slope of a tangent line

Question

The slope of the tangent line to the curve y=3/x at the point (6,12) is:  

The equation of this tangent line can be written in the form 

y=mx+b

 where m is :

b is :

Answer 1

y=3/x = 3(x^-1)

y' = -3x^-2 = -3/x^2

y'(6) = -3/(6)^2 = -3/(36) = -1/12

the tangent line is

y=mx + b where m=-1/12 and b is calculated by plugging in the point (6, 1/2)

1/2 = (-1/12)(6) + b

b = 1/2 + 1/2 = 1

y=(-1/12)x + 1

Answer 2

She probably meant to say (6,1/2) and/or the fraction bar did not get rendered...

y = 3/x = 3 * x^(-1)

y' = -3 * x^(-2) = f'(x)

slope M = f'(6) = -3 * 6^(-2) = -3/36 = -1/12

B = y - mx = 1/2 - (-1/12)*6 = 1/2 - (-1/2) = 1/2+1/2= 1

y = (-1/12)x + 1