What is the equation of the normal line of #f(x)=4x-x^2# at #x=0#?

Answer 1

Aurora Alvarado

#y=-1/4x#

Find the point the normal line will intercept:

#f(0)=0#

The normal line will intercept the point #(0,0)#.

Before you can find the slope of the normal line, find the slope of the tangent line.

Differentiate #f(x)#:

#f'(x)=4-2x#

The slope of the tangent line is:

#f'(0)=4#

Since the tangent line and normal line are perpendicular, they will have opposite reciprocal slopes.

Thus, the slope of the normal line is #-1/4#.

Since we know the line will pass through the origin, the equation of the normal line is:

#y=-1/4x#

The function and normal line graphed:

graph{(4x-x^2-y)(y+x/4)=0 [-10, 10, -5, 5]}

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Answer 2

Isabella Ackerman

The equation of the normal line of f(x)=4x-x^2 at x=0 is y = 4x.

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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.