Question

help please?!
what is the vertex form of y=-x^2+12x-4

Answer 1

vertex form is y = a(x-h)^2 + k
the vertex is at (h,k)

you need to rearrange your equation into that form. do you know how to do so?

Answer 2

yes, im stuck at the point where ive plugged the equation into y=ax^2+bx+c,
or y=1(-6)^2+12(-6) -4. im not sure why my answer isn't coming out right so I can move on to the next step..

Answer 3

okay, let's find the values of a, b, and c:

y = ax^2 + bx + c
y = -x^2 + 12x - 4

what is a=
what is b=
what is c=
?

Answer 4

a=-1, b=12 and c= -4

Answer 5

good. when you wrote "y = 1(-6)^2+12(-6)-4" I couldn't make any sense of what you were doing...

I would factor out that -1 from everything:

y = -1(x^2-12x + 4)

now "complete the square" on x^2-12x+4 so that you can write it as (x-h)^2 for the correct value of h

Answer 6

with a middle term of -12x, that means we're going to have

(x-6)^2 = (x-6)(x-6) = x^2 -6x -6x + 36 = x^2 -12x + 36
however, we have x^2 - 12x + 4, so we need to compensate for the 36-4 = 32 we add to the equation by subtracting it outside the (x-6)^2:

x^2-12x + 4 = x^2 - 12x + 4 + 32 - 32 = (x-6)^2 - 32

Answer 7

ok

Answer 8

that means our final equation becomes

y = -1((x-6)^2-32) = -(x-6)^2 + 32

can you tell where the vertex is from that?

Answer 9

the vertex would be -x(6)^2 +32 right?

Answer 10

no, you didn't copy that correctly.

y = -1(x-6)^2 + 32 is vertex form

Answer 11

-x(6)^2+32 is something completely different — it's a straight line!

Answer 12

purple line is your equation, blue curve is my parabola

Answer 13

-x(6)^2+32 = -x(6*6) + 32 = -36x + 32

Answer 14

ahhh ok now I see what you did. alright that makes sense!

Answer 15

so because parabola form is

y = a(x-h)^2 + k

and we have
y = -1(x-6)^2 + 32
that means a = -1 (parabola opens downward)
h = 6 (x-coordinate of vertex is 6)
k = 32 (y-coordinate of vertex is 32)

Answer 16

as a check, we can determine the vertex from standard form, which is y = ax^2 + bx + c

we have a = -1
b = 12
c = -4

in standard form, the vertex x-coordinate is given by x = - b / (2a)
here that gives us x = -12 / (2*-1) = -12 / -2 = 6

that agrees with what we got!

then plugging x = 6 into the standard form equation, we get

y = -1(6)^2 + 12(6) -4 = -1*36 + 12*6 - 4 = -36 + 72 - 4 = 36 -4 = 32

again, that agrees with what we got!

Answer 17

yeah I was trying to find the x coordinate of the vertex by plugging A,B and C into x=-b/2a. I think I was trying to solve it a different way. anyhow, your way made sense, so thank you greatly!

Answer 18

yeah, if the teacher asks you for vertex form, you need to know how to put it in vertex form. of course, you could find the vertex, and then write the vertex form equation from that :-)

Answer 19

looks like you posted what I was saying right as I posted that haha. but yeah, I get what your saying.

Answer 20

anyhow, the trick is to just avoid making any silly algebraic mistakes when you're doing the work. I like to think of that factoring bit as splitting 0 into its positive and negative components — they must be kept separate or they will annihilate each other, just like a story line from an old Star Trek episode :-)

Answer 21

haha great analogy, and yeah, that makes sense! thanks a ton!