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

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Answer 1 · 2018-03-11T22:01:25

#y = 1/19 x + 592/19#

The function passes through the point

Denoting the point

#(x = -3, y = f(-3))#

by

#(x_1, y_1)#

The tangent to that point has instantaneous slope
#f'(-3)#

A line perpendicular to the tangent will have slope
#-1/(f'(-3))#

The line with this slope passing through #(x_1, y_1)# will be the normal to the function at that point.

Denoting its slope by #m#, t will have equation

#(y - y_1) = m (x - x_1)#

For the normal, to the function #f(x)# at #x = -3#,

#x_1 = -3#
#y_1 = f(-3)#
#m = -1/(f'(-3))#

Noting

#f(-3) = 3(-3)^2 - (-3) +1 #
#= 27 + 3 + 1 #
#= 31#

and
#f'(x) = 6 x - 1 #

so that
#f'(-3) = 6 (-3) - 1 #
# = -19#

and
#-1/(f'(-3)) = 1/19 #

The equation of the normal to #f(x)# at #x = -3# is:

#(y - 31) = 1/19 (x - (-3))#

that is

#y = 1/19 x + 3/19 + 31#

that is

#y = 1/19 x + 592/19#