Question

math problem

Answer 1

@Florisalreadytaken

Answer 2

Do you remember what we did last time, first tell me,w hat's the vertex coordinate point

Answer 3

4, 1

Answer 4

Yes and we have \[a(x-h)^2 +k\] where (4,1) is (h,k)
can you plug h and k in? what do you get?

Answer 5

a(x-4)^2+1

Answer 6

Yes so we have \[y=a(x-4)^2 +1\]
We now have to solve for x

tell me a random coordinate point on the parabola that isn't the vertex

Answer 7

2,3

Answer 8

!@snowflake0531

Answer 9

I htink you meant (3,2)

\[2 = a(3-4)^2 +1\]
solve for a

Answer 10

a(-1)^2+1

Answer 11

1+1

Answer 12

2 = 2

Answer 13

So, a=1

Answer 14

a = 2?

Answer 15

no
a=1
1= a(-1)^2
1 = a(1)
a=1
or something like that
anyways a=1

Answer 16

So we have \[(x-4)^2 +1\] since we have to convert this to standard form, can you do (x-4)^2

remember that\[(a-b)^2 =a^2-2ab+b^2\]

Answer 17

\(\color{#0cbb34}{\text{Originally Posted by}}\) @snowflake0531
no
a=1
1= a(-1)^2
1 = a(1)
a=1
or something like that
anyways a=1
\(\color{#0cbb34}{\text{End of Quote}}\)
do you realise what you just wrote?

Answer 18

no

Answer 19

fair nuff

Answer 20

is she correct

Answer 21

positive

Answer 22

e.e i'm correct, i just said no bc floris was questioning me

Answer 23

so what is \[(x-4)^2+1\]

Answer 24

x^2-8x+17

Answer 25

yep so there's you ranswer