Question

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

Answer 1

`y-1089=-1/264(x-7)`

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 `(5+4x)^2`. By the chain rule, we know:
`f'(x)=2(5+4x)*d/(dx)[5+4x]`
`f'(x)=2(5+4x)*4`
`color(blue)(f'(x)=8(5+4x)`

We can now find the slope of the tangent line at `x=7`.
`f'(7)=8(5+4(7))=color(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(x-x_1)`

We will use the point `(7, 1089)`, 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 `(5+4x)^2`, just rewrite it as `25+40x+16x^2`, the derivative of which is `40+32x=color(blue)(8(5+4x))`.