Pre cal

determine the function f whose graph is given the vertex is (3,-13) and the y intercept is -4 how whould i work this out

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asked by shelly
  1. I am wondering what degree the funcion is: second degree, fifth degree?

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  2. second

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    posted by shelly
  3. Since you are using the word "vertex" I will assume you might be talking about a quadratic function.

    with vertex (3,-13) the parabola would be

    y = a(x-3)^2 -13
    but (0,-4) lies on it, so ....

    -4 = a(9) - 13
    9a = 9
    a = 1

    f(x) = (x-3)^2 - 13

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    posted by Reiny
  4. thank you so much but i am still not understanding how you got that

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    posted by shelly
  5. in your text or in your notes, you should have the usual form of the equation if you have the vertex

    if vertex is (h, k)
    then the equation of the quadratic is
    y = a(x-h)^2 + k
    compare the values of (h,k) with your vertex of (3,-13)

    the value of a govern which way the parabola opens and how "steep" or "shallow" it is.

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    posted by Reiny
  6. how would you find the answere if they give you a graph with two points like (-3,5) and (0-4)

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    posted by shelly
  7. how do you find A?

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    posted by shelly

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