Question

Can someone explain to me how to rewrite this equation in vertex form??
y = x^2 – 6x + 4

Answer 1

It's basically completing the square. Can you do that or do you still need help?

Answer 2

I still need help. It's in my brain somewhere but can you just take me through it step by step? :/

Answer 3

I'd be glad to. y = x^2 – 6x + 4 is in the form y = ax^2 + bx + c where a=1, b=-6, and c=4. We catch a big break here because a=1 and that greatly simplifies things. I'll show you how to work these when a=1. If you get another problem with a=some other number, I can show you that later if you like. I'm going to work with a, b, and c and you can substitute so you'll be doing something as well.

Answer 4

y=(a(x-h)^2) + k is vertex form. The vertex is (h,k). +a is smiley, -a is frowney - that tells you if the parabola opens up or down. To convert an equation to vertex form complete the square.


y=x^2+6x+4


Step 1: move the loose number (the constant term) to the y side.


y-4=x^2 + 6x


Step 2: Factor out whatever multiplies x^2. Here it's 1, so this step does nothing except place the right side in ()'s.


y-4=(x^2 + 6x)


Step 3: Make a space after the constant on the y side, place the factor before x^2 before that space and the sign before that space is the same as the sign of the number in front of x^2. Again, since it's 1, no need for that part.


y-4+()=(x^2 + 6x)


Take 1/2 the coefficient of the x term (the number in front of x). Remember its sign. Square it. Add it to both sides inside the ().


y-4+(9)=(x^2 + 6x +9)

Answer 5

So, we have y = x^2 + bx + c -> y = [x^2 + bx + (b/2)^2] + c - (b/2)^2

Answer 6

y = [x^2 + (b/2)]^2 + c - (b/2)^2 and now you have your vertex form. This is the general solution for when a=1 and can be used as a pattern for any problem of that type now. Vertex at (-[b/2], c - [b/2]^2). It's a parabola opening up and that vertex is a minimum value.

Answer 7

Is this starting to make sense to you?

Answer 8

Uhhh, I was following @Grim0519 better... No offence, I'm a visual lerner. and all the words confuse me sometimes..

Answer 9

You can now take that equation from my fourth post and substitute a=1, b=-6, and c=4 and then you're done!

Answer 10

np. Whatever way you learn better is what matters.

Answer 11

teamwork i like that...

Answer 12

I still don't get it. Where do I put them? so far I have y-4(9)=(x^2+6x+9) .... Where do I go from there?

Answer 13

Multiply out the "a times the squared coefficient" part on the left-hand side (remember, in this one a = 1 so that does nothing), and convert the right-hand side to squared form. (This is where you use that sign you kept track of earlier, putting that sign in the middle of the squared expression.)


y-4+(9)=(x+3)^2


Simplify - combine like terms.


y+5=(x+3)^2


Move the constant term from the right back to the left.


y=(x+3)^2 - 5


Write in vertex form y=(a(x-h)^2) + k. In other words if the squared term is x+h write it as x-(-h). If the k term is negative, write it as + (-k).


y=(x-(-3))^2 + (-5)


Now the values for h and k are clear.

Answer 14

Okk, so the answer is y=(x-(-3))^2+(-5) ?

Answer 15

y=x^2+6x+4
=x^2+6x +4
=x^2+6x+9+4-9 complete the square by dividing 6 by 2 and squaring the result.
=(x+3)^2-5

you should have ended with this.

Answer 16

Ok got it. Thank you! Would you be willing to help me with one more?

Answer 17

sure shoot...

Answer 18

Same thing, converting to vertex form. y=x^2+14x+50 ......

Answer 19

gimme a sec...

Answer 20

No problem

Answer 21

@Grim0519 Shouldn't that last problem solution be y = (x - 3)^2 - 5 instead of y = (x + 3)^2 - 5 ? Maybe you had a typo somewhere?

Answer 22

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

"a" represents the leading coefficient: a = 1

(h,k) are the coordinate of the vertex of parabola.

We'll create a perfect square within the given quadratic:

f(x) = (x^2 - 14x + 49) + 1

f(x) = (x - 7)^2 + 1

The coordinates of the vertex of the parabola are (7 , 1).

The vertex form of the given quadratic f(x) = x^2 - 14x + 50 is f(x) = (x - 7)^2 + 1.

Answer 23

i didnt think so...let me recheck my work...

Answer 24

So the answer to the second one is y=(x-7)^2+1 ?

Answer 25

Or f(x) rather. not y.

Answer 26

yes ma'am

Answer 27

Yay! Thank you soooo much!
Both of you!

Answer 28

Wait, that one is wrong also.

Answer 29

The second one is y=(x+7)^2+1 and the first one is y = (x - 3)^2 - 5. Grim, you're flipping the sign around. Check again.

Answer 30

Oh dear....

Answer 31

I fixed it. Thanks for catching that tcarroll! You're a life saver.

Answer 32

sorry about that the second one was wrong......

Answer 33

The confirmation is in the x^1 term. For that first equation, you have to end up with -6x. You won't get that with (x + 3)^2. The first one was wrong also.

Answer 34

the first also thanks @tcarroll010

Answer 35

np. Good working with both of you. Good effort all around.

Answer 36

Olivia, no disrespect to Grim, because he's a good hard worker. Really a good hard worker, but his method is a bit off. You might want to review my method to get the right answer. It doesn't matter to me whose method you follow, but I want you to get the right answer for these problems in the future.

Answer 37

definitely...