What is the equation of the normal line of #f(x)=-4x-1# at #x=2#?

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Answer 1 · 2016-01-10T19:18:40

#y=x/4-19/2#

If the excercise does not clearly request the derivation use, you could also think:

#y=-4x-1#

is a line, then the tangent line to it is the line itself

#y=mx+q#

with

#m=-4#

and

#q=-1#

#m_n=-1/m=-1/(-4)=1/4#

The normal line is

#(y-y_0)=m_n(x-x_0)#

with
#x_0=2#
#y_0=f(x_0)=y(x_0)=-4*2-1=-8-1=-9#

then:

the normal line becomes:

#(y-(-9))=(1/4)(x-2)#

#y+9=1/4x-1/2#
#y=x/4-1/2-9=x/4-((1+18)/2)=x/4-19/2#