find the equation of the tangent to y = x-3/x^2+1 where x = 2

The equation of the tangent line to the curve y is the
derivative of the function, at the given point.

We'll
determine the derivative of the function, using the quotient
rule:

y' = [(x-3)'*(x^2+1) - (x-3)*(x^2 + 1)']/(x^2 +
1)^2

y' = (x^2 + 1 - 2x^2 + 6x)/(x^2 +
1)^2

y' = (-x^2 + 6x + 1)/(x^2 +
1)^2

Now, we'll replace x by 2, to determine the slope of
the tangent line:

y' = m =
(-4+12+1)/25

y' = m =
9/25

We'll calculate the value of the function at x
=2:

y = (2 - 3)/(2^2+1)

y =
-1/5

We'll use the point slope formula to get the equation
of the tangent line:

y - (-1/5) = (9/25)*(x -
2)

y + 1/5 = 9x/25 - 18/25

y =
9x/25 - 18/25 - 1/5

y = 9x/25 -
23/25

The equation of the tangent line to the
given curve, at the point x = 2, is: y = 9x/25 -
23/25.