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

Question

Elijah Abram · Accepted Answer

Equation of the normal line is #x-2y-13=0#

At #x=3#, #f(3)=(3+2)/(3-4)=5/-1=-5#

Hence normal is desired at point #(3,-5)# on the curve.

Slope of tangent is given by #f'(x)#

= #((x-4)*d/dx(x+2)-(x+2)*d/dx(x-4))/(x-4)^2#

= #((x-4)*1-(x+2)*1)/(x-4)^2=(x-4-x-2)/(x-4)^2=-2/(x-4)^2#

Hence slope of tangent at #(3,-5)# is #-2/(3-4)^2=-2#.

Hence slope of normal would be #-1/-2=1/2#

And equation of the normal line using point slope form is

#(y-(-5))=1/2(x-3)# or

#2*(y+5)=(x-3)# or #x-2y-13=0#

Cameron Alexander · Answer

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

To find the slope of the tangent line at x=3, we can use the derivative of f(x). The derivative of f(x) is given by f'(x) = (6)/(x-4)^2. Evaluating f'(x) at x=3, we get f'(3) = 6/(3-4)^2 = 6.

The slope of the tangent line at x=3 is 6. To find the slope of the normal line, we take the negative reciprocal of 6, which is -1/6.

Now, we have the slope of the normal line (-1/6) and a point on the line (x=3, f(3)). Plugging x=3 into f(x), we get f(3) = (3+2)/(3-4) = -5.

Using the point-slope form of a line, the equation of the normal line is y - (-5) = (-1/6)(x - 3). Simplifying this equation gives y + 5 = (-1/6)x + 1/2.

Therefore, the equation of the normal line of f(x)=(x+2)/(x-4) at x=3 is y = (-1/6)x - 9/2.