What is the equation of the line normal to # f(x)=-x/(x+4)# at # x=6#?

Question

What is the equation of the line normal to #  f(x)=-x/(x+4)# at #  x=6#?

Christopher Allers · Accepted Answer

y = 25 x #-#753/5.

Denoting f(x) by y, the equation simplifies to #-#1 + 4 / ( x + 4 ). The point under reference is P ( 6, #-#3/5 ) #d/dx#y = #-#4 / ( x + 4 )^2. The slope of the normal at P is #-#1 / #d/dx#y at P = 25. So, its equation is y + 3/5 = 25 ( x #-#6 )..

Emily Ammerman · Answer

undefined To find the equation of the line normal to f(x)=-x/(x+4) at x=6, we need to determine the slope of the tangent line at x=6 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 at x=6, we can take the derivative of f(x) with respect to x and evaluate it at x=6.

The derivative of f(x)=-x/(x+4) is given by f'(x) = (4x+16)/(x+4)^2.

Evaluating f'(x) at x=6, we get f'(6) = (4(6)+16)/(6+4)^2 = 40/100 = 0.4.

The slope of the tangent line at x=6 is 0.4.

To find the slope of the normal line, we take the negative reciprocal of 0.4, which is -2.5.

Now, we have the slope of the normal line, -2.5, and the point of tangency, (6, f(6)).

Using the point-slope form of a line, the equation of the line normal to f(x)=-x/(x+4) at x=6 is y - f(6) = -2.5(x - 6).