What is the equation of the normal line of #f(x)= -2x^3-10x # at #x=2 #?

Question

Luke Alvarez · Accepted Answer

The equation of the normal line is:

#y = x/34 -36-1/17# The equation of the line normal to the graph #y=f(x)# in the point #(bar x, f(bar x))# is given by: #y=f(bar x)-1/(f'(barx)) (x - barx)# In our case #bar x = 2# and: #f(x) = -2x^3-10x => f(2) = -36# #f'(x) = -6x^2-10 => f'(2) = -34# So the equation of the normal line is: #y = -36 +1/34(x-2) = x/34 -36-1/17#

Charles Arocha · Answer

undefined To find the equation of the normal line at x=2 for the function f(x)=-2x^3-10x, 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 take the derivative of the function f(x). The derivative of f(x)=-2x^3-10x is f'(x)=-6x^2-10.

Evaluating f'(x) at x=2, we get f'(2)=-6(2)^2-10=-26.

The slope of the tangent line at x=2 is -26.

To find the slope of the normal line, we take the negative reciprocal of -26, which is 1/26.

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 find f(2)=-2(2)^3-10(2)=-36.

Therefore, the point of intersection is (2, -36).

Using the point-slope form of a line, we can write the equation of the normal line as y-(-36)=(1/26)(x-2).

Simplifying this equation, we get y+36=(1/26)x-(1/13).

Rearranging the equation, we have y=(1/26)x-(1/13)-36.

Thus, the equation of the normal line of f(x)=-2x^3-10x at x=2 is y=(1/26)x-(1/13)-36.