What is the equation of the line tangent to #f(x)=-x^2 -2x - 1 # at #x=7#?

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Answer 1 · 2016-11-18T17:08:11

Equation of tangent is #16x+y-48=0#

For finding equation of a tangent to a curve, defined by #f(x)# (here #f(x)=-x^2-2x-1#), at #x_0#,

we first identify the point at which tangent is desired, which is #(x_0,f(x_0))#

and then slope of the tangent at that point i.e. #f'(x_0)#.

Then the tangent using point slope form is

#(y-f(x_0))=f'(x_0)(x-x_0)#

Here we have to find tangent at #x=7#,

#f(7)=-7^2-2xx7-1=-49-14-1=-64# and

#f'(x)=-2x-2# and #f'(7)=-2xx7-2=-16#

Hence equation of tangent is

#(y-(-64))=-16(x-7)# or #y+64=-16x+112#

i.e. #16x+y-48=0#
graph{(y+x^2+2x+1)(16x+y-48)=0 [-76.4, 83.6, -59, 21]}