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

Mar 26, 2016

Equation of normal at $x = - 1$ is $27 x + 6 y + 29 = 0$

#### Explanation:

As $f \left(x\right) = {x}^{2} / \left(1 + 4 x\right)$, at $x = - 1$, we have $f \left(- 1\right) = {\left(- 1\right)}^{2} / \left(1 + 4 \left(- 1\right)\right) = \frac{1}{-} 3 = - \frac{1}{3}$

And hence normal passes through $\left(- 1 , - \frac{1}{3}\right)$

Now as $f \left(x\right) = {x}^{2} / \left(1 + 4 x\right)$,

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\left(1 + 4 x\right) \times 2 x - 4 \times {x}^{2}}{1 + 4 x} ^ 2 = \frac{\left(2 x + 8 {x}^{2}\right) - 4 {x}^{2}}{1 + 4 x} ^ 2$

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{2 x \left(1 + 2 x\right)}{1 + 4 x} ^ 2$

and hence slope of curve i.e. tangent at $x = - 1$ is

$\frac{2 \left(- 1\right) \left(1 + 2 \left(- 1\right)\right)}{1 + 4 \left(- 1\right)} ^ 2$ or $\frac{\left(- 2\right) \times \left(- 1\right)}{- 3} ^ 2 = \frac{2}{9}$

and slope of normal would be $- \frac{1}{\frac{2}{9}} = - \frac{9}{2}$

Hence, equation of normal at $x = - 1$ is given by $\left(y + \frac{1}{3}\right) = - \frac{9}{2} \times \left(x + 1\right)$

or $6 \left(y + \frac{1}{3}\right) = - 9 \times \frac{6}{2} \times \left(x + 1\right)$ or $6 y + 2 = - 27 x - 27$ or

$27 x + 6 y + 29 = 0$