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Quadratic Function

by viet_jon
Tags: function, quadratic, solved
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viet_jon
#1
Nov28-07, 08:40 PM
P: 135
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?
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Kurdt
#2
Nov28-07, 08:52 PM
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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
#3
Nov28-07, 10:25 PM
HW Helper
P: 6,202
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
#4
Nov28-07, 10:28 PM
P: 135
Quadratic Function

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
#5
Nov28-07, 10:36 PM
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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
#6
Nov28-07, 10:58 PM
P: 135
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
#7
Nov29-07, 02:56 AM
P: 8
You use LaTeX. This tread explains in more detail.

http://www.physicsforums.com/showthread.php?t=8997
viet_jon
#8
Nov29-07, 02:16 PM
P: 135
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
#9
Nov29-07, 02:21 PM
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Quote Quote by viet_jon View Post
test:

<tex>x^2</tex>

<tex>(x)=-3 x^2+9x+1/4</tex>
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
If you want to test then there are plenty of preview websites. Here is one of them.

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


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