Question

Can you put this in standard form please?
y = 1(x-11)^2 + 30

Answer 1

you mean y = ax^2 +bx + c ?

Answer 2

\[y=(x-11)^{2}+30\]

Is that the equation you start with (you have an extra 1 in your first post...)

Answer 3

Yes

Answer 4

?

Answer 5

y = 2x + 8*

Answer 6

so take the bracket \[(x-11)^{2}\]

and expand it

Answer 7

y = 2x - 22, which simplifies to y = 2x + 8 if you include the 30 correct?

Answer 8

NO

the bracket is 'squared' - you are simply multiplying it by 2

Answer 9

So it'd be x^2 + 121? since the 2 negatives cancel

Answer 10

slow down

do oyu know how to expand (a+b)^2 ?

Answer 11

No, isn't it (a+b)(a+b)?

Answer 12

and you use the distributive property?

Answer 13

That is correct - but that is not what you wrote above

Answer 14

Oh wow, Idk what I just did lol
so it'd be y = (x-11)(x-11) + 30
?

Answer 15

Yes

Answer 16

y = 2x -11x -11x + 121 + 30?

Answer 17

NO

You are still multiplying by 2 rather than squaring

Answer 18

your last post was close - but 1 mistake

Answer 19

oh x^2?

Answer 20

that's it - so complete the sum - group all the factors together ... and voila - standard form

Answer 21

y = x^2 -22x + 151?

Answer 22

That's correct - just check your original though - why did oyu write a 1 in front of the bracket?

Answer 23

Wow thanks.
Um, that's what A was substituted for in the original equation, she wants me to show all work, even the 1.

Answer 24

Appreciate your help!

Answer 25

NP - outta here now