What is the equation of the line normal to #f(x)=9x^2 - 28x - 34 # at #x=-1#?

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Answer 1 · 2016-12-29T15:54:25

#46y-x-139=0#

let #" "y=f(x)#

1) find the gradient of the normal

This is done in two parts

(i) find gradient of tangent #((dy)/(dx))_(x=-1)#

#y=9x^2-28x-34=>(dy)/(dx)=18x-28#

#((dy)/(dx))_(x=-1)=18xx-1-28=-18-28=-46#

(ii) the normal is perpendicular to the tangent .

#m_txxm_n=-1#

#-46xxm_n=-1=>m_n=1/46#

2) find the eqn. of normal

use #" "y-y_1=m_n(x-x_1)#

need to find #" " y_1" "#first

#y(-1)=9xx(-1)^2-28xx(-1)-34#

#y(-1)=9+28-34=3#

#(x_1,y_1)=(-1,3)#

#:.y-3=1/46(x--1)#

#46y-138=x+1#

#46y-x-139=0#