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Determine the standard form of a quadratic function whose graph has a y intercept of 8 and a vertex at (-2,-4). The answer is f(x)= 3x^2 + 12x +8 but I can't figure out how to get the answer.

Solve x^2 + x=6, by factoring. The answer is x=-3,2 but I can't figure out how to get the answer.

Solve 2x^2 - 5= 3x, using the quadratic formula. The answer is x=-1, 2.5 but I can't figure out how to get the answer.

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2 answers

  1. This question and the other one I just answered for you are fundamental questions that you just have to know how to do in this topic.
    To simply say, "I can't figure out how to get the answer" tells me you are not even in the course.
    Anyway....
    since you know the vertex, the equation must be
    y = a(x+2)^2 - 4
    since (0,8) lies on it ....
    8 = a(2^2) - 4

    12 = 4a
    a = 3

    f(x) = 3(x+2)^2 - 4
    I will leave it up to you to expand it and get the desired answer

    x^2 + x=6
    x^2 + x - 6 = 0
    (x+3)(x-2) = 0
    and ......

    2x^2 - 5= 3x
    2x^2 - 3x - 5 = 0

    a = 2, b = -3, c = -5
    now just grind it out

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  2. Than you Reiny for your help. And yes, I am in this course. I figured out the answer, I wanted to make sure that I was doing the correct steps.

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