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anonymous

  • 5 years ago

Fill in the missing term so that the quadratic equation has a graph that opens up, with a vertex of (– 3, – 25), and x intercepts at x = -8 and x = 2. (Do not include the negative sign in your answer.) y = x2 + 6x − ___

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  1. anonymous
    • 5 years ago
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    what ?

  2. anonymous
    • 5 years ago
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    you know it looks like \[a(x+8)(x-2)=a(x^2+6x-16)\] so you know the constant is -16

  3. anonymous
    • 5 years ago
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    i said TMI because the first coordinate is an unnecessary piece of information

  4. anonymous
    • 5 years ago
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    ohkay thanks and are you sure that is the correct answer ?

  5. anonymous
    • 5 years ago
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    you are asked to enter it without the minus sign, so put in "16"

  6. anonymous
    • 5 years ago
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    ok thanks

  7. anonymous
    • 5 years ago
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    the point is if the vertex was not (-3,-25) there would be no solution to this, so it is a rather dumb question. you are given more information that the poor quadratic can handle. you might want to mention this to your math teacher

  8. anonymous
    • 5 years ago
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    since the vertex is (-3,-25) you know it looks like \[y=a(x+3)^2-25\] and now all you need is one more point to find out that a = 1 in other words: the vertex and one other point would give the answer!

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