# Thread: Equation of a tangent line at a certain point.

1. ## Equation of a tangent line at a certain point.

"Find the equation of the tangent line to the graph of $\displaystyle y=(2x-5)/(x+1)$ at the point at which x=0."

I thought I had started off doing this right, but I must've messed up somewhere.

Don't I need to find the derivative of the function and then evaluate it at x=0?

2. The equation of a tangent for this function is

$\displaystyle y-f(0) = f'(0)\times (x-0)$

Does this help?

3. Well, I can form the tangent line once I have the values. But I think I'm coming out with the wrong values.

Could you check my work below?

To get the derivative, you must use the quotient rule.
So you end up with $\displaystyle f'(x)=((2(x+1))-(2x+5)(1))/(x+1)^2$, which simplifies to $\displaystyle f'(x)=((2x+1)-(2x+5))/(x+1)^2$ and then $\displaystyle f'(x)=-4/(x+1)^2$. Is this right?

4. You have expanded the function incorrectly.

I get

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

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

$\displaystyle f'(x) = \frac{7}{(x+1)^2}$