Find an equation of the tangent line to the graph of f(x) = -1/x at a) (-1,1); b) (2, -1/2); c) (-5, 1/5)

I've had success with completing similar problems but I am having a lot of trouble getting a simplified difference quotient since the given component (-1/x) is negative. How does it simplify and what are the steps? Thanks :)

f(x) = -1/x
f'(x) = 1/x^2

a)
x = -1
y = 1

f'(-1) = 1 / (-1)^2 = 1/1 = 1

1 = 1 * (-1) + b
1 = -1 + b
2 = b

y = x + 2


b)

f'(2) = 1 / 2^2 = 1/4

y = (1/4) * x + b
y = -1/2
x = 2

-1/2 = (1/4) * 2 + b
-1/2 = 1/2 + b
-1 = b

y = (1/4) * x - 1


c)

f'(-5) = 1 / 25

(1/5) = (1/25) * (-5) + b
1/5 = -1/5 + b
2/5 = b
y = (1/25) * x + (2/5)



Do you see a general formula? We could solve for some value (a , -1/a)

f'(a) = 1/a^2

y = (1/a^2) * a + b
-1/a = 1/a + b
-2/a = b

y = (1/a^2) * x - 2/a

And you could just plug in the value for a. For instance, a = -1

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

a = 2
y = (1/4) * x - 2/2
y = (1/4) * x - 1

a = -5
y = (1/25) * x - 2/5


EDIT:
Using the difference quotient

f(x) = -1/x
f(x + h) = -1/(x + h)

f(x + h) - f(x) =>
-1 / (x + h) - (-1 / x) =>
1/x - 1/(x + h) =>
(x + h) / (x * (x + h)) - x / (x * (x + h)) =>
(x + h - x) / (x * (x + h)) =>
h / (x * (x + h))

Divide that all by h

h / (h * x * (x + h)) =>
1 / (x * (x + h))

Take the limit as h goes to 0

1 / (x * (x + 0)) =>
1 / (x * x) =>
1 / x^2