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uhh...little help here!

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  • uhh...little help here!

    Well i'm stuck on this problem which i have to get an answer for...related to quadratics

    y=-3(x+1)^2 + 4x^2 + 5(x+1)

    Vertex: _________________

    Axis of Symmetry: ______________

    Any help appreciated!
    ***My Blog***

  • #2
    Simplified, the equation becomes 7x^2+11x+11. To find the vertex, we take the derivative of the function and get:

    y = 14x+11
    now we need to maximize/minimize that value by finding the x that gives us 0:
    14x + 11 = 0
    14x = -11
    ---- ----
    14 14

    x = -11/14 (this is your vertex X coordinate)

    Now, to get the y-coordinate of the vertex, simply plug in your x-coordinate into the original function:

    7*(-11/14)^2+11(-11/14)+11 = 6.67

    Hence, the vertex is (-11/14, 20/3)

    We know we have a parabola that opens up. So the axis of symetry is now easy to find. The axis of symetry must be x = -11/14 .

    So your answer:
    VERTEX: (-11/14, 20/3)
    AXIS: x=-11/14

    Edit: now that I think about it, part of this solution required calculus (the derivative part). I don't know what math class this is for exactly, but if you're not expected to know calculus, then I'm sure there's another way to solve this... are you supposed to put it in standard (h,k) form and then solve it?


    • #3
      yes, it does require some knowledge of Calc....

      Thanks for the great help there! and of course my answer came out way wrong...hahaha Well back to work, about 20 more to go!
      ***My Blog***


      • #4
        My head hurts after reading this...