If I have f(x)=(x^2)/3x+2, what is an equation of the tangent line to f(x) at x=1?

Sorry about the delay, but I just managed to get on the PC and answer it.
I assume that is supposed to be f(x) = x^2/(3x+2).

Given that f(x) = g(x)/h(x), the quotient rule says that f'(x) = (hg' -gh')/h^2.

Since g(x) = x^2, g'(x) = 2x.
Since h(x) = 3x+2, h'(x) = 3.

That gives [(3x+2)2x - (x^2)3]/(3x+2)^2.
That works out to (6x^2 + 4x - 3x^2)/(3x+2)^2.
Combining terms and factoring out an x on top gives x(3x+4)/(3x+2)^2.

Sorry about that ... now to continue to answer.

The equation of a line at x=1 would be y = m(x-1) + b where
m was the slope of the line and b was the value of y at 1.

To find m, put 1 into the derivative.
To find the value for b, put 1 into the equation for f(x) and see what you get.  


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