# Equation of Tangent Line

Can someone check if I did this right:

Find the equation of the tangent line to the graph of $$f(x) = \frac {2x-5}{x+1}$$ at the point at which x = 0.

$$f(x) = \frac {2(0) - 5}{(0) + 1} = -5$$

So I got y = -5

Then I wanted to find the derivative of the original using the quotient rule to find the slope.

$$f'(x) = \frac{(2x-5)(1) - (2)(x+1)}{(x+1)^2} = \frac{2x-5 - 2x -2}{(x+1)^2}$$

Is the slope -7? I plugged in 0 for x.

If so, I get :

y -y1 = m(x-x1)

y - -5 = m(x-0)

y + 5 = -7x + 0

y = -7x -5

Last edited:

StatusX
Homework Helper
I think you got the quotient rule wrong. It's low dee high minus high dee low.

Not sure what you mean. I was told you can do it either of these two ways:

$$F'(\frac{f}{g}) = \frac{fg' - f'g}{g^2} or \frac {f'g - fg'}{g^2}$$

Last edited:
StatusX
Homework Helper
They can't both be right: one is the negative of the other. You want the second one.

Ok so now I get: $$\frac{2x+1 - 2x+5}{(x+1)^2}$$

Do I plug x = 0 into that to find my slope? Which would give me m = 6

StatusX
Homework Helper
Yes, everything else you did in the first post looks right.