We are given:

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

#f'(x)=d/dx[f(x)]#

#color(white)(f'(x))=d/dx[-((x+1)(x-2))^(1/2)]#

#color(white)(f'(x))=-d/dx[((x+1)(x-2))^(1/2)]#

#color(white)(f'(x))=-((x+1)(x-2))^(1/2-1)*1/2*d/dx[(x+1)(x-2)]#

#color(white)(f'(x))=-((x+1)(x-2))^(1/2-1)* 1/2*((x+1)d/dx[x-2]+(x-2)d/dx[x+1])#

#color(white)(f'(x))=-((x+1)(x-2))^(-1/2)/2*((x+1)(1)+(x-2)(1))#

#color(white)(f'(x))=-((x+1)(x-2))^(-1/2)/2*(x+1+x-2)#

#color(white)(f'(x))=-((x+1)(x-2))^(-1/2)/2*(2x-1)#

#color(white)(f'(x))=-((2x-1)((x+1)(x-2))^(-1/2))/2#

#f'(6)=-((2(6)-1)((6+1)(6-2))^(-1/2))/2=-(11sqrt(7))/28#

However, for the normal, #"gradient"=-1/(f'(x))=-1/(-(11sqrt(7))/28)=28/(11sqrt(7))#

The normal will be in the form of #y=mx+c#,

#m=28/(11sqrt(7))#

#x=6#

#y=-((6+1)(6-2))^(1/2)=-sqrt(7*4)=-2sqrt(7)#

#-2sqrt(7)=6*28/(11sqrt(7))+c#

#c=-2sqrt(7)-6*28/(11sqrt(7))#

#color(white)(c)=-(46sqrt(7))/11#

#y=(28x)/(11sqrt(7))-(46sqrt(7))/11#

#y=(4sqrt(7)x)/11-(46sqrt(7))/11#

#y=(4sqrt(7)x-46sqrt(7))/11#

#11y=4sqrt(7)x-46sqrt(7)#

#4sqrt(7)x-11y-46sqrt(7)=0#