m}_{2} \left(x - \left(- 1\right)\right)$

$y + \frac{2}{7} = \frac{49}{6} \cdot \left(x + 1\right) = \frac{49}{6} x + \frac{49}{6}$

$y - \frac{49}{6} x - \frac{331}{42} = 0$ is the required equation of normal.