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anonymous

  • 5 years ago

The vertex of the graph of a quadratic functon is (3,1) and the y intercept is (0,10). What is the equations of the function?

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  1. anonymous
    • 5 years ago
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    \[\text{Let } f(x) = k(x-a)^2 + c\] (general quadratic with square completed) (3,1) is a minimum => a = .... and c = .... When x = 0, f(x) = 10, so k = ... We can now expand and find the solution in the form f(x) = px^2 + qx + r

  2. anonymous
    • 5 years ago
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    No idea what to do after that. But thank you!

  3. anonymous
    • 5 years ago
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    Well, if a quadratic is at a minimum, x = a in the bracket (as the bracket is >= 0 for all x) So the only value is the + c This occurs at the point x = 3, y = 1, so we can deduce that the "a" = 3, and the c = 1 => f(x) = k(x-3)^2 + 1 When x = 0, f(x) = 10. So k* (-3)^2 + 1 = 10 -> 9k + 1 = 10 -> k = 1 So f(x) = (x-3)^2 + 1, which we could expand (if we want) to f(x) = ....

  4. anonymous
    • 5 years ago
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    ¬_¬

  5. anonymous
    • 5 years ago
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    I didn't get it correct :/

  6. anonymous
    • 5 years ago
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    What answer did you give?

  7. anonymous
    • 5 years ago
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    (x-3)^2 + 1 = x^2 - 6x + 10, which is the answer

  8. anonymous
    • 5 years ago
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    I don't remember! :/ it's an onlne class so I can't go back and change my anwers.

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