## anonymous 3 years ago can anyone help me with this problem find vertex and equation, line of symmetry and graph the function. f(x)= 1/3 x^2

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1. anonymous

f(x)=$\frac{ x ^{2} }{ 3 }$

2. anonymous

I might be reading it wrong I am sorry

3. anonymous

the parabola or quadratic equation is f(x)=ax^2 +bx +c, vertex x=-b/2a if you have a formula of f(x)= 1/3 x^2+bx +c --> here b=0 and c=0 vertex x=-b/2a x=-0/2(1/3)=0 sub this to f(x)= y=1/3 x^2=0 then what is the vertex V(x,y)=___? is it V(0,0) yes or no? for graphing use values of x=0,+-1,+-2 +-3 etc..etc....:D good luck now

4. anonymous

You had it right I think I was confused, but you really helped me understand the question...

5. anonymous

ok good,,, good luck now and have fun ....:D

6. anonymous

I will once someone looks at my problems lol..... One person is looking I just want to ensure I am on the right track....

7. anonymous

ok where is the prob? i may be able to help a little bit :D

8. anonymous

How would I show you?

9. anonymous

If anyone can help me check my work I would be very thankful. So it can be more understandable I attached my work in a word file. Thank you so very much I just want to ensure I am on the right track.

10. anonymous

Thats the question

11. anonymous

hmm go to your prob site and notify my name there then i will click it and be there :D

12. anonymous

like this hi andjie

13. anonymous

14. anonymous

yes copy and paste my name there then ill just click on it

15. anonymous

ok I did that

16. anonymous

hmm i didnt have a notification,, did you highlight and copy then paste my name there?

17. anonymous

Paste it where exactly lol

18. anonymous

19. anonymous

on where you posted your problem

20. anonymous

ok go back to where you posted your problem then paste my name there

21. anonymous

I posted your name in the question

22. anonymous

So confusing

23. anonymous

hmm i dont know why i didnt get a notification?

24. anonymous

is it still open? why dont you close it hen repost them new

25. anonymous

ok