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

Feb 6, 2016

12y + x - 109 = 0

#### Explanation:

To obtain equation of normal , y-b = m(x-a) . Require to find it's gradient m and a point on it (a , b )

differentiating f(x) will give gradient of tangent and evaluating f(x) for x = 1 will give a point on it.

differentiate using the $\textcolor{b l u e}{\text{ chain rule }}$

f'(x) = 2(1 + 2x)$\frac{d}{\mathrm{dx}} \left(1 + 2 x\right) = 2 \left(1 + 2 x\right) .2 = 4 \left(1 + 2 x\right)$

and f'(1) 4(1+2) = 12 = m of tangent

now tangent and normal are perpendicular and so the product of their gradients will equal -1

If ${m}_{1} \textcolor{b l a c k}{\text{ is gradient of normal}}$

then $m . {m}_{1} = - 1$

 rArr 12 xx m_1 = -1 → m_1 = -1/12

f(1) = ${\left(1 + 2\right)}^{2} = 9 \Rightarrow \textcolor{b l a c k}{\text{ (a , b ) = ( 1 , 9 ) }}$

equation of normal: y- 9 =$- \frac{1}{12} \left(x - 1\right)$
multiplying by 12 to eliminate fraction ) gives

12y - 108 = - x + 1

$\Rightarrow 12 y + x - 109 = 0$