hhopke
  • hhopke
What is the range of function 3x^2+6x-1 in interval notation?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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hhopke
  • hhopke
Also, according to the HW program, \[(-4,\infty)\] is wrong.
anonymous
  • anonymous
the domain of all polynomial function is all real numbers.
hhopke
  • hhopke
Yeah. Domain is \[(-\infty,\infty)\]

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anonymous
  • anonymous
:)
hhopke
  • hhopke
But what is the range?
hhopke
  • hhopke
If you plug it into a graphing calculator, it looks like this:
hhopke
  • hhopke
You still there, @Photon336 ?
Photon336
  • Photon336
@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^2-4ac < 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)^2-4(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. \[\frac{ dy }{ dx } 3x ^{2}+6x-1 = 6x+6 \] that's continuous for x value.
hhopke
  • hhopke
Thank 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
  • Photon336
@hhopke yeah, sorry about that lol I thought you were asking for the range
hhopke
  • hhopke
I was asking for range. Everything else I search on the internet gives me all sorts of weird stuff using set notation and such... :(
Photon336
  • Photon336
so for a parabola the general formula is this: \[ax ^{2}+bx+c\]
Photon336
  • Photon336
@hhopke the sign in front of the a value tells us whether a prabola will concave up or down
hhopke
  • hhopke
Right. This one is up, b/c 3>0 right?
Photon336
  • Photon336
look at this |dw:1444541020427:dw|
hhopke
  • hhopke
Yeah. When I graphed it on my calculator, it gave me a concave-up 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. ???
hhopke
  • hhopke
Photon336
  • Photon336
@hhopke Domain = all x values
Photon336
  • Photon336
while range is all y values
hhopke
  • hhopke
All possible Y-values for the function is the range of that function. The lowest possible y-value for this particular function is -4. Interval notation for the range should be (-4, oo) right? (oo is infinity)
Photon336
  • Photon336
yeah I just plotted your graph on wolfram alpha. -4 is the lowest y value
hhopke
  • hhopke
So what am I doing wrong? :(
Photon336
  • Photon336
@ganeshie8
Photon336
  • Photon336
i'm not sure, let me tag someone who knows better
hhopke
  • hhopke
Alrighty. Same thing happened w/ function 3x^2. Plugged it in, lowest y-value was 0. Plugged in (0, oo) for range and it was marked wrong. :(
hhopke
  • hhopke
@ganeshie8
hhopke
  • hhopke
This happened with all the problems, if that helps.
ganeshie8
  • ganeshie8
\[3x^2+6x-1\\~\\=3(x^2-2x) - 1 \\~\\= 3(x^2-2x+1)-4 \\~\\= 3(x-1)^2-4\\~\\\ge -4\] therefore the range is \([-4, \infty)\)
hhopke
  • hhopke
ooooooooooooooohhhhhhhhhhhhhhh.... I formatted it wrong. Thank you sosososososososososososososo much!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
ganeshie8
  • ganeshie8
change the left parenthesis, "\((\)", to square backet "\([\)" and you will be okay
ganeshie8
  • ganeshie8
np, other than that ur work looks good ! nice job!
hhopke
  • hhopke
THANK YOU SO MUCH TO BOTH @ganeshie8 and @Photon336
Photon336
  • Photon336
in the third line, how did you get -4 @ganeshie8 ?

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