Question

math problem

Answer 1

@Florisalreadytaken

Answer 2

@snowflake0531

Answer 3

Hopefully you know how to do this lol

Answer 4

You have the vertex, and so we have y=a(x-h)^2 + k
(h,k) is the vertex, while (x,y) can just be a random point
we need to solve for a

Answer 5

You did a couple of problems before this, hopefully, you'll know hwo to solve for a, after you get that
expand the form of a(h-k)^2 + k
to get standard form -> ax^2 + bx-c

Answer 6

let the eq. of parabola be
y=ax^2+bx+c
it passes through (-1,2),(3,2) and (1,-2)
2=a-b+c ...(1)
2=9a+3b+c ...(2)
-2=a+b+c ...(3)
(3)-(1) gives
-2-2=b+b
2b=-4,
b=-2
(2)-(1) gives
0=8a+4b
0=2a+b
2a=-b=-(-2)
a=2/2=1
from (1)
2=1+2+c
c=-1
so
y=x^2-2x-1
https://www.desmos.com/calculator/xjzkiefn3p
y=x^2-2x-1

Answer 7

you put it in wrong

Answer 8

-2x, you put 2x

Answer 9

snoflake

Answer 10

snoflake?

Answer 11

@snowflake0531

Answer 12

yes, i think that's it