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8 October, 01:17

Graph the function g (x) = x^2+4x+1

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  1. 8 October, 03:42
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    Step-by-step explanation:

    Let's "complete the square," which will give us the vertex of this vertical, opens-up parabola:

    g (x) = x^2+4x+1 can be rewritten as g (x) = x^2+4x + 4 - 4 + 1, where that + 4 comes from squaring half of the coefficient of x.

    Then we have g (x) = x^2+4x + 4 - 4 + 1 = > g (x) = (x + 2) ^2 - 3.

    Comparing this to y = (x - h) ^2 + k,

    we see that the vertex, (h, k), is located at (-2, - 3).

    Plot this vertex. Also, plot the y-intercept (0, g (0)), which is (0, 1).

    This information is enough to permit graphing the function roughly.
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