Question

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

Answer 1

`-1/264x+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'(7)`

Power rule:

`x^n=nx^(n-1)` if `n` is a constant.

Chain rule:

`d/dx[f(g(x))]=f'(g(x))*g'(x)`

`=>2*(5+4x)^(2-1)*d/dx[5+4x]`

`=>2(5+4x)^(1)*(5*0*x^(0-1)+4*1*x^(1-1))`

`=>2(5+4x)*(0+4)`

`=>8(5+4x)`

When `x=7...`

`=>f'(7)=8(5+4*7)`

`=>f'(7)=8(33)`

`=>f'(7)=264`

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

`=>f(7)=(5+4*7)^2`

`=>f(7)=(33)^2`

`=>f(7)=1089`

The normal line will have the slope of `-1/264` (Because it is perpendicular to the tangent line)

We use the slope-intercept form:

`m(x-x_1)=y-y_1`

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

`=>-1/264x+7/264+1089=y`

`=>-1/264x+287503/264=y`