Originally Posted by
masters Hello Magentarita,
If you expand and simplify what you thought was the right answer, you would see quickly that your result would not match the original function.
Standard form (vertex form) here is
$\displaystyle f(x)=a(x-h)^2+k$
You have:
$\displaystyle f(x)=-x^2+6x-4$ which can be converted to:
$\displaystyle f(x)=-1(x^2-6x)-4$ by factoring out -1.
Complete the square.
$\displaystyle f(x)=-1(x^2-6x+9)-4+9$
$\displaystyle f(x)=-(x-3)^2+5$
The vertex is at (3, 5) and the parabola opens downward because a < 0. We can't lose the negative because it was part of your original function.
$\displaystyle f(x)=ax^2+bx+c$
$\displaystyle f(x)=-x^2+6x-4$