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

Answer 1

Emilia Aguilar

#:.x=-1#

The slope #m_n=-1/m_(x_0)#

#m_(x_0)=f'(x_0)#

and the generic line equation is:

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

with #y_0=f(x_0)#

#f'(x)=2(x+1)^(2-1)*1=2(x+1)#

#f'(x_0)=f'(-1)=2(-1+1)=0#

#:.x_0=-1# it's a critical point of #f(x)#

#m_n=-1/m_(x_0)=-1/0 =>cancel EE#

then in #x_0# the normal line has no slope and it's

#x=x_0#

#:.x=-1#

Alternatively:

#f(x)=(x+1)^2#

It's a parabola with two cincident real roots.

The roots are #x=-1=x_0#

Then the vertex of the parabola it's #V(-1,0)#. The vertex it's a local min (because if we rewrite #f(x)# as #y=ax^2+bx+c# we have #a>0# ), where #f'(x)=0#. The tangent line it's #y=0# therefore the normal it's #x=-1#

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Answer 2

Evelyn Agard

The equation of the line normal to f(x)=(x+1)^2 at x=-1 is y = -2x - 1.

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Answer from HIX Tutor

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