What is the equation of the normal line of #f(x)=(x-3)/(x+4)# at #x=-1#?

1 Answer
May 27, 2018

#21y+27x+55=0#

Explanation:

#f(x)=(x-3)/(x+4)#

Recall: the quotient rule is given by #(dy)/(dx)=(vu'-uv')/v^2# if #y=u/v#

#f'(x)=((x+4)(1)-(x-3)(1))/(x+4)^2#

#f'(x)=(x+4-x+3)/(x+4)^2#

#f'(x)= 7/(x+4)^2#

When #x=-1#

#f'(-1)=7/(-1+4)^2#

#f'(-1)=7/9#

You just found the gradient of the tangent but we want the normal. Recall that there is a rule #m_1m_2=-1# where #m_1# is a gradient and #m_2# is another gradient and when you multiply it together, it equals to #-1#

ie. #7/9m_2=-1#

#m_2=-9/7#

When #x=-1#, y is equal to #f(-1)=(-1-3)/(-1+4)=-4/3#
#(-1,-4/3)#

Using the point gradient formula,

#(y+4/3)=-9/7(x+1)#

#7y+28/3=-9x-9#

#7y+9x+55/3=0#

#21y+27x+55=0#