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

1 Answer
Nov 14, 2015

#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))#.