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

Apr 15, 2018

$- \frac{1}{264} x + \frac{287503}{264} = y$

#### Explanation:

A normal line is a line that is perpendicular to the tangent line of a graph at the point of tangency.

Let's find $f ' \left(7\right)$

${x}^{n} = n {x}^{n - 1}$ if $n$ is a constant.

$\frac{d}{\mathrm{dx}} \left[f \left(g \left(x\right)\right)\right] = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$

$\implies 2 \cdot {\left(5 + 4 x\right)}^{2 - 1} \cdot \frac{d}{\mathrm{dx}} \left[5 + 4 x\right]$

$\implies 2 {\left(5 + 4 x\right)}^{1} \cdot \left(5 \cdot 0 \cdot {x}^{0 - 1} + 4 \cdot 1 \cdot {x}^{1 - 1}\right)$

$\implies 2 \left(5 + 4 x\right) \cdot \left(0 + 4\right)$

$\implies 8 \left(5 + 4 x\right)$

When $x = 7. . .$

$\implies f ' \left(7\right) = 8 \left(5 + 4 \cdot 7\right)$

$\implies f ' \left(7\right) = 8 \left(33\right)$

$\implies f ' \left(7\right) = 264$

With this in mind, let's find the point of tangency.

$\implies f \left(7\right) = {\left(5 + 4 \cdot 7\right)}^{2}$

$\implies f \left(7\right) = {\left(33\right)}^{2}$

$\implies f \left(7\right) = 1089$

The normal line will have the slope of $- \frac{1}{264}$ (Because it is perpendicular to the tangent line)

We use the slope-intercept form:

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

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

$\implies - \frac{1}{264} x + \frac{7}{264} + 1089 = y$

$\implies - \frac{1}{264} x + \frac{287503}{264} = y$