Question

Help with Changing Equations of Conics from Standard to Vertex Form

Answer 1

x2 − 6x + 2y + 9 = 0

Answer 2

welp you can move 9 to the other side \[x^2-6x+2y=-9\]

Answer 3

\[ (x^2 − 6x) + 2y = -9\]
yes you can now complete the square method
(x^2-6x)

Answer 4

Would you divide by 9?

Answer 5

or -9 i mean?

Answer 6

so we can get =1

Answer 7

nope
btw that's a important part
is it hyperbola
circle
ellipse
or parabola ?

Answer 8

AC=2 so 2>0 so ellipses?

Answer 9

nope
for ellipse
hyperbola
and circle y and x should have square
x^2 and y^2
it's parabola bec only x is raising to the 2nd power

Answer 10

so doesn't matter what number should be at right side :-)

Answer 11

oh because we are missing Bxy and Ey?

Answer 12

oh wait no we are missing cy^2

Answer 13

not Ey S:

Answer 14

That would make sense because parabolas have (x-h)^2 and (y-k)

Answer 15

o_0ohh alright

Answer 16

x2 − 6x +Cy^2 + 2y + 9 = 0
like this ?

Answer 17

I am confused lool

Answer 18

me too what's ur original question ?

Answer 19

I was just thinking out loud it seems. I was saaying that is why it is a parabola

Answer 20

because we were missing those parts

Answer 21

LOL

Answer 22

yes right bec parabola equation in standard from
(x-h)^2 = 4p(y-k) as u can see y isn't not squared

Answer 23

so that means we need to subtract y to the other side

Answer 24

\[(x^2-6x)=4a(-2y-9)\]

Answer 25

not yet first we have to apply complete the square method

Answer 26

compete the square
divide b/2