What is the equation of the normal line of #f(x)=2x^3-x^2-3x+9# at #x=-1#?

Question

Isabella Ackerman · Accepted Answer

Let's first find what point the tangent passes through, given #x = a#.

#f(x) = 2(-1)^3 - (-1)^2 - 3(-1) + 9# #f(x) = -2 - 1 + 3 + 9# #f(x) = 9# Therefore, the tangent passes through #(-1, 9)#. Now that we know this, we must differentiate the function. By the power rule: #f'(x) = 6x^2 - 2x - 3# The slope of the tangent is given by evaluating #f(a)# inside the derivative, a being #x = a#. #f'(-1) = 6(-1)^2 - 2(-1) - 3# #f'(-1) = 6 + 2 - 3# #f'(-1) = 5# The slope of the tangent is #5#. The normal is always perpendicular to the tangent, so the slope will be the negative reciprocal of that of the tangent. This means the slope of the normal line is #-1/5#. By point slope form, we can find the equation of the normal line: #y - y_1 = m(x - x_1)# #y - 9 = -1/5(x - (-1))# #y - 9 = -1/5x - 1/5# #y = -1/5x - 1/5 + 9# #y = -1/5x + 44/5# #:.# The equation of the normal line is #y = -1/5x + 44/5#. Hopefully this helps!

Aurora Alvarado · Answer

undefined To find the equation of the normal line at a given point, we need to determine the slope of the tangent line at that point and then find the negative reciprocal of that slope.

To find the slope of the tangent line, we take the derivative of the function f(x).

The derivative of f(x) = 2x^3 - x^2 - 3x + 9 is f'(x) = 6x^2 - 2x - 3.

To find the slope of the tangent line at x = -1, we substitute x = -1 into the derivative: f'(-1) = 6(-1)^2 - 2(-1) - 3 = 6 + 2 - 3 = 5.

The slope of the tangent line at x = -1 is 5.

To find the slope of the normal line, we take the negative reciprocal of the slope of the tangent line: -1/5.

Now, we have the slope of the normal line, and we also have a point on the normal line, which is (-1, f(-1)).

Substituting x = -1 into the original function f(x), we find f(-1) = 2(-1)^3 - (-1)^2 - 3(-1) + 9 = -2 - 1 + 3 + 9 = 9.

Therefore, the point on the normal line is (-1, 9).

Using the point-slope form of a linear equation, we can write the equation of the normal line:

y - y1 = m(x - x1), where (x1, y1) is the point on the normal line and m is the slope of the normal line.

Substituting the values, we get:

y - 9 = -1/5(x - (-1))

Simplifying further:

y - 9 = -1/5(x + 1)

Expanding:

y - 9 = -1/5x - 1/5

Rearranging the equation:

y = -1/5x + 44/5

Therefore, the equation of the normal line of f(x) = 2x^3 - x^2 - 3x + 9 at x = -1 is y = -1/5x + 44/5.