Find an equation of the tangent line to the graph of the fun

Find an equation of the tangent line to the graph of the function at the given point.

y=	6x	, (3,9)
?x+1

Solution

y= 6x/v(x+1) (3,9) y\' = v(x+1) (3x+6) / (x2 + 2x +1) So slope of tangent = Put in x=3 in y\' = 2*15/16 = 15/8 So eqn of line is y-9 = 15/8 (x-3) ie 8y-72 = 15x - 45 15x-8y = -27 x interccept is -ve and y intercept is positive. SO figure is 4th one Hope you like it :)