Question

help pleaseeee

Answer 1

looks like \(4p(x-2)=(y+2)^2\)

Answer 2

what is the original equation?

Answer 3

\[4p(x-h)=(y-k)^2\] if it opens to the right

Answer 4

center is \((h,k)\)

Answer 5

you still have to find \(4p\)

Answer 6

how do you do that

Answer 7

replace \(x\) by \(3\) and \(y\) by \(-1\) and see what it has to be

Answer 8

\[4p(x-2)=(y+2)^2\]
\[4p(3-2)=(-1+2)^2\]

Answer 9

this give \(4p=1\) conveniently enough

Answer 10

so final answer is
\[x-2=(y-2)^2\]

Answer 11

oooo ok thanks for explaining

Answer 12

oooh i made a typo!!

Answer 13

where is it

Answer 14

\[\huge x-2=(y+2)^2\]

Answer 15

i put \(y-2\) by mistake on one line , it is \(y+2\)

Answer 16

its ok thank you:)

Answer 17

yw