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What is the equation of the normal line of #f(x)=x^3 + 3x^2 + 7x - 1 # at #x=-1 #?

Answer 1

Christopher Allers

#y=x/4+23/4 #

#f(x)=x ^3+3x^2+7x-1# The gradient function is the first derivative #f'(x)=3x^2+6x+7#

So the gradient when X=-1 is 3-6+7=4 The gradient of the normal , perpendicular, to the tangent is #-1/4#

If you are not sure about this draw a line with gradient 4 on squared paper and draw the perpendicular.

So the normal is #y=-1/4x+c#

But this line goes through the point (-1,y) From original equation when X=-1 y=-1+3-7-1=6

So 6=#-1/4*-1+c# #C=23/4#

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Answer 2

Evelyn Agard

The equation of the normal line of f(x)=x^3 + 3x^2 + 7x - 1 at x=-1 is y = -4x - 1.

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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.