Question

Help me write this graph please in general form:
y=a(x-h)^2+k

(Graph will be shown below)

Answer 1

Firstly not that this graph will be in the form: \(\bf x=a(y-h)^2+k\). Do you agree? @Loujoelou

Answer 2

R u asking if I agree that it'll be written in the form: x=a(y-h)^2+k I think so, I'm not entirely sure tho

Answer 3

Wait isn't the vertex in this graph (0,0)?

Answer 4

Yup

Answer 5

Pretty sure then since vertex = (0,0) we're using different formula then

Answer 6

@Loujoelou what do you mean? Since the vertex is 0 then \(\bf k = h = 0\). This allows us to simplify the graph to be \(\bf x=ay^2\).

Answer 7

oh ok, how would i find a tho?

Answer 8

hmm that's what I'm wondering. We know that a > 0 because of how the graph opens..

Answer 9

I know a= 1/4c, but how would i know what's c?

Answer 10

i think a= 1/12. The focus in this equation is (3,0) like it shows in the picutre and in this equation (c,0) so C=3 And using a=1/4(3)= a=1/12 so maybe the answer is x=1/12y^2 ? :3

Answer 11

Wait isn't 'c' the focus? Which is at (3, 0)?

Answer 12

and yeah using x=ay^2 Focus = (c,0)

Answer 13

Because the focus is at (3, 0) then \(\bf a=\frac{1}{4(3)}=\frac{1}{12}\)

Answer 14

yeah that's what i was thinking :) but it says 2 put it in general form, shoudl i just show it in both ways? x=1/12y^2 and y=1/12(x-0)+0?

Answer 15

Yes but in general form the h and k would cancel out since they're both 0. So you could show in general form with k = h = 0 but that would be pointless..

Answer 16

Our final simplified form would be \(\bf x=\frac{1}{12}y^2\). Since h = k = 0, we don't need to show. If you're heart so desires however then you can show it like this: \(\bf x=\frac{1}{12}(y-0)^2+0\).

Answer 17

ok lol thx :D

Answer 18

yw ;]