hipie

The domain is (−∞,6)∪(6,∞). The reason is that x cannot equal 6. So to express it in Set-Builder Notation, it will be x≠6 

Hey can you put the parenthesis in the right spots?

If we simplify and use special products. and then combine like terms, and then you'll get 2x^2 + 22x + 65. The equation for solving the axis of symmetry is -b/2a In ax^2 + bx + c. I think you can do the rest. 

^0.5 means ^(1/2), and that basically means square root of. You can do the rest. Just remember to solve the parenthesis first

The set is (-2, 4) 

2*2 = 4. 4+2 = 6

Because there are absolute value bars inside, which makes it positive. Likewise you don't see a negative sign behind the square root I put. Teachers will remember to test you so have to remember to only use the plus or minus in case of quadratics or something like that. But to put simply square root itself must be positive in the real domain. It doesn't feel right to just say that the sqrt(9) = -3.

no I think it's 1

x^2 + 6x + 9 = 24 can be written as x^2 + 6x - 15. If we plug in the quadratic formula, we get -3+- sqrt(24). But since it's the smallest real value of x, it's really the minus sign so the real answer is -3 - sqrt(24)

Use the quadratic form in this case, which is ax^2 + bx + c. ( = 3, = -5, = 7)

Substituting in 0 gives us c = 3. Keep substituting for x and for c. If you do this for the rest, you'll end up having sets of equations

which comes out as

a + b = -8

4a+2b = 4

I recommend using the addition method but you can do whatever method you like for this. I chose to multiply both sides of the first equation by -4 , and I got 

-2b=36 and b=-18. I think you know where this is going.