Question

How to convert standard form to vertex form?

Answer 1

Im having trouble grasping the concept. Any chance someone could help me understand the concept?

Answer 2

Well y = x^2 - 6x + 4 is the actual problem im working on. They are asking me to rewrite it in vertex form. I know its something to do with factoring, but i never really payed attention during factoring class...

Answer 3

Well, \[x = -\frac{ b }{ 2a }\]

Answer 4

For instance: \[y=(x-1)^2\] then expand it and use the formula given above, almost like completee the square

Answer 5

Hmm.. Im not sure im following. I know x = - b/2a is the formula to factor a number but i get confused...

Answer 6

The standard form is:

y = ax^2 + bx + c

Here's how to go from that to the vertex form:

y = ax^2 + bx + c = a[x^2 + (b/a)x] + c

= a[x^2 + (b/a)x + (b^2)/(4a^2)] + c - (b^2)/(4a)

= a[x + (b/2a)]^2 + c - (b^2)/(4a)

vertex at (-(b/2a), c - (b^2)/(4a))

Answer 7

The first line puts "a" outside of the brackets for the upcoming completing of the square. The second line takes half of the coefficient of "x" and squares it within the brackets.. The last term on line 2 does not have a^2 in the denominator. The third line gets the exponent outside of the brackets.

Answer 8

I know that first post of mine is full of variables, but if you go through it step-by-step, that will always get you to the vertex form.

Answer 9

Ok.. I sort of get it..
Im going to give it a try with my equation. Let me know if i do something wrong

Answer 10

I also can do the equation for you and you can use it as a study example if you want.

Answer 11

No i want to do it myself. I cant learn if i dont.

Answer 12

I applaud your spirit. That is very admirable and you are absolutely right. The best way to learn is by doing. Go for it!

Answer 13

y = x^2 - 6x + 4
a[x^2 - (6/a)x] + 4
I dont understand where the a is coming from tho.. Ill continue anyway
a[x^2 - (6/a)x - (6^2)/(4a^2)] + 4 - (b^2)/(4a)
Ok iv gotten here Is everything correct so far?

Answer 14

Would it be easier if we did an online whiteboard?

Answer 15

This typing text math stuff is really overwhelming.

Answer 16

For this problem, a = 1 (the implied coefficient on x^2) so you don't have to work with "a" in this problem. The 3rd term in the brackets should be +(3/a)^2. I don't know how to do an online whiteboard, but if you set it up, I'll see if I can follow you.

Answer 17

Ok could you try and join the room. It would be much easier for me to follow if i could draw it out
http://www.scriblink.com/index.jsp?act=phome&roomid=416&KEY=D1A14A0A3C419D4BB28F64E134E0FE6B

Answer 18

The biggest key to doing these is to take half of the coefficient of the "x" term and then square that. That's the big hurdle.

Ok, I'll go there if my computer will let me.

Answer 19

I have a window open, but I don't know what to do next.

Answer 20

Hmm.. Ok ill do my best to type in here then. Let me restart

Answer 21

Given that this will be a lot of typing, maybe I should do it just this once.

Answer 22

Ok.. Please do

Answer 23

Do you still need my help?

Answer 24

The more help the better

Answer 25

http://www.twiddla.com/980248
Come here

Answer 26

y = x^2 - 6x + 4 y = (x^2 - 6x + 3^2) + 4 - 3^2

y = (x - 3)^2 - 5

Answer 27

So, that's it. Notice how for the second step all I did was take half of -6 and square it.

Answer 28

And the vertex becomes (3, -5)

Answer 29

Ahh.. Ok This is much more simple to understand... You just insert the formula (b/2a) after the b term. But Im a little lost on the third step...

Answer 30

When "a" is 1, the problem becomes easier than if it is not 1. The third step is getting the exponent out of the parentheses. Along with doing the subtraction at the end of the right-hand side.

Answer 31

Thank you for your help sir this really helped! Im going to try a few more to get comfortable^^

Answer 32

Great working with you! Thx for the recognition!