1. ## Proving functions

Hey all, I just have a homework problem example that I can't even begin.

it says:

Prove {-x^2+6x-8 x is in Reals} = (- , 1]

I'm not even sure what to do, even looking in my notes and online didn't really help me, thought I'd give this place a try, thanks guys!

2. Originally Posted by teacast
Hey all, I just have a homework problem example that I can't even begin.

it says:

Prove {-x^2+6x-8 x is in Reals} = (- , 1]

I'm not even sure what to do, even looking in my notes and online didn't really help me, thought I'd give this place a try, thanks guys!
You are asked to find the range of the function:

$y = -x^2+6x-8 = -(x^2-6x+8) = -(x^2-6x+9-1)= -((x-3)^2-1)=-(x-3)^2+1$

The graph of the function is a parabola opening down, thus the vertex V(3, 1) is the maximum point and all values of y are equal or smaller than 1. Therefore $y\in (-\infty,\ 1]$

3. Oh okay, just one question... why did you add a -1 to $-(x^2-6x+9-1)$? Just a way to say adding 8 so you could make it factorable?

Thanks for the help!

4. Originally Posted by teacast
Oh okay, just one question... why did you add a -1 to $-(x^2-6x+9-1)$? Just a way to say adding 8 so you could make it factorable?

Thanks for the help!
He didn't' add -1 to anything. Her wrote the 8 as 9- 1 (which he can do because 8= 9- 1 so that he would have a perfect square, $x^2- 6x+ 9= (x- 3)^2$

What he did is called "completing the square".

5. ohhh okay okay, that's sort of what I meant, thanks for the clarification!