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hhopke

  • one year ago

What is the range of function 3x^2+6x-1 in interval notation?

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  1. hhopke
    • one year ago
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    Also, according to the HW program, \[(-4,\infty)\] is wrong.

  2. anonymous
    • one year ago
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    the domain of all polynomial function is all real numbers.

  3. hhopke
    • one year ago
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    Yeah. Domain is \[(-\infty,\infty)\]

  4. anonymous
    • one year ago
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    :)

  5. hhopke
    • one year ago
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    But what is the range?

  6. hhopke
    • one year ago
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    If you plug it into a graphing calculator, it looks like this:

  7. hhopke
    • one year ago
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    You still there, @Photon336 ?

  8. Photon336
    • one year ago
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    @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.

  9. hhopke
    • one year ago
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    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. :)

  10. Photon336
    • one year ago
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    @hhopke yeah, sorry about that lol I thought you were asking for the range

  11. hhopke
    • one year ago
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    I was asking for range. Everything else I search on the internet gives me all sorts of weird stuff using set notation and such... :(

  12. Photon336
    • one year ago
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    so for a parabola the general formula is this: \[ax ^{2}+bx+c\]

  13. Photon336
    • one year ago
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    @hhopke the sign in front of the a value tells us whether a prabola will concave up or down

  14. hhopke
    • one year ago
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    Right. This one is up, b/c 3>0 right?

  15. Photon336
    • one year ago
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    look at this |dw:1444541020427:dw|

  16. hhopke
    • one year ago
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    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. ???

  17. hhopke
    • one year ago
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  18. Photon336
    • one year ago
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    @hhopke Domain = all x values

  19. Photon336
    • one year ago
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    while range is all y values

  20. hhopke
    • one year ago
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    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)

  21. Photon336
    • one year ago
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    yeah I just plotted your graph on wolfram alpha. -4 is the lowest y value

  22. hhopke
    • one year ago
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    So what am I doing wrong? :(

  23. Photon336
    • one year ago
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    @ganeshie8

  24. Photon336
    • one year ago
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    i'm not sure, let me tag someone who knows better

  25. hhopke
    • one year ago
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    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. :(

  26. hhopke
    • one year ago
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    @ganeshie8

  27. hhopke
    • one year ago
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    This happened with all the problems, if that helps.

  28. ganeshie8
    • one year ago
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    \[3x^2+6x-1\\~\\=3(x^2-2x) - 1 \\~\\= 3(x^2-2x+1)-4 \\~\\= 3(x-1)^2-4\\~\\\ge -4\] therefore the range is \([-4, \infty)\)

  29. hhopke
    • one year ago
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    ooooooooooooooohhhhhhhhhhhhhhh.... I formatted it wrong. Thank you sosososososososososososososo much!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

  30. ganeshie8
    • one year ago
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    change the left parenthesis, "\((\)", to square backet "\([\)" and you will be okay

  31. ganeshie8
    • one year ago
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    np, other than that ur work looks good ! nice job!

  32. hhopke
    • one year ago
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    THANK YOU SO MUCH TO BOTH @ganeshie8 and @Photon336

  33. Photon336
    • one year ago
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    in the third line, how did you get -4 @ganeshie8 ?

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