What is the equation of the normal line of #f(x)=x^2# at #x=-5#?

1 Answer
Oct 28, 2016

#y=-10x-25#

Explanation:

The equation of a straight line is determined by computing its #color(red)(slope) # and the point it passes through name it #(color(purple)(x_1,y_1))#

This line is normal to #f(x)# at #color(purple)(x=-5)# so it passes through the point#(color(purple)(-5,f(-5)))#

#color(purple)(f(-5))=(-5)^2=25#

This line passes through the point #(color(purple)(-5,25))#

The #color(red)(slope) # at #x=-5# is determined by computing #color(red)(f'(-5))#
#f(x)# is differentiated by using power rule differentiation
#(x^n)'=n(x^(n-1))#

#f'(x)=2x#
# color(red)(f'(-5)=2(-5)=-10)#

The equation is:

#y-y_1=color(red)(slope)(x-x_1)#

#y-f(-5)=-10(x-(-5))#

#y-25=-10(x+5)#

#y=-10x-50+25#

#y=-10x-25#