# Homework Help: At what points on this curve is the tangent line horizontal?

1. Mar 6, 2013

### physphys

1. The problem statement, all variables and given/known data
At what points on the curve y = (x^2)/(2x+5) is the tangent line horizontal?

2. Relevant equations
Quotient rule

3. The attempt at a solution
I figured out the derivative which is
2x(2x+5) - 2x^2
-----------------
(2x+5)^2

I also know that for the equation of the tangent to be horizontal, it needs to have a slope of 0, so y' = 0

Therefore I tried to do 2x(2x+5) - 2x^2 = 0, and I got 2x^2 + 10x = 0 which means that x must be -5. However when I put it into the calculator, it doesn't agree with me, so I went wrong somewhere.

2. Mar 6, 2013

### HallsofIvy

I have no idea what you mean by "put it into the calculator". However, "2x^2+ 10x= 0" does NOT mean "that x must be -5". How did you get that?

3. Mar 6, 2013

### physphys

I got a ti84, and i can draw a tangent line and it'll tell me the formula, so when I drew a tangent line through -5, the line was not horizontal completely. If you plug in -5 into 2x^2 + 10x= 0, it works out, but when you plug it back into the first one you don't get 0.

4. Mar 6, 2013

### Staff: Mentor

There are two critical numbers; x = -5 is one of them.

If you substitute x = -5 into y = x2/(2x + 5), you'll get the y value at one of the points where the tangent is horizontal.

5. Mar 6, 2013

### physphys

But the calculator disagrees with the fact that x = -5 is one of them, and also how can I find the other point?

6. Mar 6, 2013

### Staff: Mentor

I have no idea what you're entering into the calculator, so I can't say that its objections are valid.

To find the other critical number, find both solutions of 2x2 + 10x = 0.

I hope you aren't doing this:

2x2 = - 10x

and then dividing both sides by 2x.

7. Mar 6, 2013