Mark M. answered • 09/03/16

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Slope of tangent line = (-3-3)/(6-(-2)) = -3/4

Equation of tangent line: y - 3 = (-3/4)(x - (-2))

y - 3 = (-3/4)x - 3/2

y = (-3/4)x + 3/2

At the point of tangency (the point where the tangent line intersects the curve), the y-coordinate of the point on the tangent line is the same as the y-coordinate of the point on the graph of y = b/(x+1)

^{2}.So, the x-coordinate of the point of tangency must satisfy the equation (-3/4)x + 3/2 = b/(x+1)

^{2}(*)From Calculus, the slope of the tangent line to a curve is the derivative evaluated at the x-coordinate of the point of tangency.

So, y' = -2b/(x+1)

^{3}= -3/4Solving for b/(x+1)

^{2}, we have b/(x+1)^{2}= (3/4)[(x+1)/2]Simplify to get b/(x+1)

^{2}= (3/8)(x+1) (**)Using equations (*) and (**), we now have two different expressions for b/(x+1)

^{2}.Equating these expressions, (3/8)(x+1) = (-3/4)x + 3/2

(3/8)x + 3/8 = (-3/4)x + 3/2

Multiply by 8 to obtain 3x + 3 = -6x + 12

9x = 9

So, x = 1

Finally, since x = 1 and b/(x+1)

^{2}= (3/8)(x+1), we have b/4 = 3/4.Therefore b = 3.