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

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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.

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