Question

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Answer 1

It's already in general form, when you have ax^2 + x = y

Answer 2

Right. I meant standard form only! ahah, sorry its been a long day

Answer 3

Wait, I might be wrong..

Answer 4

Oh, uh, this in the standard form, http://puu.sh/rncB

Answer 5

So now, you need the general form, which is something that looks like y = a(x-h)^21+k

Answer 6

Isnt it already in general form?

Answer 7

Um, standard form is y = a(x-h)^2 + k
and the general form it what the equation is right now.

Answer 8

I'm terribly sorry, it's kinda late so, is a little bit confused >.>

Answer 9

haha no i was just confused cause you just said that equation was general form above then you said it was standard.. so whats the standard form for my equation? can you help me find it?

Answer 10

Alright, we know that the standard form is y = a(x+h)^2+k

We know that to find the x value of the vertex, we do -b/2a

y = -x^2 + 12x - 28

-12/2*-1
-12/-2
6

So x=6, plug it in the equation

y = -(6)^2 + 12(6) - 28
y = -36 + 72 - 28
y = 8

We know that -1 is the slope, 6 is h and 8 is k
y = -1(x-6)^2+8

Answer 11

I know that there's a faster way to find it, but what I did here is to find the x and y intercepts, and simply plugged it in the standard form equation, with the slope. :)