What is the equation of the line normal to #f(x)=3x^2 +x-5 # at #x=1#?

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Answer 1 · 2016-03-28T08:50:46

#x+7y=-6#

We need to obtain first the slope m which is negative reciprocal of the slope of the tangent line at x=1 on the curve #f(x)=3x^2+x-5#

Compute the first derivative #f'(x)# and #m=f' (1)#

#f(x)=3x^2+x-5#

#f' (x)=d/dxf(x)=d/dx(3x^2+x-5)=6x+1#

#f' (x)=6x+1#

at #x=1#

#f' (1)=6(1)+1=7#

Compute #m#

#m=-1/(f' (1))=-1/7#

Determine the ordinate for the given abscissa #x=1# using the original equation #f(x)=3x^2+x-5#

#f(1)=3(1)^2+1-5=-1#

The point is at #(1, -1)#

Determine the Normal Line now using Point-Slope Form

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

#y--1=-1/7(x-1)#

#7(y+1)=-x+1#

#7y+7=-x+1#
#x+7y=-6#

See the graph of #f(x)=3x^2+x-5# and #x+7y=-6#

graph{(y-3x^2-x+5)(x+7y+6)=0[-22,22,-10,10]}