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