# Equation of a tangent line at a certain point.

• Nov 15th 2009, 01:37 PM
Rumor
Equation of a tangent line at a certain point.
"Find the equation of the tangent line to the graph of $y=(2x-5)/(x+1)$ 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?
• Nov 15th 2009, 01:44 PM
pickslides
The equation of a tangent for this function is

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

Does this help?
• Nov 15th 2009, 01:54 PM
Rumor
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 $f'(x)=((2(x+1))-(2x+5)(1))/(x+1)^2$, which simplifies to $f'(x)=((2x+1)-(2x+5))/(x+1)^2$ and then $f'(x)=-4/(x+1)^2$. Is this right?
• Nov 15th 2009, 02:02 PM
pickslides
You have expanded the function incorrectly.

I get

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

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

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