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Spartan_Of_Ares
 2 years ago
Best ResponseYou've already chosen the best response.0what is the domain of the graph?

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1The 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..

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1There 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.

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1Since 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.

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

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1So 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.

Spartan_Of_Ares
 2 years ago
Best ResponseYou've already chosen the best response.0as the vertex maybe

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1Oh man yer a lot faster at that than I am XD haha

Spartan_Of_Ares
 2 years ago
Best ResponseYou've already chosen the best response.0i use graphing tech

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

Spartan_Of_Ares
 2 years ago
Best ResponseYou've already chosen the best response.0i am they give us graphing calculators

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1So 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

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1The 2 represents the X value of our vertex. (2,1).

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