What is the equation of the line tangent to #f(x)=(x+6)^2-7x # at #x=-1#?

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Answer 1 · 2016-05-04T23:16:14

Tangent Line #y=3x+35#

Start from the given function

#f(x)=(x+6)^2-7x# at #x=-1#

find the ordinate first using x=-1

#f(-1)=(-1+6)^2-7(-1)#

#f(-1)=32#

Let #(x_1, y_1)=(-1, 32)#

slope for the slope #m# by obtaining the first derivative of #f(x)#

#f(x)=(x+6)^2-7x#
#f' (x) =2(x+6)^(2-1)-7#

slope #m=f' (-1)=2(-1+6)-7=10-7=3#

#m=3#

Solve for the tangent line

#y-y_1=m(x-x_1#

#y-32=3(x- -1)#

#y=3x+3+32#

#y=3x+35" "#the tangent line

kindly see the graph of #f(x)=(x+6)^2-7x# and the tangent line #y=3x+35#

desmos

God bless....I hope the explanation is useful.