## bev199 2 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. andjie

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

2. andjie

I might be reading it wrong I am sorry

3. mark_o.

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. andjie

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

5. mark_o.

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

6. andjie

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. mark_o.

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

8. andjie

How would I show you?

9. andjie

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. andjie

Thats the question

11. mark_o.

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

12. mark_o.

like this hi andjie

13. andjie

14. mark_o.

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

15. andjie

ok I did that

16. mark_o.

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

17. andjie

Paste it where exactly lol

18. andjie

19. mark_o.

on where you posted your problem

20. mark_o.

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

21. andjie

I posted your name in the question

22. andjie

So confusing

23. mark_o.

hmm i dont know why i didnt get a notification?

24. mark_o.

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

25. andjie

ok