anonymous
  • anonymous
help
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
how to graph x^2=6y
NotTim
  • NotTim
Rearrange so that "y" is 'alone'
NotTim
  • NotTim
do you understand?

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anonymous
  • anonymous
no
anonymous
  • anonymous
so it would be x^2+6y
NotTim
  • NotTim
No.
NotTim
  • NotTim
make it so it is like: y=....
anonymous
  • anonymous
ohh
anonymous
  • anonymous
y=x^2+6
NotTim
  • NotTim
Almost, but not quite.
NotTim
  • NotTim
you have to divide both sides by 6 so that 6y is just y.
anonymous
  • anonymous
then what is it
anonymous
  • anonymous
anonymous
  • anonymous
how to graph x^2=6y
NotTim
  • NotTim
you have to re-arrange it frist.
NotTim
  • NotTim
divide both sides by 6.
Michele_Laino
  • Michele_Laino
it is simple: the graph of y=x^2 is a parabole which passes at the origin of the coordinate system: |dw:1433527737795:dw|
anonymous
  • anonymous
it is asking x^2=6y
Michele_Laino
  • Michele_Laino
the function: x^2=6y can be rewritten as below: \[\Large y = \frac{1}{6}{x^2}\]
Michele_Laino
  • Michele_Laino
where I have used the suggestion of @NotTim
anonymous
  • anonymous
how do u rewrite
Michele_Laino
  • Michele_Laino
I have divide both sides by 6, as @NotTim well said
anonymous
  • anonymous
ok
Michele_Laino
  • Michele_Laino
Now the graph of your function: \[\Large y = \frac{1}{6}{x^2}\] is similar to the graph of the function y= x^2. We have the subsequent drawing: |dw:1433528004632:dw|
anonymous
  • anonymous
i need accurate one
anonymous
  • anonymous
like u have to find focus, dirretrix
Michele_Laino
  • Michele_Laino
it is another parabola, which passes at the origin of the coordinate system
anonymous
  • anonymous
like u have to find focus, dirretrix
Michele_Laino
  • Michele_Laino
here are the formulas to find focus and directrix:
anonymous
  • anonymous
ok
anonymous
  • anonymous
where are they
Michele_Laino
  • Michele_Laino
we have to start from this equation: \[y = a{x^2} + bx + c\]
Michele_Laino
  • Michele_Laino
by comparison with our equation, we get: a= 1/6, b=0, and c=0, right?
anonymous
  • anonymous
yes
Michele_Laino
  • Michele_Laino
now the general formula for focus and directrix are: \[\Large \begin{gathered} Focus = \left( { - \frac{b}{{2a}},\frac{{1 - {b^2} + 4ac}}{{4a}}} \right) \hfill \\ \hfill \\ y = - \frac{{1 + {b^2} - 4ac}}{{4a}},\quad directrix \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
please substitute the parameters: a=1/6, b=c=0
Michele_Laino
  • Michele_Laino
what do you get?
anonymous
  • anonymous
|dw:1433528576564:dw|
Michele_Laino
  • Michele_Laino
please, c=0
anonymous
  • anonymous
so we can eliminate that
Michele_Laino
  • Michele_Laino
\[\Large y = - \frac{{1 + {b^2} - 4ac}}{{4a}} = - \frac{{1 + 0 - 4 \times \left( {1/6} \right) \times 0}}{{4/6}}\]
anonymous
  • anonymous
that is foucs
Michele_Laino
  • Michele_Laino
\[\large y = - \frac{{1 + {b^2} - 4ac}}{{4a}} = - \frac{{1 + 0 - 4 \times \left( {1/6} \right) \times 0}}{{4/6}}\]
Michele_Laino
  • Michele_Laino
that is the directrix
anonymous
  • anonymous
ohh
anonymous
  • anonymous
how would u find that
Michele_Laino
  • Michele_Laino
the focus is: \[\Large Focus = \left( { - \frac{0}{{2/6}},\frac{{1 - 0 + 4/6 \times 0}}{{4/6}}} \right)\]
Michele_Laino
  • Michele_Laino
those are general formulas, which can be derived
Michele_Laino
  • Michele_Laino
so after a simplification, we get: \[\Large \begin{gathered} Focus = \left( {0,\;\frac{3}{2}} \right) \hfill \\ \hfill \\ directrix:\quad y = - \frac{3}{2} \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
please check my result above
anonymous
  • anonymous
ok

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