1. ## Completing the square

Hi Guys,

If i've got the completed square : $f(x) = \frac{1}{9}(x - 3)^2 - 4$

How do i get this back into the form $f(x) = ax^2 + bx + c$

having a slow afternoon,

Thanks

2. You would:

$
\frac{1}{9}(x^2 - 6x + 9) - 4$

$\frac{1}{9}x^2 - \frac{2}{3}x + 1 - 4$

$\frac{1}{9}x^2 - \frac{2}{3}x - 3$

Now complete the square.

Hi Guys,

If i've got the completed square : $f(x) = \frac{1}{9}(x - 3)^2 - 4$

How do i get this back into the form $f(x) = ax^2 + bx + c$

having a slow afternoon,

Thanks

HI

Expand the whole things again .

$f(x)=\frac{1}{9}(x^2-6x+9)-4$

$=\frac{1}{9}x^2-\frac{2}{3}x+1-4$

then simplify from here .

4. I'm trying to find the x coordinates, would it be easier to factorise rather than use completing the squares? by graphing i know the answers to be -3 and 9, its just showing that algebraically

I'm trying to find the x coordinates, would it be easier to factorise rather than use completing the squares? by graphing i know the answers to be -3 and 9, its just showing that algebraically
I find the quadratic formula easiest - especially when $a \neq 1$. To find the x intercepts set f(x) to 0

$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

However if you've got it in completed square form it would be simpler to use that form.

$

f(x) = \frac{1}{9}(x - 3)^2 - 4 = 0
$

$(x-3)^2 = 36$

$x-3 = \pm 6$

$x = 9$ or $x=-3$ as you said