Equation of tangent line to f(x)=(2x -5) / (x+1) at x=0

equation of tangent line to f(x)=(2x -5) / (x+1) at x=0
first i need to find the derivative and i got : 7/(x+1)^2
and then i need to plug 0 in? if so then it will equal 0 how would i
get the answer ; y= 7x -5 ?

f '(x) = ((x+1)(2) - (2x-5)) / (x+1)²
f '(x) = 7/(x+1)²

f '(0) = 7/(0+1)² = 7/1 = 7, not 0.

f(0) = -5 (plugging 0 into f(x))

y - (-5) = 7(x - 0)
y + 5 = 7x
y = 7x - 5

YOUR DERIVATIVE IS QUITE RIGHT.
THE SLOPE OF THE TANGENT AT X = 0 IS 7/(X + 1) = 7/1 = 7

YOUR EQUATION IS Y = 7X + C

C IS THE CONSTANT WHEN X = 0 IN F(X) = (2X - 5)/( X + 1) I.E. 5

SO EQUATION IN Y = 7X -5

f(x) = (2x - 5)/(x+1)

f(0) = -5

f ' (x) = [(x + 1)(2) - (2x - 5) ] /(x + 1)^2

f ' (0) = ( 2 + 5) /1 = 7

eqn of tangent is

y = f(0) + f '(0) (x - 0)

y = -5 + 7x