hollerback1
Junior Member
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- Dec 21, 2005
- Messages
- 80
Hello hollerback here, I'm just learned how to put the quadratics in vertex form...one problem. I DONT KNOW HOW TO DO IT :!: Please send me a link...explanation..what ever thanks guys.
Quadratics....vertex form? wow plz help lol
- Thread starter hollerback1
- Start date
It is difficult for me to guess what kind of help you are looking for. The best way is for you to fire up your favorite search engine and look for something like
quadratic vertex form math
And pick one that fits your needs. My best guess is that
http://www.valleyview.k12.oh.us/vvhs/de ... thes3.html
might help.
quadratic vertex form math
And pick one that fits your needs. My best guess is that
http://www.valleyview.k12.oh.us/vvhs/de ... thes3.html
might help.
- Joined
- Feb 4, 2004
- Messages
- 16,582
If you mean "rearranging a quadratic so you can read the vertex off the equation", then the basic process is as follows:
. . . . .y = ax<sup>2</sup> + bx + c
. . . . .Factor out whatever is multiplied on the squared term,
. . . . .but factor it only from the variable terms.
. . . . .y = a[x<sup>2</sup> + (b/a)x] + c
. . . . .Take half of whatever is now multiplied on the linear
. . . . .("x") term, and square it. To "complete the square",
. . . . .you'll be adding this inside the brackets; to keep that
. . . . .side "even" (unchanged), you'll also be subtracting it
. . . . .from the same side.
. . . . .y = a[x<sup>2</sup> + (b/a)x + (b/2a)<sup>2</sup>] + c - a[(b/2a)<sup>2</sup>]
. . . . .Now convert to "complete the square" form:
. . . . .y = a[x + (b/2a)]<sup>2</sup> + c - (b<sup>2</sup>/4a)
Simplify, and read off the values for h and k.
Eliz.
. . . . .y = ax<sup>2</sup> + bx + c
. . . . .Factor out whatever is multiplied on the squared term,
. . . . .but factor it only from the variable terms.
. . . . .y = a[x<sup>2</sup> + (b/a)x] + c
. . . . .Take half of whatever is now multiplied on the linear
. . . . .("x") term, and square it. To "complete the square",
. . . . .you'll be adding this inside the brackets; to keep that
. . . . .side "even" (unchanged), you'll also be subtracting it
. . . . .from the same side.
. . . . .y = a[x<sup>2</sup> + (b/a)x + (b/2a)<sup>2</sup>] + c - a[(b/2a)<sup>2</sup>]
. . . . .Now convert to "complete the square" form:
. . . . .y = a[x + (b/2a)]<sup>2</sup> + c - (b<sup>2</sup>/4a)
Simplify, and read off the values for h and k.
Eliz.
hollerback1
Junior Member
- Joined
- Dec 21, 2005
- Messages
- 80
Confused...
hmmm...I tried translating the equations, and I just don't get what to divide half by(/2) or to square. Please rephrase it again in a easier way for me if u can, thanks alot.
Back1 :wink:
hmmm...I tried translating the equations, and I just don't get what to divide half by(/2) or to square. Please rephrase it again in a easier way for me if u can, thanks alot.
Back1 :wink:
hollerback1
Junior Member
- Joined
- Dec 21, 2005
- Messages
- 80
TO gene
hey I looked over all the equations and i think the first one i need...but iam still confused on it.
y = a[x2 + (b/a)x] + c
hey I looked over all the equations and i think the first one i need...but iam still confused on it.
y = a[x2 + (b/a)x] + c
Look at the site I posted or what Staple did. (They say the same thing.)
Start at
y = a[x^2 + (b/a)x] + c
And look at what happens with the multiplier of the x term, b/a.
It is what is divided by 2 then squared, added inside the bracket and subtracted outside.
y = a[x^2 + (b/a)x + (b/2a)^2] + c - a[(b/2a)^2]
Start at
y = a[x^2 + (b/a)x] + c
And look at what happens with the multiplier of the x term, b/a.
It is what is divided by 2 then squared, added inside the bracket and subtracted outside.
y = a[x^2 + (b/a)x + (b/2a)^2] + c - a[(b/2a)^2]