Equation of a tangent line at a certain point.

**Rumor** · November 15th 2009, 12:37 PM

"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?

**pickslides** · November 15th 2009, 12:44 PM

The equation of a tangent for this function is

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

Does this help?

**Rumor** · November 15th 2009, 12:54 PM

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?

**pickslides** · November 15th 2009, 01:02 PM

You have expanded the function incorrectly.

I get

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

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

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

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

LinkBack

Thread Tools

Display

Equation of a tangent line at a certain point.

Similar Math Help Forum Discussions

Equation of the tangent line at that point

equation of the tangent line to the curve at the given point

Find equation of tangent line to the function at given point.

Find the equation of the tangent line to the following curve at the given point P. Il

Find an equation of the tangent line to the curve at the point (7, 2).

Search Tags