Mark M. answered 09/03/16
Tutor
4.9
(926)
Retired college math professor. Extensive tutoring experience.
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 = -3/4
Solving 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.