What is the equation of the line tangent to # f(x)=-(x+2)^2-5-3# at # x=5#?

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Answer 1 · 2016-03-18T18:27:52

#y=-12x+3#

#f(x)=-[(x+2)^2]-5-3=-[(x+2)^2]-8#

Applying the chain rule:

#f'(x)=-[2(x+2)^1xx1]#

#=-2x-2#

The tangent line is of the form:

#y=mx+c#

#:.m=-2x-2#

When #x=5# then #m=-(2xx5)-2=-12#

To get the #y# coordinate:

#f(5)=-[(5+2)^2]-8=-57#

#:.-57=-12xx5+c#

#c=60-57=3#

So the equation of the tangent line#rArr#

#y=-12x+3#

graph{(-12x+3-y)(-(x+2)^(2)-8-y)=0 [-285.2, 285.4, -142.6, 142.6]}