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anonymous

  • one year ago

For the function f(x) = –2(x + 3)2 − 1, identify the vertex, domain, and range.

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  1. anonymous
    • one year ago
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    use desmos.com

  2. SolomonZelman
    • one year ago
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    f(x) is a polynomial function (why? cuz it has whole number powers of x only, and no other complex terms), and a polynomial is always continuous on the interval (-∞,+∞). that is for your domain.

  3. SolomonZelman
    • one year ago
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    the vertex of a parabola (your function is a parabola) in a form of: f(x)=a(x-h)²+k is the point: (h,k)

  4. SolomonZelman
    • one year ago
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    can you identify the vertex here? \(f(x)=-2(x+3)^2-1\)

  5. anonymous
    • one year ago
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    Is it (3, -1)?

  6. SolomonZelman
    • one year ago
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    almost.... \(f(x)=-2(x+3)^2-1\) \(f(x)=-2(x-\color{red}{-3})^2-\color{red}{1}\)

  7. SolomonZelman
    • one year ago
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    it is (-3,1)

  8. SolomonZelman
    • one year ago
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    and the domain is all real numbers (as I have reviously explained)

  9. SolomonZelman
    • one year ago
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    Now, your leading coefficient is negative, that means your parabola opens down and goes into -∞..... BUT, the range will be limited, and it is limited by the vertex because vertex is the maximum point (in any case when parabola opens down)

  10. SolomonZelman
    • one year ago
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    what is the y-value of your vertex? that is where the range ends.... and it starts from negative infinity

  11. anonymous
    • one year ago
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    Wouldn't the range be y is equal to or lesser than -3

  12. SolomonZelman
    • one year ago
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    -3 is the x-coordinate of the vertex....

  13. SolomonZelman
    • one year ago
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    so not to -3, but to?

  14. anonymous
    • one year ago
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    This it is -1

  15. SolomonZelman
    • one year ago
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    yes, the range is y≤-3

  16. SolomonZelman
    • one year ago
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    oh sorry

  17. SolomonZelman
    • one year ago
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    I mean y≤1

  18. anonymous
    • one year ago
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    ok thank you

  19. SolomonZelman
    • one year ago
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    yw

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