Question

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

Answer 1

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

Answer 2

I might be reading it wrong I am sorry

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

How would I show you?

Answer 9

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.

Answer 10

Thats the question

Answer 11

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

Answer 12

like this

hi andjie

Answer 13

I say your name

Answer 14

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

Answer 15

ok I did that

Answer 16

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

Answer 17

Paste it where exactly lol

Answer 18

in the type your reply area, because I did that

Answer 19

on where you posted your problem

Answer 20

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

Answer 21

I posted your name in the question

Answer 22

So confusing

Answer 23

hmm i dont know why i didnt get a notification?

Answer 24

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

Answer 25

ok