What is the equation of the line tangent to # f(x)=(-x^2-1)/(x+4) # at # x=5 #?

1 Answer
Timber Lin
Jan 9, 2018

#y+26/9=-64/81(x-5)#

Explanation:

the slope of the tangent line at #x=5# is the derivative of the function evaluated at #x=5#
use quotient rule or #d/dx(f(x)/g(x))=(f'(x)g(x)-g'(x)f(x))/((g(x))^2)# to find the derivative:

#d/dx((-x^2-1)/(x+4))#
#=(-2x*(x+4)-1*(-x^2-1))/((x+4)^2)#

when #x=5#:

#=(-2(5)*(5+4)-1*(-5^2-1))/((5+4)^2)#

#=(-90+26)/(81)#
#=-64/81#

find the point #(5,f(5))# to complete the equation of the tangent line:
#f(5)=(-5^2-1)/(5+4)#
#f(5)=(-26)/(9)#

use point slope form to find the equation of the line:
#y-(-26/9)=-64/81(x-5)#
#y+26/9=-64/81(x-5)#

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