# Thread: Equation of a horizontal tangent line..

1. ## Equation of a horizontal tangent line..

Let f be the function defined by $f(x)=3x^5-5x^3+2$

How would I find the horizontal tangent line..? ><"

Let f be the function defined by $f(x)=3x^5-5x^3+2$

How would I find the horizontal tangent line..? ><"
The tangent to the curve is horizontal at the stationary points.

So you need to calculate the y-coordinate of all stationary points. I suggest using calculus to first get the x-coordinates.

3. I don't quite understand.

I don't quite understand.
Find the y-coordinates of the stationary points. Can you do that?

5. nope.

nope.
Start by finding the x-coordinates of the stationary points. Can you do that?

7. If by stationary points you mean critical numbers, I have x=0 and x=+1

If by stationary points you mean critical numbers, I have x=0 and x=+1
Good. Now substitute these values into y = f(x) to get the y-coordinates of the critical points.

9. Originally Posted by mr fantastic
Good. Now substitute these values into y = f(x) to get the y-coordinates of the critical points.
(0,2) (1,0) and (-1, 4)

(0,2) (1,0) and (-1, 4)
Good. So the horizontal tangents have equation y = 2, y = ..... and .....

11. y=2, y=4, and y=0?

y=2, y=4, and y=0?
Yes.

13. Originally Posted by mr fantastic
Yes.

Thank you so muchhh.
It's kind of a dumb question, and I really appreciate your patience with me.
I've been working on calculus since I got out of school today. u___u
Again, thanks so much.