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

Question

Vegljamzt6 · Accepted Answer

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 chain rule

f^{'} (x) = 2 (1 + 2 x) \frac{d}{d x} (1 + 2 x) = 2 (1 + 2 x) \cdot 2 = 4 (1 + 2 x)

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}

is gradient of normal
then

m \cdot m_{1} = - 1

\Rightarrow 12 \times m_{1} = - 1 \to m_{1} = - \frac{1}{12}

f (1) = {(1 + 2)}^{2} = 9 \Rightarrow (a, b) = (1, 9)

equation of normal:

y - 9 = - \frac{1}{12} (x - 1)

multiplying by 12 to eliminate fraction gives
12y-108=-x+1

\Rightarrow 12 y + x - 109 = 0

What is the equation of the normal line

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