What is the equation of the line that is normal to #f(x)=-2x^3+4x-2 # at # x=-2 #?

Answer 1

#y-6=1/20(x+2)#

The normal line is perpendicular to the tangent line.

#f'(x)=-6x^2+4# so #f'(-2)=-20#, so the slope of the normal line is #1/20#, the opposite reciprocal.
#f(-2) = -2(-2)^3+4(-2)-2=6#.
The equation of the normal line is #y-6=1/20(x+2)#
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Answer 2

#y-6= 1/20 (x+2)#

At x=-2, the y co-ordinate of the point would be #-2(-2)^3 +4(-2) -2 =6#

the given point through which the required normal would pass is (-2,6)

Slope of the curve is #dy/dx =-6x^2+4#. The slope at point (-2,6) would be #-6(-2)^2 +4 = -20#
The slope of the normal passing through this point would thus be#1/20# (Recollect the formula for slopes of two perpendicular lines #m_1 m_2 =-1#)
The point slope form of the required line would be #y-6= 1/20 (x+2)#
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Answer 3

To find the equation of the line that is normal to the function f(x) = -2x^3 + 4x - 2 at x = -2, we need to determine the slope of the tangent line at x = -2 and then find the negative reciprocal of that slope to obtain the slope of the normal line.

To find the slope of the tangent line, we can take the derivative of the function f(x) with respect to x and evaluate it at x = -2.

The derivative of f(x) = -2x^3 + 4x - 2 is f'(x) = -6x^2 + 4.

Evaluating f'(x) at x = -2, we get f'(-2) = -6(-2)^2 + 4 = -6(4) + 4 = -24 + 4 = -20.

The slope of the tangent line at x = -2 is -20.

To find the slope of the normal line, we take the negative reciprocal of -20, which is 1/20.

Now, we have the slope of the normal line, and we also know that it passes through the point (-2, f(-2)).

Substituting x = -2 into the original function f(x), we get f(-2) = -2(-2)^3 + 4(-2) - 2 = -2(-8) - 8 - 2 = 16 - 8 - 2 = 6.

Therefore, the point of intersection is (-2, 6).

Using the point-slope form of a linear equation, y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope, we can substitute the values into the equation.

y - 6 = (1/20)(x - (-2))

Simplifying, we get y - 6 = (1/20)(x + 2).

Expanding, we have y - 6 = (1/20)x + 1/10.

Rearranging the equation, we get y = (1/20)x + 61/10.

Therefore, the equation of the line that is normal to f(x) = -2x^3 + 4x - 2 at x = -2 is y = (1/20)x + 61/10.

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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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