What is the equation of the line that is normal to #f(x)= sqrt( x^2+3x+2) # at # x=1 #?

1 Answer
Feb 8, 2018

#y=sqrt6/5(-2x+7)#
See explanation below

Explanation:

The value of derivative of #f(x)# at #x=1# is the slope of tangent line to #f(x)# at #x=1#.

As per, #f'(x)=1/2(2x+3)/sqrt(x^2+3x+2)#

#f'(1)=1/2(2·1+3)/sqrt(1^2+3·1+2)#=#5/(2sqrt6)#

By other hand: if the straigh line general equation is

#y=mx+b# (where #m# is the slope and #b# is the intersection with #y# axis) then, the normal line equation is #y=-1/mx+c#

So: the normal line equation requested is #y=-2sqrt6/5x+b#

The value b can be found by the fact: the value of #f(1)# and by normal line must be the same (because they intercept in #x=1#

Thus: #f(1)=sqrt6= -2sqrt6/5+b# trasposing elements we found

#b=7sqrt6/5#

Finally, the normal line equation is #y=-2sqrt6/5x+7sqrt6/5# or

#y=sqrt6/5(-2x+7)#