bev199
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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andjie
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f(x)=\[\frac{ x ^{2} }{ 3 }\]
andjie
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I might be reading it wrong I am sorry
mark_o.
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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
andjie
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You had it right I think I was confused, but you really helped me understand the question...
mark_o.
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ok good,,, good luck now and have fun ....:D
andjie
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I will once someone looks at my problems lol..... One person is looking I just want to ensure I am on the right track....
mark_o.
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ok where is the prob? i may be able to help a little bit :D
andjie
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How would I show you?
andjie
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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.
andjie
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Thats the question
mark_o.
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hmm go to your prob site and notify my name there then i will click it and be there :D
mark_o.
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like this
hi andjie
andjie
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I say your name
mark_o.
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yes copy and paste my name there then ill just click on it
andjie
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ok I did that
mark_o.
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hmm i didnt have a notification,, did you highlight and copy then paste my name there?
andjie
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Paste it where exactly lol
andjie
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in the type your reply area, because I did that
mark_o.
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on where you posted your problem
mark_o.
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ok go back to where you posted your problem then paste my name there
andjie
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I posted your name in the question
andjie
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So confusing
mark_o.
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hmm i dont know why i didnt get a notification?
mark_o.
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is it still open? why dont you close it hen repost them new
andjie
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ok