# Thread: Parabola in Standard Form

1. ## Parabola in Standard Form

Write f(x) = -x^2 + 6x - 4 in the standard form for a parabola.

I came up with (x - 3)^2 + 5 but the right answer is
-(x - 3)^2 + 5.

Why must we place the minus infront of the quantity
(x - 3) in this case?

2. Originally Posted by magentarita
Write f(x) = -x^2 + 6x - 4 in the standard form for a parabola.

I came up with (x - 3)^2 + 5 but the right answer is
-(x - 3)^2 + 5.

Why must we place the minus infront of the quantity
(x - 3) in this case?
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.

$\displaystyle f(x)=ax^2+bx+c$
$\displaystyle f(x)=-x^2+6x-4$

3. ## ok but.............

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.

$\displaystyle f(x)=ax^2+bx+c$
$\displaystyle f(x)=-x^2+6x-4$

I don't understand why we must factor out -1.

4. Originally Posted by magentarita
I don't understand why we must factor out -1.
Because to complete the square, the $\displaystyle x^2$ coefficient is always 1, not -1.

5. ## yes, I..........

Originally Posted by Prove It
Because to complete the square, the $\displaystyle x^2$ coefficient is always 1, not -1.
Yes, I forgot about that rule for completing the square.