[SOLVED] Quadratic Function

viet_jon

1. The problem statement, all variables and given/known data


solve (f(x)=-3 x^2 + 9 x + 1/4 )
fx=(-3)(x^2-3 x)+1/4

fx=(-3)(x^2-3 x+9/4)+1/4+27/4

fx=(-3)(x^2-3 x+9/4)+28/4
fx=(-3) (x-3/2)^2+7



2. Relevant equations



3. The attempt at a solution


the solution is already given in the book, but I don't understand why 9/4 was inserted in the third line (bolded). I know at the end of the equation, 27/4 is added to balance out the 9/4.......but why put in 9/4 there?

Kurdt

What they've done is simply completed the square. If you have an equation of the form: [itex] x^2+2ax [/itex], you can write it as: [itex] x^2+2ax = (x+a)^2 - a^2 [/itex].

rock.freak667

f(x)=-3 x^2 + 9 x + 1/4

they just put 27/4 and -27/4 to make completion easier

f(x)=-3 x^2 + 9 x -27/4+ 1/4 +27/4
=-3(x^2-3x+9/4)+7
=-3(x-3x/2)^2 +7

expand (x-3x/2)^2 and check it yourself

viet_jon

yea bro...

the thing is, how do you come up with 9/4 from to make it easier?

I'm on purplemath.com right now, trying to figure it out without bothering you guys... hopefully I can figure this one out.

Kurdt

Well heres how I would have done that problem. You're given:

[tex] f(x) = -3x^2 + 9x + \frac{1}{4} [/tex]
[tex] f(x) = -3( x^2 -3x) + 1/4 [/tex]

Now we notice that the term in brackets is of the form [itex] x^2+2ax [/itex], with [itex]a=\frac{-3}{2}[/itex], and so we complete the square [itex] x^2+2ax = (x+a)^2 - a^2 [/itex]:

[tex] f(x) = -3\left(\left(x-\frac{3}{2}\right)^2 - \frac{9}{4} \right) + \frac{1}{4} [/tex]
[tex] f(x) = -3\left(x - \frac{3}{2}\right)^2 + \frac{27}{4} + \frac{1}{4} [/tex]
[tex] f(x) = -3\left(x - \frac{3}{2}\right)^2 + 7 [/tex]

viet_jon

okie.........i got it.


so the point of adding -9/4 inside the brackets is so that you can make it factorable. And since you added a (-), now you have to minus it to even it out.


thnkx.....can't believe it took me so long.

anyhow, how do you type the math out like that?

jaime2000

You use LaTeX. This tread explains in more detail.

http://www.physicsforums.com/showthread.php?t=8997

viet_jon

test:

[tex]x=-3 x^2+9x+1/4[/tex]

[tex]fx=(-3)(x^2┤-3 x)+1/4\\[/tex]

[tex]fx=(-3)(x^2┤-3 x+9/4)+1/4+27/4\\[/tex]

[tex]fx=(-3)(x^2┤-3 x+9/4)+28/4\\[/tex]

[tex]fx=(-3) (x┤-3/2)^2+7\\[/tex]

Kurdt

If you want to test then there are plenty of preview websites. Here is one of them.

http://at.org/~cola/tex2img/index.php