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hhopke
 one year ago
What is the range of function 3x^2+6x1 in interval notation?
hhopke
 one year ago
What is the range of function 3x^2+6x1 in interval notation?

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hhopke
 one year ago
Best ResponseYou've already chosen the best response.2Also, according to the HW program, \[(4,\infty)\] is wrong.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the domain of all polynomial function is all real numbers.

hhopke
 one year ago
Best ResponseYou've already chosen the best response.2Yeah. Domain is \[(\infty,\infty)\]

hhopke
 one year ago
Best ResponseYou've already chosen the best response.2If you plug it into a graphing calculator, it looks like this:

hhopke
 one year ago
Best ResponseYou've already chosen the best response.2You still there, @Photon336 ?

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1@hhopke since it's quadratic you could use part of the quadratic formula, and see what value it is whether if it's greater than zero, this will give you a sense of it's real or imaginary. i feel that this is important because if b^24ac < 0 then your function is going to be imaginary, so i guess not all real numbers will apply to this. \[b ^{2}4ac \] > 0 b = 6 a = 3 c = 1 (6)^24(3)(1) = 48 it's all real numbers, it's going to be continuous everywhere, every x value has a distinct y v value. my math skills are weak, but usually if a function is continuous everywhere it's also differentiable on that particular interval. if we take the derivative. \[\frac{ dy }{ dx } 3x ^{2}+6x1 = 6x+6 \] that's continuous for x value.

hhopke
 one year ago
Best ResponseYou've already chosen the best response.2Thank you, but I have absolutely no clue what you just said. We just started this. All the information I have on the subject (and supposedly, all I need) is that if a parabola concaves up, then the range is anything greater than the Y value of the vertex, and if it concaves down, then the range is anything less than the Y value of the vertex. Really basic stuff no quadratic formula yet. :)

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1@hhopke yeah, sorry about that lol I thought you were asking for the range

hhopke
 one year ago
Best ResponseYou've already chosen the best response.2I was asking for range. Everything else I search on the internet gives me all sorts of weird stuff using set notation and such... :(

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1so for a parabola the general formula is this: \[ax ^{2}+bx+c\]

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1@hhopke the sign in front of the a value tells us whether a prabola will concave up or down

hhopke
 one year ago
Best ResponseYou've already chosen the best response.2Right. This one is up, b/c 3>0 right?

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1look at this dw:1444541020427:dw

hhopke
 one year ago
Best ResponseYou've already chosen the best response.2Yeah. When I graphed it on my calculator, it gave me a concaveup parabola. The vertex was at (1,4). But when I tried to do the range in interval notation, I typed (4,oo) and the program said it was wrong. ???

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1@hhopke Domain = all x values

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1while range is all y values

hhopke
 one year ago
Best ResponseYou've already chosen the best response.2All possible Yvalues for the function is the range of that function. The lowest possible yvalue for this particular function is 4. Interval notation for the range should be (4, oo) right? (oo is infinity)

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1yeah I just plotted your graph on wolfram alpha. 4 is the lowest y value

hhopke
 one year ago
Best ResponseYou've already chosen the best response.2So what am I doing wrong? :(

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1i'm not sure, let me tag someone who knows better

hhopke
 one year ago
Best ResponseYou've already chosen the best response.2Alrighty. Same thing happened w/ function 3x^2. Plugged it in, lowest yvalue was 0. Plugged in (0, oo) for range and it was marked wrong. :(

hhopke
 one year ago
Best ResponseYou've already chosen the best response.2This happened with all the problems, if that helps.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1\[3x^2+6x1\\~\\=3(x^22x)  1 \\~\\= 3(x^22x+1)4 \\~\\= 3(x1)^24\\~\\\ge 4\] therefore the range is \([4, \infty)\)

hhopke
 one year ago
Best ResponseYou've already chosen the best response.2ooooooooooooooohhhhhhhhhhhhhhh.... I formatted it wrong. Thank you sosososososososososososososo much!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1change the left parenthesis, "\((\)", to square backet "\([\)" and you will be okay

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1np, other than that ur work looks good ! nice job!

hhopke
 one year ago
Best ResponseYou've already chosen the best response.2THANK YOU SO MUCH TO BOTH @ganeshie8 and @Photon336

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1in the third line, how did you get 4 @ganeshie8 ?
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