View Full Version : can i have some help with quadratic equations

CmptrGk

12-13-2004, 07:10 PM

i have an algebra test tomorrow and i really don't understand how to convert standard form (y=ax^2+bx+c) to vertex (y=a(x-h)^2+k). and vice versa

this is what i have for notes on the subject but i don't get it

vertex -> standard

y=2(x+1)^2-3

2(x+1)(x+1)

2[x^2+2x+1]-3

2x^2+4x+2-3

2x^2+4x-1

y=2x^2+4x-1

i really have no clue how this process works, i would like it if some1 could explain it to me

also standard -> vertex

f(x)=2x^2-4x+1

1.)Find vertex

(-b/2a, f(-b/2a)) vertex formula???

a=2 (-b/2a f(1)=2(1)^2-4(1)+1

b=-4 (-(-4)/2(2) = 2-4 +1

c=1 (4/4 = -2-1

=-1

vertex (1,-1)

h k

y=2(x-1)^2-1

i also have no clue how this works also

Kevin Sevcik

12-13-2004, 07:45 PM

Well.... the standard to vertex thing is straight factoring. I'll demonstrate so you can see where those formulae come from:

We're trying to do this:

ax^2 + bx + c --> a(x-h)^2 + k

From here I'm just working with the left side:

(1) a(x^2 + b/a*x + c/a)

Now, we're looking for a perfect square. (x-h)^2 = x^2 - 2hx + h^2

so you match up the similar termsand get:

-2hx = b/a*x

h=-b/2a

plug this into (1) to get the squared term:

a(x^2 - 2hx + h^2 - h^2 + c/a)

note: the -2hx goes in cause we made it equivalent to b/a*x. the h^2 and -h^2 cancel out so you can add them in. you've got a perfect square in there now, so you factor it.

a((x - h)^2 - h^2 + c/a)

a(x - h)^2 + a(-h^2 + c/a)

a(x - h)^2 - ah^2 + c

now you just have that weird term out at the end. that's where k goes, so you set them equal to each other.

k = -a(h)^2 + c

k = -a(-b/2a)^2 + c

k = -b^2/4a + c

so there's your formula for h and k and how you get them.

h=-b/2a

k = -b^2/4a + c

This makes perfect sense to me, which undoubtedly means it's completely incomprehensible. so let me know what specifically isn't clear.

sanddrag

12-13-2004, 07:54 PM

I think you have more of a clue than you think. I went through everything and as far as I can tell you are doing it all correctly. The only thing is that you are using different equations for standard->vertex and vertex->standard. I wasn't sure if you intended that or not.

CmptrGk

12-13-2004, 07:54 PM

From here I'm just working with the left side:

(1) a(x^2 + b/a*x + c/a)

Now, we're looking for a perfect square. (x-h)^2 = x^2 - 2hx + h^2

so you match up the similar termsand get:

-2hx = b/a*x

h=-b/2a

plug this into (1) to get the squared term:

a(x^2 - 2hx + h^2 - h^2 + c/a)

now where do these come from, and what is this perfect square think you speak of?

Mike Betts

12-13-2004, 08:00 PM

OK, so now I'm feeling old... We used to call this "completing the square".

Your first question: Multiply it out...

y=2(x+1)^2-3

y=2[(x+1)(x+1)]-3

y=2[x^2+2x+1]-3

y=2x^2+4x+2-3

y=2x^2+4x-1

Now for the reverse (completing the square). First, make a=1 and (temporarily) get rid of c.

y=2x^2+4x-1

y=2[x^2+2x]-1 ***

Now, we concentrate on the equation in brackets. What would be have to add/subtract to factor this?

[x^2+2x] ==> [x^2+2x+1] = [(x+1)(x+1)]

Note that we added 1 to complete the square. Now:

[x^2+2x] = [(x^2+2x+1)-1] = [(x+1)(x+1)-1]

And we substitute the bracketed equation back into *** as follows:

y=2[x^2+2x]-1 (I repeated *** for clarity)

y=2[(x+1)(x+1)-1]-1

y=[2(x+1)^2-2]-1

y=2(x+1)^2-3

Does this make more sense?

Kevin Sevcik

12-13-2004, 08:02 PM

(1) comes from dividing ax^2 + bx + c by a:

ax^2 + bx + c = a*(ax^2/a + b/a*x + c/a)

a perfect square is a perfect square factor, I just picked a name. (x - h)^2 is a square, as opposed to (x - h) * (x + q) which is not a square. you're looking for a term like (x - h)^2, so you expand that out to what it equals, x^2 - 2hx + h^2

at that point you say, "hey, -2hx and b/a*x are both multiplied by x. I'll make them equal"

Edit: Mike, I knew there was a better name, but I've long since forgotten.

CmptrGk

12-13-2004, 08:14 PM

ok, althoug several problems, my teacher never told us about "completing the square",and i vageuly inderstand what you are saying mike betts.

also

Now for the reverse (completing the square). First, make a=1 and (temporarily) get rid of c.

y=2x^2+4x-1

y=2[x^2+2x]-1 ***

where does a come into the picture

Mike Betts

12-13-2004, 08:24 PM

ok, althoug several problems, my teacher never told us about "completing the square",and i vageuly inderstand what you are saying mike betts.

I understand where they get "vertex" from and, yes, it will be important later in your education.

Just try that method on a few problems and see if it is easier to think that way. Then go back to your class notes and learn the way your teacher is presenting it... They are not too different.

Now look at some of the other posts here and you will see that they are all slightly different ways of skinning the same cat.

Don't tell PETA...

CmptrGk

12-13-2004, 08:27 PM

thanks fo the help evey1 i beilve that i may be able to figure out this whole process

Don't tell PETA...

this is completly unrealated, but, isnt PETA "People for the Eating of Tasty Animals" (according to friends shirt at school)

CmptrGk

12-13-2004, 08:33 PM

geuss what everyone, i figured out why i couldnt get this to work, my teacher put the wrong formula pieces on the board

vBulletin® v3.6.4, Copyright ©2000-2017, Jelsoft Enterprises Ltd.