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

Question

1385 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-18T07:26:31

#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#.