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

Answer 1

Charles Anctil

#x+10y=3#

Given:
#color(white)("XXX")f(x)=(x-1)(x-3)(x+2)#

#f(3)=0# (since #(x-3)=0# when #x=3#)

So the point of intersection is going to be at #(x,y)=(3,0)#

Expanding the given equation we get
#color(white)("XXX")f(x)=x^3-2x^2-5x+6#

So
#color(white)("XXX")f'(x)=3x^2-4x-5#

And at #(3,0)# the tangent slope will be
#color(white)("XXX")f'(3)= 3(3^2)-4(3)-5 = 10#

Therefore the slope of the normal will be #(-1/10)#

Using the slope-point form for the normal we have
#color(white)("XXX")(y-0)=(-1/10)(x-3)#

which can be simplified as
#color(white)("XXX")x+10y=3#

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Answer 2

Isabella Ackerman

The equation of the normal line of f(x)=(x-1)(x-3)(x+2) at x=3 is y = -4x + 15.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.