How do you find the equation of a line tangent to the function #y=-3/(x^2-4)# at (1, 1)?

Question

2255 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-26T19:01:33

#y=2/3x+1/3#

We need to find the derivative of this function. This will allow us to find the gradient of the tangent line.

Notice we can write this as:

#3(x^2-4)^(-1)#

Using the chain rule:

#3(x^2-4)^(-2)*(2x)=(6x)/(x^2-4)^2#

Plugging in 1 will give us the gradient of the line.

#(6(1))/((1)^2-4)^2= 6/9=2/3#

We now have the gradient #2/3# and two points on the line #(1 , 1)#
equation of a line #y=mx+b#

#1=2/3(1) +b=> b= 1/3#

Equation of line is:

#y=2/3x+1/3#

enter image source here