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andreascott

What is the maximum or minimum value of the function? What is the range? y = –2x^2 + 32x –12

  • one year ago
  • one year ago

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  1. cwrw238
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    find f'(x) and equate it to 0 solve for x

    • one year ago
  2. andreascott
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    I don't understand that ?

    • one year ago
  3. cwrw238
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    oh - i've assumed you know some calculus

    • one year ago
  4. andreascott
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    Would it be like y=-2(0)^2+32(0)-12 ?

    • one year ago
  5. cwrw238
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    no what you have there is the intercept on y-axis

    • one year ago
  6. cwrw238
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    you need to change it to the vertex form y = –2x^2 + 32x –12 = -2(x^2 - 16x + 6) = -2( x - 8)^2 -64 + 6) = -2[(x-8)^2 - 58)] so the x coordinate at maximum = 8 a negative coefficient before x^3 means there is a maximum value

    • one year ago
  7. andreascott
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    oh , but it says the maximum and range is only 116 . its either 116 or -116 ?

    • one year ago
  8. cwrw238
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    at maximum the function has a value -2*-58 = 116

    • one year ago
  9. andreascott
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    ok , i see that . . .

    • one year ago
  10. cwrw238
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    the range is f(x)<= 116

    • one year ago
  11. andreascott
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    less than or greater than ?

    • one year ago
  12. cwrw238
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    |dw:1350681832911:dw|

    • one year ago
  13. cwrw238
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    below the graph less or equal to 116

    • one year ago
  14. andreascott
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    Thats what i thought it was (: . Thank you soo much for helping me ! (;

    • one year ago
  15. cwrw238
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    yw

    • one year ago
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