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## Homework Statement

For the following questions, find the Domain and Range.

Use either the graph or turning point method.

## Homework Equations

1. y= -x+6x-8

2. y= x^2-3x-5

3. y= -x^2+4x+4

And a question on the matter, when you complete the square, as in when I half the middle term square it plus take away etc... If before i did that, there is a negative, would i use that in my final answer?

Very hard to write, check my attempts below, might be a bit more descriptive.

## The Attempt at a Solution

I won't show the beginning of my working out, but what's near the end or it'll take forever.

1. y= -x^2+6x-8

-(x^2-6x+9)-9+8 = 0

-(x-3)^2-1=0

So what i was asking before was, the negative in front, does that affect the turning point? I wrote it as (3,-1), because the first is the opposite of what's inside and the other is what is outside the brackets exactly, but does that negative change what is inside the brackets that we use for the turning point? And also x is part of the real field right? because it is infinity.

2. y= x^2-3x-5

(x^2-3x-5)=0

(x^2-3x+2.25)-2.25-5=0

(x-1.5)^2-7.25=0

Turning point is = (1.5,-7.25)?

3. y= -x^2+4x+4

-(x^2-4x-4)=0

-(x^2-4x+4)-4-4=0

-(x-2)^2-8=0

Turning point is = (2,-8) and again, does that negative in front affect anything?

And also with this last one, Since it is y=-x^2+4x+4, does that not mean that the parabola will be negative? if it does, the Range would be that {y:y<-8} which means that there would be no X intercept, except for when you use the discriminant, it says that there are 2 answers?

D = b^2-4ac

= 4^2-(4x-1x4)

= 16 + 16

= 32

If the discriminant > 0 doesn't that mean there are 2 x answers?

Thanks for helping out!