# What is the equation of the normal line of f(x)=(5+4x)^2 at x=7?

Nov 14, 2015

$y - 1089 = - \frac{1}{264} \left(x - 7\right)$

#### Explanation:

The normal line to a tangent is perpendicular to the tangent at a point. Therefore, we must first find the slope of the tangent using the derivative.

We can use the chain rule to differentiate ${\left(5 + 4 x\right)}^{2}$. By the chain rule, we know:
$f ' \left(x\right) = 2 \left(5 + 4 x\right) \cdot \frac{d}{\mathrm{dx}} \left[5 + 4 x\right]$
$f ' \left(x\right) = 2 \left(5 + 4 x\right) \cdot 4$
color(blue)(f'(x)=8(5+4x)

We can now find the slope of the tangent line at $x = 7$.
$f ' \left(7\right) = 8 \left(5 + 4 \left(7\right)\right) = \textcolor{red}{264}$

Now, since we want the normal line, which is perpendicular to the tangent line, we want the opposite reciprocal slope: color(indigo)(-1/264

We can use point-slope form to quickly find the equation of the normal line: $y - {y}_{1} = m \left(x - {x}_{1}\right)$

We will use the point $\left(7 , 1089\right)$, which can be determined by plugging $7$ into the original equation. Thus, the equation of the normal line is color(green)(y-1089=-1/264(x-7).

Aside: if you haven't yet learned the chain rule, or want another way to differentiate ${\left(5 + 4 x\right)}^{2}$, just rewrite it as $25 + 40 x + 16 {x}^{2}$, the derivative of which is $40 + 32 x = \textcolor{b l u e}{8 \left(5 + 4 x\right)}$.