What is the equation of the normal line of #f(x)=2x^3-8x^2+2x-1# at #x=-1#?

Answer 1

#y+13=-1/24(x+1)#

First, find the point the normal line will intercept by finding the function value at #x=-1#.
#f(-1)=2(-1)^3-8(-1)^2+2(-1)-1=-2-8-2-1=ul(-13#
The normal line will pass through the point #(-1,-13)#.
Before we can find the slope of the normal line, we must first find the slope of the tangent line. The slope of the tangent line is equal to the value of the function's derivative at #x=-1#.

Find the derivative of the function through the power rule:

#f(x)=2x^3-8x^2+2x-1#
#f'(x)=6x^2-16x+2#

The slope of the tangent line is

#f'(-1)=6(-1)^2-16(-1)+2=6+16+2=ul(24#
Since the tangent line and normal line are perpendicular, their slopes will be opposite reciprocals. The opposite reciprocal of #24# is #-1/24#.
We can relate the information we know about the normal line as a linear equation in point-slope form, which takes a point #(x_1,y_1)# and slope #m#:
#y-y_1=m(x-x_1)#
Since the normal line passes through #(-1,-13)# and has slope #-1/24#, its equation is
#y+13=-1/24(x+1)#

Graphed are the function and its normal line:

graph{(2x^3-8x^2+2x-1-y)(y+13+(x+1)/24)=0 [-5, 7, -18.16, 2.12]}

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Answer 2

To find the equation of the normal line of f(x) at x=-1, we need to find the slope of the tangent line at that point and then determine the negative reciprocal of that slope to find the slope of the normal line.

First, we find the derivative of f(x) to get the slope of the tangent line. The derivative of f(x)=2x^3-8x^2+2x-1 is f'(x)=6x^2-16x+2.

Next, we substitute x=-1 into the derivative to find the slope of the tangent line at x=-1. f'(-1)=6(-1)^2-16(-1)+2=24.

The slope of the normal line is the negative reciprocal of the slope of the tangent line. So, the slope of the normal line is -1/24.

Now, we have the slope of the normal line and the point (-1, f(-1)) on the curve. We can use the point-slope form of a line to find the equation of the normal line.

Using the point-slope form, the equation of the normal line is y - f(-1) = (-1/24)(x - (-1)).

Simplifying this equation, we get y - f(-1) = (-1/24)(x + 1).

Therefore, the equation of the normal line of f(x)=2x^3-8x^2+2x-1 at x=-1 is y - f(-1) = (-1/24)(x + 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.

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.

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.

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.

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