Question

can someone check my work please?

Answer 1

check your work on what?

Answer 2

@iGreen

Answer 3

is 4+3x-4=0 standard form for this ?

Answer 4

@Xd1521 @K_V8

Answer 5

Oh sorry I was lagging and didn't see your comments

Answer 6

so am i correct ?

Answer 7

I'm still working it out

Answer 8

is that a parabola?

Answer 9

if so you drawing is not accurate
it is not symmetrical
there should be a axis of symmetry y=a

Answer 10

yes , its a parabola

Answer 11

Oh I see the parabola is not symmetrical

Answer 12

your info are incorrect! there is no symmetry what so ever
parabola is always symmetrical along y=a
determined by the y coordinate of the vertex
in your drawing the vertex is (0.-4) so y=-4 should be the axis of symmetry
but your drawing does not agree with this

Answer 13

@Pawtpie
parabola is symmetrical i said no the other way around
parabolas give even functions

Answer 14

well you can't if your info are incorrect! do you understand?

Answer 15

best scenario for you is to attach you homework, the graph that is given to you
post is here accurately

Answer 16

the standard form is
\(\large f(x)=a(x-h)^2+k\)
where \((h.k)\) is the vertex

Answer 17

f(x)=ax^2+bx+c is the general form

Answer 18

here is the thing
if the vertex is (0,-4)
since the graph cut the x axis at 3 in the right
then it should cut the x axis on the left at -3 not -2
it should be same distance from the origin to 3 and to -3

Answer 19

otherwise we would not call it a parabola anymore