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Spartan_Of_Ares Group TitleBest ResponseYou've already chosen the best response.0
what is the domain of the graph?
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
The domain is the set of X values that this function can take on. So it appears we have a quadratic function, opening downward. The function that is graphed I think actually would look like this, \(f(x)=(x2)^2+3\) So the question is, what X values are you allowed to plug into that function? :O Hmmm..
 one year ago

Spartan_Of_Ares Group TitleBest ResponseYou've already chosen the best response.0
x ≥ 2 x ≤ −1
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
There is no square root, no dividing by x, no fractional exponent. There is nothing that we have to worry about when plugging values in for X. If we plug in 1 bajillion, we'll get a really big negative number for f(x), but it's still a real number.
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Since it's simply a quadratic, there are NO restrictions on the X term. The domain is ALL real numbers. :) You won't run into domain restrictions when you have nice polynomials like this  just simple powers of X.
 one year ago

Spartan_Of_Ares Group TitleBest ResponseYou've already chosen the best response.0
oh ok can you help me with this too? What is the range of the graph of y = −2x^2 − 8x − 7?
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
So this one is similar to the last one. It will be a parabola opening DOWNWARD. Unfortunately, they didn't give it to us in vertex form. So let's convert it so it's a little easier to see what's going on.
 one year ago

Spartan_Of_Ares Group TitleBest ResponseYou've already chosen the best response.0
(2,1)
 one year ago

Spartan_Of_Ares Group TitleBest ResponseYou've already chosen the best response.0
as the vertex maybe
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Oh man yer a lot faster at that than I am XD haha
 one year ago

Spartan_Of_Ares Group TitleBest ResponseYou've already chosen the best response.0
i use graphing tech
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Oh hah c: Well I wouldn't rely TOO MUCH on that, unless of course you're allowed to use it on a test ^^
 one year ago

Spartan_Of_Ares Group TitleBest ResponseYou've already chosen the best response.0
i am they give us graphing calculators
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Oh good good ^^
 one year ago

Spartan_Of_Ares Group TitleBest ResponseYou've already chosen the best response.0
:)}]
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
So we have the vertex of our parabola at (2,1) and it's opening downward (because of the negative sign on the x^2 term). So similar to the last problem, if we put in 2 bajillion for x, we'll get a really huge negative number for Y. But in this problem, if you graphed it, you may have noticed that we can't get any Y values larger than our vertex. The range represents the Y values the function can take on. So what would your range be? :O
 one year ago

Spartan_Of_Ares Group TitleBest ResponseYou've already chosen the best response.0
y ≤ −2?
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
The 2 represents the X value of our vertex. (2,1).
 one year ago

Spartan_Of_Ares Group TitleBest ResponseYou've already chosen the best response.0
oh so y ≤ 1
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
So Y can get as large as 1, and as negative as it wants? Yah that sounds right c: good job.
 one year ago

Spartan_Of_Ares Group TitleBest ResponseYou've already chosen the best response.0
ok thanks :)
 one year ago
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