# CALCULUS

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An equation of the line tangent to the graph of (5x - 1)/(3x + 1) at the point where x = 1 is? Any ideas?

• CALCULUS -

sure, I have an idea. Find the derivative to get the slope of the tangent line! Then use the point-slope form of the line's equation:

y = (5x-1)/(3x+1)
y(1) = 4/4 = 1

y' = 8/(3x+1)^2
y'(1) = 8/16 = 1/2

tangent line is thus

(y-1) = 1/2 (x-1)

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