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

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Answer 1 · 2018-04-15T21:35:02

#-1/264x+287503/264=y#

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)#

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

#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#