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mathtard

  • 4 years ago

Find the vertex f(x)= 3x^2-18x+3 please help me. Awards will be given

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  1. satellite73
    • 4 years ago
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    use the all mighty formula \[x= -\frac{b}{2a}\] to find the first coordinate

  2. satellite73
    • 4 years ago
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    in this case \[a=3,b=-18\] so \[-\frac{b}{2a}=-\frac{-18}{2\times3}=3\]

  3. satellite73
    • 4 years ago
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    that is the first coordinate and the second coordinate is what you get when you replace x by 3. you get \[3\times 3^2-18\times 3+3=27-54+3=-27+3=-24\] check my arithmetic because it is late

  4. satellite73
    • 4 years ago
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    also because i didn't really compute. as soon as i saw 27 i knew the next one would be -54. always works that way

  5. satellite73
    • 4 years ago
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    vertex is (3,-24) unless i messed up

  6. mathtard
    • 4 years ago
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    so (3,-24) ? and that would make it a maximum right?

  7. satellite73
    • 4 years ago
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    heck no

  8. satellite73
    • 4 years ago
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    this is a parabola that faces up. no max, but a min

  9. satellite73
    • 4 years ago
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    |dw:1314675140476:dw|

  10. satellite73
    • 4 years ago
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    the minimum value is -24

  11. mathtard
    • 4 years ago
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    Ah I got it mixed up!

  12. mathtard
    • 4 years ago
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    I always do that with the minimum maximum.

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