# Axis of a Parabola

1. Dec 4, 2006

### chaoseverlasting

If someone could tell me how to go about it, I'd really appreciate it. Just point me in the right direction please. Kinda stuck.

The vertex of a parabola is (-3,1) and it touches the x axis at the origin. Find the equation of the axis of the parabola.

2. Dec 4, 2006

Did you mean, the parabola intersects the x axis at the origin? Or am I missing something?

3. Dec 4, 2006

### chaoseverlasting

No, the x-axis is the tangent to the parabola at the origin. I dont know what to do with it.

4. Dec 4, 2006

I dont think that possible.

5. Dec 4, 2006

### chaoseverlasting

Couldnt it be a tilted parabola of the type $$y^2=4ax$$ so that its lower arm touches the x-axis at the origin?

6. Dec 4, 2006

yes it could. In that case, it would be opening to the right.

7. Dec 4, 2006

### chaoseverlasting

So how would you find its axis?

8. Dec 4, 2006

it would simply be y = 1.

9. Dec 4, 2006

### chaoseverlasting

Why would it be y=1?

10. Dec 4, 2006

### arildno

Since the axis has to go through the parabola's vertex, and must be parallell to the x-axis.

11. Dec 4, 2006

### HallsofIvy

Why would the axis of the parabola have to parallel to the x-axis? We are told that the x-axis is tangent to the parabola at (0,0). No line parallel to the axis of a parabola is ever tangent to it.

12. Dec 5, 2006

### arildno

THe x-axis is tangent to the parabola at the origin.

The symmetry axis of the parabola is parallell to the x-axis, because the the tangent to the vertex is parallell to the y-axis.
The symmetry axis of a parabola is perpendicular to the tangent at the vertex.

13. Dec 5, 2006

### chaoseverlasting

It doesnt say the tangent at the vertex is paralell to the y-axis. You dont know the equation so you cannot assume that. the parabola could be tiled so that the axis of the parabola makes some angle with the x-axis and one of the arms just touches the x-axis at the origin.

14. Dec 5, 2006

### HallsofIvy

Yes, that is given.

Where was that said?

15. Dec 5, 2006

### chaoseverlasting

Where do you begin on something like this?

16. Dec 5, 2006

### arildno

I'm sorry, I was wrong.

17. Dec 5, 2006

### chaoseverlasting

18. Dec 5, 2006

### Hubert

Well, if it is a parabola, the equation is of the form y = f(x) = ax^2+bx+c

Now, f(-3)=1 and

f(0)=0

Finally, f'(0)=0

Also, since f(x)= ax^2+bx+c and f(0)=0 c must also equal zero.

EDIT: I think I'm wrong again (as usual). I guess the equation of a parabola does not have to be of the form ax^2+bx+c=f(x)

Last edited: Dec 5, 2006
19. Dec 6, 2006

### chaoseverlasting

Could it be that the question is incomplete?

20. Dec 6, 2006

### kishtik

I think the equation should be y=a(x-h)^2+k
Since (-3,1) is the vertex h=-3 and k=1
we get y=a(x+3)^2+1
If x=0 and y=0
9a+1=0 a=-1/9 y=1-((x+3)^2)/9