Hi Guys,
If i've got the completed square : $\displaystyle f(x) = \frac{1}{9}(x - 3)^2 - 4$
How do i get this back into the form $\displaystyle f(x) = ax^2 + bx + c$
having a slow afternoon,
Thanks
Hi Guys,
If i've got the completed square : $\displaystyle f(x) = \frac{1}{9}(x - 3)^2 - 4$
How do i get this back into the form $\displaystyle f(x) = ax^2 + bx + c$
having a slow afternoon,
Thanks
I find the quadratic formula easiest - especially when $\displaystyle a \neq 1$. To find the x intercepts set f(x) to 0
$\displaystyle x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$
However if you've got it in completed square form it would be simpler to use that form.
$\displaystyle
f(x) = \frac{1}{9}(x - 3)^2 - 4 = 0
$
$\displaystyle (x-3)^2 = 36$
$\displaystyle x-3 = \pm 6$
$\displaystyle x = 9$ or $\displaystyle x=-3$ as you said