How do you find the equation of a line tangent to the function #y=x/(1-3x)# at (-1,-1/4)?

Question

Isabella Ackerman · Accepted Answer

#x-16y-3=0#. See the tangent-inclusive Socratic graph.

#y(1-3x)=x#. So, y/(1-3x)-3y=1# At the point of contact #P(-1. -1/4) #, 4y' +3=1, giving the slope of the tangent # y'=1/16#. Now, the equation to the tangentat P is #y+1/4-1/16(x+1)#, giving #x-16y-3=0# graph{(x/(1-3x)-y)(x-16y-3)=0 [-4.967, 4.967, -2.483, 2.484]} The graph is a rectangular hyperbola, with perpendicular asymptotes # y =-1/3 and x = 1/3#.

Ezekiel Alexander · Answer

undefined To find the equation of a line tangent to a function at a given point, you need to find the derivative of the function and evaluate it at the given point.

First, find the derivative of the function y = x/(1-3x) using the quotient rule.

The derivative of y with respect to x is given by:
dy/dx = (1-3x)(1) - x(-3)/(1-3x)^2

Simplifying this expression, we get:
dy/dx = (1-3x + 3x)/(1-3x)^2
dy/dx = 1/(1-3x)^2

Next, evaluate the derivative at the given point (-1, -1/4). 
Substitute x = -1 into the derivative expression:
dy/dx = 1/(1-3(-1))^2
dy/dx = 1/(1+3)^2
dy/dx = 1/16

So, the slope of the tangent line at (-1, -1/4) is 1/16.

Now, we can use the point-slope form of a line to find the equation of the tangent line. 
Using the point (-1, -1/4) and the slope 1/16, the equation of the tangent line is:
y - y1 = m(x - x1)
y - (-1/4) = (1/16)(x - (-1))
y + 1/4 = (1/16)(x + 1)
y + 1/4 = (1/16)x + 1/16
y = (1/16)x + 1/16 - 1/4
y = (1/16)x - 3/16