Please help with graphing a quadratic function. y=(x-1)^2 Got stuck at y=(x-1)(x-1) and p=1, q=1 Which doesn't work, because you can't fix a vertex with one point on a graph.

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asked by Kelly
  1. What is Alberbra?

    I don't know what your p and q are upposed to be.

    Define a new variable x' = x-1. Your equation becomes
    y = x'^2

    The vertex of your parabola is at x=1, y=0.

    That is one point of the graph. Other points are easily plotted.

    The parameter p is often used to designate the distance from the vertex to the focus. In tis case, p = 1/4, since the standard parabola formula is y = 4px^2 , with p being the focal distance.

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    posted by drwls
  2. The standard quadratic is

    where (h,k) is the vertex, and x=h is the axis of symmetry.
    The zeroes (p,q) are at -h±√(k/a).
    If a>0, the quadratic is concave upwards, i.e. the vertex is at the minimum.
    If a<0, the graph is concave downwards, or the vertex is at the maximum.

    In the case of y=(x-1)²,
    a=1, h=1, k=0
    The zeroes are coincident because k=0, and coincide with the vertex. This also means that the graph is tangent to the x-axis.
    I will leave it to you to figure out the concavity and the maximum/minimum.
    Post if you need more information.

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    posted by MathMate

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